MCQ 11 Mark
If $\sin \alpha$ and $\cos \alpha$ are the roots of the equations $a x^2+b x+c=0$, then $b^2=$
- A
$a^2-2 a c$
- ✓
$a^2+2 a c$
- C
$a^2-a c$
- D
$a 2+a c$
AnswerCorrect option: B. $a^2+2 a c$
The given quadric equation is $ax^2 + bx + c = 0$, and $\sin\alpha$ and $\cos\beta$ are roots of given equation.
And, $a = a, b = b$ and $c = c$
Then, as we know that sum of the roots
$\sin\alpha+\cos\beta=\frac{-\text{b}}{\text{a}}\ ...(\text{i})$
And the product of the roots
$\sin\alpha\cdot\cos\beta=\frac{\text{c}}{\text{a}}\ ....(\text{ii})$
Squaring both sides of equation $(i)$ we get
$(\sin\alpha+\cos\beta)^2=\Big(\frac{-\text{b}}{\text{a}}\Big)^2$
$\sin^2\alpha+\cos^2\beta+2\sin\alpha\cos\beta=\frac{\text{b}^2}{\text{a}^2}$
Putting the value of $\sin\alpha+\cos\beta=1$ we get
$1+2\sin\alpha\cos\beta=\frac{\text{b}^2}{\text{a}^2}$
$\text{a}^2(1+2\sin\alpha\cos\beta)=\text{b}^2$
Putting the value of $\sin\alpha\cdot\cos\beta=\frac{\text{c}}{\text{a}}$ we get
$\text{a}^2\Big(1+2\frac{\text{c}}{\text{a}}\Big)=\text{b}^2$
$\text{a}^2\Big(\frac{\text{a}+2\text{c}}{\text{a}}\Big)=\text{b}^2$
$\text{a}^2+2\text{ac}=\text{b}^2$
Threfore, the value of $b^2=a^2+2 a c$
Thus, the correct answer is $(b)$
View full question & answer→MCQ 21 Mark
If $2$ is a root of the equation $x^2+b x+12=0$ and the equation $x^2+b x+q=0$ has equal roots, then $q=$
Answer$2$ is the common roots given quadric equation are $x^2+b x+12=0$, and $x^2+b x+q=0$
Then find the value of $q$.
Here, $x^2+b x+12=0$
$x^2+b x+q=0$
Putting the value of $x=2$ in equation $(i)$ we get
$2^2+b \times 2+12=0$
$4+2 b+12=0$
$2 b=-16$
$b=-8$
Now, putting the value of $b=-8$ in equation $(ii)$ we get
$x^2-8 x+q=0$
Then,
$a_2=1, b_2=-8 \text { and } c_2=q$
As we know that $D_1=b^2-4 a c$
Putting the value of $a_2=1, b_2=-8$ and $c_2=q$
$=(-8)^2-4 \times 1 \times q$
$=64-4 q$
The given equation will have equal roots, if $D=0$
$64-4 q=0$
$4 q=64$
$q=\frac{64}{4}$
$q=16$
Thus, the correct answer is $(c)$
View full question & answer→MCQ 31 Mark
If the equation $ax^2 + 2x + a = 0$ has two distinct roots, if:
- ✓
$\text{a}=\pm1$
- B
$a = 0$
- C
$a = 0, 1$
- D
$a = -1, 0$
AnswerCorrect option: A. $\text{a}=\pm1$
In the equation $a x^2+2 x+a=0$
$\Rightarrow D=b^2-4 a c$
$\Rightarrow D=(2)^2-4 \times a \times a$
$\Rightarrow D=4-4 a^2$
Roots are real and equal
$\Rightarrow D=0$
$\Rightarrow 4-4 a^2=0$
$\Rightarrow 4=4 a^2$
$\Rightarrow 1=a^2$
$\Rightarrow a^2=1$
$\Rightarrow\text{a}^2=(\pm1)^2$
$\Rightarrow\text{a}=\pm1$
View full question & answer→MCQ 41 Mark
If the equation $x^2+4 x+k=0$ has real and distinct roots, then:
- ✓
$\text{k}<4$
- B
$\text{k}>4$
- C
$\text{k}\geq4$
- D
$\text{k}\leq4$
AnswerCorrect option: A. $\text{k}<4$
In the equation $x^2+4 x+k=0$
$a=1, b=4, c=k$
$\Rightarrow D=b^2-4 a c$
$\Rightarrow D=(4)^2-4 \times 1 \times k$
$\Rightarrow D=16-4 k$
Roots are real and distinct
$\Rightarrow D>0$
$\Rightarrow 16-4 k>0$
$\Rightarrow 16>4 k$
$\Rightarrow 4>k$
$\Rightarrow k<4$
View full question & answer→MCQ 51 Mark
The value of $c$ for which the equation $ax^2 + 2bx + c = 0$ has equal roots is:
- ✓
$\frac{\text{b}^2}{\text{a}}$
- B
$\frac{\text{b}^2}{4\text{a}}$
- C
$\frac{\text{a}^2}{\text{b}}$
- D
$\frac{\text{a}^2}{4\text{b}}$
AnswerCorrect option: A. $\frac{\text{b}^2}{\text{a}}$
$\Rightarrow a x^2+2 b x+c=0$
$\Rightarrow D=b^2-4 a c$
$\Rightarrow D=(2 b)^2-4 \times a \times c$
$\Rightarrow D=4 b^2-4 a c $ Roots are equal
$\Rightarrow D=0$
$\Rightarrow 4 b^2-4 a c=0$
$\Rightarrow 4 a c=4 b^2$
$\Rightarrow c=\frac{4 b^2}{4 a}$
$\Rightarrow c=\frac{b^2}{a}$
View full question & answer→MCQ 61 Mark
The positive value of $k$ for which the equation $x^2+k x+64=0$ and $x^2-8 x+k=0$ will both have real roots, is :
AnswerThe given quadric equation are $x^2+k x+64=0$ and $x^2-8 x+k=0$ roots are real.
Then find the value of $a$.
Here, $x^2+k x+64=0$
$x^2-8 x+k=0$
$a_1=1, b_1=k $ and $ c_1=64$
$a_2=1, b_2=-8 $ and $ c_2=k$
As we know that $D_1=b^2-4 a c$
Putting the value of $a_1=1, b_1=k$ and $c_1=64$
$=(k)^2-4 \times 1 \times 64$
$=k^2-256$
The given equation will have real and distinct roots, if $D>0$
$k^2-256=0$
$k^2=256$
$k=\sqrt{256}$
$k= \pm 16$
Therefore, putting the value of $k=16$ in equation $(ii)$ we get
$x^2-8 x+16=0$
$(x-4)^2=0$
$x-4=0$
$x=4$
The value of $k=16$ satisfying to both equations.
Thus, the correct answer is $(d)$
View full question & answer→MCQ 71 Mark
The values of $k$ for which the quadratic equation $16 x^2+4 k x+9=0$ has real and equal roots are:
AnswerCorrect option: C. $6,-6$
The given quadratic equation $16 x ^2+4 kx +9=0$, has equal roots.
Here, $a=16, b=4 k$ and $c=9$
As we know that $D=b^2-4 a c$
Putting the value of $a=16, b=4 k$ and $c=9$
$\Rightarrow D=(4 k)^2-4(16)(9)$
$\Rightarrow D=16 k^2-576$
The given equation will have real and equal roots, if $D=0$
Thus, $16 k ^2-576=0$
$\Rightarrow k^2-36=0$
$\Rightarrow(k+6)(k-6)=0$
$\Rightarrow k+6=0 \text { or } k=6$
Therefore, the value of $k$ is $6,-6$
Hence, the correct option is $(c)$
View full question & answer→MCQ 81 Mark
If the equation $x^2 - ax + 1 = 0$ has two distinct roots, then :
- A
$|a| = 2$
- B
$|a| < 2$
- ✓
$|a| >2$
- D
AnswerCorrect option: C. $|a| >2$
The given quadric equation is $x^2-a x+1=0$, and roots are dostinct.
Then fond the value of $a$.
Here, $a=1, b=a$ and $c=1$
As we know that $D=b^2-4 a c$
Putting the value of $a=1, b=a$ and $c=1$
$=(a)^2-4 \times 1 \times 1$
$=a^2-4$
The given equation will have real and distinct roots, if $D>0$
$a^2-4>0$
$a^2>4$
$a>\sqrt{4}$
$a> \pm 2$
Therefore, the value of $|a|>2$
Thus, the correct answer is $(c)$
View full question & answer→MCQ 91 Mark
If one root of the equation $a x^2+b x+c=0$ is three times the other, then $b^2: a c=$
- A
$3 : 1$
- B
$3 : 16$
- ✓
$16 : 3$
- D
$16 : 1$
AnswerCorrect option: C. $16 : 3$
Quad equation is $ax^2 + bx + c = 0$
Let first root $=\alpha,$ Then
Second root $=3\alpha$
$\therefore$ Sum of root $=\alpha+3\alpha=\frac{-\text{b}}{\text{a}}$
$\Rightarrow4\alpha=\frac{-\text{b}}{\text{a}}$
$\Rightarrow\alpha=\frac{-\text{b}}{4\text{a}}\ ....(\text{i})$
and produt of roots $=\alpha\times3\alpha=\frac{\text{c}}{\text{a}}$
$\Rightarrow3\alpha^2=\frac{\text{c}}{\text{a}}$
$\Rightarrow\alpha^2=\frac{\text{c}}{3\text{a}}$
$\Rightarrow\Big(\frac{-\text{b}}{4\text{a}}\Big)^2=\frac{\text{c}}{3\text{a}} \ [$From $(i)]$
$\Rightarrow\frac{\text{b}^2}{16\text{a}^2}=\frac{\text{c}}{3\text{a}} \ ($Dividing by a$)$
$\Rightarrow\frac{\text{b}^2}{16\text{a}}=\frac{\text{c}}{3}$
$\Rightarrow\frac{\text{b}^2}{\text{ac}}=\frac{16}{3}$
$\Rightarrow\text{b}^2:\text{ac}=16:3$
View full question & answer→MCQ 101 Mark
If the roots of the equations $\left(a^2+b^2\right) x^2-2 b(a+c) x+\left(b^2+c^2\right)=0$ are equal, then:
- A
$2 b=a+c$
- ✓
$b^2=a c$
- C
$b=\frac{2 a c}{a+c}$
- D
$b=a c$
AnswerCorrect option: B. $b^2=a c$
The given quadric equation is $\left(a^2+b^2\right) x^2-2 b(a+c) x+\left(b^2+c^2\right)=0$, and roots equal. Here, $a=\left(a^2+b^2\right), b=-2 b(a+c)$ and $c=b^2+c^2$
As we know that $D=b^2-4 a c$
Putting the value of $a=\left(a^2+b^2\right), b=-2 b(a+c)$ and $c=b^2+c^2$
$=\{-2 b(a+c)\}^2-4 \times\left(a^2+b^2\right) \times\left(b^2+c^2\right)$
$=4 a^2 b^2+4 b^2 c^2+8 a b^2 c-4\left(a^2 b^2+a^2 c^2+b^4+b^2 c^2\right)$
$=4 a^2 b^2+4 b^2 c^2+8 a b^2 c-4 a^2 b^2-4 a^2 c^2-4 b^4-4 b^2 c^2$
$=+8 a b^2 c-4 a^2 c^2-4 b^4$
$=-4\left(a^2 c^2+b^4-2 a b^2 c\right)$
The given equation will have equal roots, if $D=0$
$-4\left(a^2 c^2+b^4-2 a b^2 c\right)=0$
$a^2 c^2+b^4-2 a b^2 c=0$
$\left(a c-b^2\right)^2=0$
$a c-b^2=0$
$a c=b^2$
Thus, the correct answer is $(b)$
View full question & answer→MCQ 111 Mark
If the sum and product of the roots of the equation $kx^2+ 6x + 4k = 0$ are equal, then $k =$
- A
$-\frac{3}{2}$
- ✓
$\frac{3}{2}$
- C
$\frac{2}{3}$
- D
$-\frac{2}{3}$
AnswerCorrect option: B. $\frac{3}{2}$
$ k x^2+6 x+4 k=0 $
Here $a=k, b=6, c=4 k $
$ \Rightarrow D=b^2-4 a c $
$ \Rightarrow D=(6)^2-4 \times k \times 4 k $
$ \Rightarrow D=36-16 k^2$
Roots are equal
$ \Rightarrow D=0 $
$ \Rightarrow 36-16 \mathrm{k}^2=0 $
$ \Rightarrow 16 \mathrm{k}^2=36$
$\Rightarrow \text{k}^2=\frac{36}{16}$
$\Rightarrow \text{k}^2=\Big(\frac{6}{4}\Big)^2$
$\Rightarrow \text{k}=\frac{6}{4}$
$\Rightarrow \text{k}=\frac{3}{2}$
View full question & answer→MCQ 121 Mark
If $a$ and $b$ are roots of the equation $x^2+ ax + b = 0,$ then $a + b =$
Answer$a$ and $b$ are the roots of the equation $x^2+ ax + b = 0$
Sum of roots $= -a$ and product of roots $= b$
Now $a + b = -a$
and $ab = b$
$\Rightarrow a = 1 ....(i)$
$\Rightarrow 2a + b = 0$
$\Rightarrow 2 \times 1 + b = 0$
$\Rightarrow b = -2$
Now $a + b = 1 - 2 = -1$
View full question & answer→MCQ 131 Mark
If the sum of the roots of the equation $x^2- (k + 6)x + 2(2k - 1) = 0$ is equal to half of their product, then $k =$
AnswerThe given quadric equation is $x^2- (k + 6)x + 2(2k - 1) = 0$, and roots are equal
Then find the value of $k$.
Let $\alpha$ and $\beta$ be two roots of given equation
And, $a = 1, b = -(k + 6)$ and of the roots
$\alpha+\beta=\frac{-\text{b}}{\text{a}}$
$\alpha+\beta=\frac{-\{-(\text{k}+6)\}}{1}$
$\alpha+\beta=(\text{k}+6)$
And the product of the roots
$\alpha\cdot\beta=\frac{\text{c}}{\text{a}}$
$\alpha\beta=\frac{2(2\text{k}-1)}{1}$
$\alpha\beta=2(2\text{k}-1)$
According to question, sum of the roots $=\frac{1}{2}\times$ product of the roots
$(\text{k}+6)=\frac{1}{2}\times2(2\text{k}-1)$
$\text{k}+6=2\text{k}-1$
$6+1=2\text{k}-\text{k}$
$7=\text{k}$
Therefore, the value of $k = 7$
Thus, the correct answer is $(b)$
View full question & answer→MCQ 141 Mark
If the sum of the roots of the equation $\text{x}^2-\text{x}=\lambda(2\text{x}-1)$ is zero, then $\lambda=$
- A
$-2$
- B
$2$
- ✓
$-\frac{1}{2}$
- D
$\frac{1}{2}$
AnswerCorrect option: C. $-\frac{1}{2}$
$\Rightarrow\text{x}^2-\text{x}=\lambda(2\text{x}-1)$
$\Rightarrow\text{x}^2-\text{x}=2\lambda\text{x}-\lambda$
$\Rightarrow\text{x}^2-\text{x}-2\lambda\text{x}+\lambda=0$
$\Rightarrow\text{x}^2-(1+2\lambda)\text{x}+\lambda=0$
Sum of roots $=\frac{-\text{b}}{\text{a}}$
$=\frac{1+2\lambda}{1}$
$\frac{1+2\lambda}{1}=0$
$\Rightarrow2\lambda=-1$
$\lambda=-\frac{1}{2}$
View full question & answer→MCQ 151 Mark
A quadratic equation whose one root is $2$ and the sum of whose roots is zero, is:
- A
$ x^2+a^4=0 $
- ✓
$ x^2-4=0 $
- C
$ 4 x^2-1=0 $
- D
$ x^2-2=0$
AnswerCorrect option: B. $ x^2-4=0 $
Let $\alpha$ and $\beta$ be the roots of quadratic equation in such a way that $\alpha=2$
Then, according to question sum of the roots
$\alpha+\beta=0$
$2+\beta=0$
$\beta=-2$
And the product of the roots
$\alpha\cdot\beta=2\times(-2)$
$\alpha\cdot\beta=-4$
As we know that the quadratic equation
$\text{x}^2-(\alpha+\beta)\text{x}+\alpha\beta=0$
Putting the value of $\alpha$ and $\beta$ in above
Therefore, the require be
$\text{x}^2-0\times\text{x}+(-4)=0$
$\text{x}^2-4=0$
Thus, the correct answer is $(b)$
View full question & answer→MCQ 161 Mark
If one root of the equation $x^2+ ax + 3 = 0$ is $1,$ then its other root is:
AnswerThe quad equation is $x^2+ ax + 3 = 0$
One root $= 1$
and product of roots $=\frac{\text{c}}{\text{a}}=\frac{3}{1}=3$
Second root $=\frac{3}{1}=3$
View full question & answer→MCQ 171 Mark
If $a$ and $b$ can take values $1, 2, 3, 4$. Then the number of the equations of the from $ax^2+ bx + 1 = 0$ having real roots is:
AnswerGiven that the equation $ax^2+ bx + 1 = 0$
For given equation to have real roots, discriminant $(\text{D})\geq0$
$\Rightarrow\text{b}^2-4\text{a}\geq0$
$\Rightarrow\text{b}^2\geq4\text{a}$
$\Rightarrow\text{b}\geq2\sqrt{\text{a}}$
Now, it is given that $a$ and $b$ can take values of $1, 2, 3 $ and $4$
The above condition $\text{b}\geq2\sqrt{\text{a}}$ can be satisfied when
- $b = 4$ and $a = 1, 2, 3, 4$
- $b = 3$ and $a = 1, 2$
- $b = 2$ and $a = 1$
So, there will be a maximum of $7$ equations for the values of $(a, b) = \{(1, 4), (2, 4), (3, 4), (4, 4), (1, 3), (2, 3)\}$ and $(1, 2)$
Thus, the correct option is $(b)$ View full question & answer→MCQ 181 Mark
If $p$ and $q$ are the roots of the equation $x^2- px + q + 0,$ then :
- ✓
$p = 1, q = -2$
- B
$p = 0, q = 1$
- C
$p = -2, q = 0$
- D
$p = -2, q = 1$
AnswerCorrect option: A. $p = 1, q = -2$
Given that $p$ and $q$ be the roots of the equation $x^2- Px + q + 0$
Then find the value of $p$ and $q$.
Here $, a = 1, b = -p$ and $c = q$
$p$ and $q$ be the roots of the given equation
Therefore, sum of the roots
$\text{p}+\text{q}=\frac{-\text{b}}{\text{a}}$
$\text{p}+\text{q}=\frac{-\text{p}}{1}$
$\text{p}+\text{q}=-\text{p}$
$\text{q}=-\text{p}-\text{p}$
$\text{q}=-2\text{p}\ ....(\text{i})$
Product of the roots
$\text{p}\times\text{q}=\frac{\text{q}}{1}$
As we know that
$\text{p}=\frac{\text{q}}{\text{q}}$
$\text{p}=1$
Putting the value of $p = 1$ in equation $(i)$
$q = -2 × 1$
$q = -2$
Therefore, the value of $p = 1, q = -2$
Thus, the correct answer is $(a)$
View full question & answer→MCQ 191 Mark
If the equation $\left(a^2+b^2\right) x^2-2(a c+b d) x+\left(c^2+d^2\right)=0 $ has equal roots, then:
AnswerCorrect option: B. $\text{ad}=\text{bc}$
In the equation
$ \Rightarrow\left(a^2+b^2\right) x^2-2(a c+b d) x+\left(c^2+d^2\right)=0 $
$ \Rightarrow D=B^2-4 A C $
$ \Rightarrow D=[-2(a c+b d)]^2-4\left(a^2+b^2\right)\left(c^2+d^2\right) $
$ \Rightarrow D=4\left[a^2 c^2+b 2 d^2+2 a b c d\right]-4\left[a^2 c^2+a^2 d^2+b^2 c^2+b^2 d^2\right] $
$ \Rightarrow D=4 a^2 c^2+4 b^2 d^2+8 a b c d-4 a^2 c^2-4 a^2 d^2-4 b^2 c^2-4 b^2 d^2 $
$ \Rightarrow D=8 a b c d-4 a^2 d 2-4 b^2 c^2 $
$ \Rightarrow D=-4\left[a^2 d^2+b^2 c^2-2 a b c d\right] $
$ \Rightarrow D=-4(a d-b c)^2$
$\because$ Roots are equal
$\therefore D = 0$
$⇒ -4(ad - bc)^2= 0$
$⇒ ad - bc = 0$
$⇒ ad = bc$
View full question & answer→MCQ 201 Mark
$\text { If }\left(a^2+b^2\right) x^2+2(a b+b d) x+c^2+d^2=0 $ has no real roots, then:
- A
$ab = bc$
- B
$ab = cd$
- C
$ac = bd$
- ✓
$\text{ad}\neq\text{bc}$
AnswerCorrect option: D. $\text{ad}\neq\text{bc}$
The given quadric equation is $\left(a^2+b^2\right) x^2+2(a b+b d) x+c^2+d^2=0$, and roots are equal.
Here, $a=\left(a^2+b^2\right), b=2(a b+b d)$ and, $c=c^2+d^2$
As we know that $D=b^2-4 a c$
Putting the value of $a=\left(a^2+b^2\right), b=2(a b+b d)$ and, $c=c^2+d^2$
$ =\{2(a b+b d)\}^2-4 \times\left(a^2+b^2\right) \times\left(c^2+d^2\right) $
$ =4 a^2 b^2+4 b^2 d^2+8 a b^2 d-4\left(a^2 c^2+a^2 d^2+b^2 c^2+b^2 d^2\right) $
$ =4 a^2 b^2+4 b^2 d^2+8 a b^2 d-4 a^2 c^2-4 a^2 d^2-4 b^2 c^2-4 a^2 d^2 $
$ =4 a^2 b^2+8 a b^2 d-4 a^2 c^2-4 a^2 d^2-4 b^2 c^2 $
$ =4\left(a^2 b^2+2 a b^2 d-a^2 c^2-a^2 d^2-b^2 c^2\right)$
The given equation will have no real roots, if $D<0$
$ 4\left(a^2 b^2+2 a b^2 d-a^2 c^2-a^2 d^2-b^2 c^2\right)<0 $
$ a^2 b^2+2 a b^2 d-a^2 c^2-a^2 d^2-b^2 c^2<0$
$\text{ad}\neq\text{bc}$
Thus, the correct answer is $(d)$
View full question & answer→MCQ 211 Mark
If one root the equation $2x^2+ kx + 4 = 0$ is $2,$ then the other root is :
AnswerLet $\alpha$ and $\beta$ be the roots of quadratic equation $2x^2+ kx + 4 = 0$ in such a way that $a = 2$
Here $, a = 2, b = k$ and $c = 4$
Then, according to question sum of the roots
$\alpha+\beta=\frac{-\text{b}}{\text{a}}$
$2+\beta=\frac{-\text{k}}{2}$
$\beta=\frac{-\text{k}}{2}-2$
$\beta=\frac{-\text{k}-4}{2}$
And the product of the roots
$\alpha\cdot\beta=\frac{\text{c}}{\text{a}}$
$\alpha\cdot\beta=\frac{4}{2}$
$\alpha\cdot\beta=2$
Putting the value $\beta=\frac{-\text{k}-4}{2}$ in above
$2\times\frac{(-\text{k}-4)}{2}=2$
$(-\text{k}-4)=2$
$\text{k}=-4-2$
$\text{k}=-6$
Putting the value of $k$ in $\beta=\frac{-\text{k}-4}{2}$
$\beta=\frac{-(-6)-4}{2}$
$\beta=\frac{6-4}{2}$
$\beta=\frac{2}{2}$
$\beta=1$
Therefore, value of other root be $\beta=1$
Thus, the correct answer is $(d)$
View full question & answer→MCQ 221 Mark
if $x^2+k(4 x+k-1)+2=0$ has equal rrots, then $k =$
- A
$-\frac{2}{3},1$
- ✓
$\frac{2}{3},-1$
- C
$\frac{3}{2},\frac{1}{3}$
- D
$-\frac{3}{2},-\frac{1}{3}$
AnswerCorrect option: B. $\frac{2}{3},-1$
The given quadric equation is $x^2+k(4 x+k-1)+2=0$, and roots are equal Then find the value of $k$.
$ x^2+k(4 x+k-1)+2=0 $
$ x^2+4 k x+\left(k^2-k+2\right)=0$
Here, $a=1, b=4 k$ and $c=k^2-k+2$
As we know that $D=b^2-4 a c$
Putting the value of $a=1, b=4 k$ and $c=k^2-k+2$
$=(4 k)^2-4 \times 1 \times\left(k^2-k+2\right) $
$ =16 k^2-4 k^2+4 k-8 $
$ =12 k^2+4 k-8 $
$ =4\left(3 k^2+k-2\right)$
The given equation will have real and distinct roots, if $D=0$
$ 4\left(3 k^2+k-2\right)=0 $
$ 3 k^2+k-2=0 $
$ 3 k^2+3 k-2 k-2=0 $
$ 3 k(k+1)-2(k+1)=0 $
$ (k+1)(3 k-2)=0 $
$ (k+1)=0 \text { or }(3 k-2)=0 $
$ k=-1 \text { or } k=\frac{2}{3}$
Therefore, the value of $\text{k}=\frac{2}{3},-1$
Thus, the correct answer is $(b)$
View full question & answer→MCQ 231 Mark
The value of $\sqrt{6+\sqrt{6+\sqrt{6+}}}...$ is:
AnswerLet $\text{x}=\sqrt{6+\sqrt{6+\sqrt{6+}}}...$
$\Rightarrow\text{x}=\sqrt{6+\text{x}}$
$ \Rightarrow x^2=6+x $
$ \Rightarrow x^2-x-6=0 $
$ \Rightarrow x^2-3 x+2 x-6=0 $
$ \Rightarrow x(x-3)+2(x-3)=0 $
$ \Rightarrow(x-3)(x+2)=0$
Either $x-3=0$, then $x=3$ Or
$x+2=0$, then $x=-2$
Now if $x = 3$, then
$3=\sqrt{6+\sqrt{6+\sqrt{6+}}}...$
$=\sqrt{6+ 3}=\sqrt{9}$
$=3$
If $x = -2,$ then
$\Rightarrow\text{x}=\sqrt{6+\text{x}}$
$\Rightarrow-2=\sqrt{6-2}$
$\Rightarrow-2=\sqrt{4}$
$\Rightarrow-2\neq2$
Which is not possible $x = 3$ is correct.
View full question & answer→MCQ 241 Mark
If one root of the equation $4\text{x}^2-2\text{x}+(\lambda-4)=0$ be the reciprocal of the other, then $\lambda=$
AnswerLet $\alpha$ and $\beta$ be the roots of quadratic equation$4\text{x}^2-2\text{x}+(\lambda-4)=0$ in such a way
Then, $\alpha=\frac{1}{\beta}$
Here, $\text{a}=4,\text{b}=-2$ and $\text{c}=(\lambda-4)$
Then, according to question sum of the roots
$\alpha+\beta=\frac{-\text{b}}{\text{a}}$
$\frac{1}{\beta}+\beta=\frac{-(-2)}{4}$
$\frac{1+\beta^2}{\beta}=\frac{1}{2}$
$2+2\beta^2=\beta$
$2\beta^2-\beta+2=0$
And the product of the roots
$\alpha\cdot\beta=\frac{\text{c}}{\text{a}}$
$\frac{1}{\beta}\times\beta=\frac{\lambda-4}{4}$
$1=\frac{\lambda-4}{4}$
$\lambda-4=4$
$\lambda=4+4$
$\lambda=8$
Therefore, value of $\lambda=8$
Thus, the correct answer is (a)
View full question & answer→MCQ 251 Mark
The number of quadratic equations having real roots and which do not change by squaring their roots is:
AnswerAs we know that the number of quadratic equations having real roots and which do not change by squaring their roots is 2.
Thus, the correct answer is (c)
View full question & answer→MCQ 261 Mark
If the equation $ 9 x^2+6 k x+4=0 $, has equal roots, then the roots are both equal to.
- ✓
$\pm\frac{2}{3}$
- B
$\pm\frac{3}{2}$
- C
$0$
- D
$\pm3$
AnswerCorrect option: A. $\pm\frac{2}{3}$
In the equation
$ 9 x^2+6 k x+4=0 $
$ a=9, b=6 k, c=4 $ then
$ \Rightarrow D=b^2-4 a c $
$ \Rightarrow D=(6 k)^2-4 \times 9 \times 4 $
$ \Rightarrow D=36 k^2-144$
Roots are equal
$ \Rightarrow D=0 $
$ \Rightarrow 36 k^2-144=0 $
$ \Rightarrow 36 k^2=144$
$\Rightarrow\text{k}^2=\frac{144}{36}$
$\Rightarrow\text{k}^2=4$
$\Rightarrow\text{k}^2=(\pm2)^2$
$\therefore\text{k}=\pm2$
$\therefore$ Roots are $=\frac{-\text{b}}{2\text{a}}$
$=\frac{\pm2\times6}{2\times9}$
$=\pm\frac{2}{3}$
View full question & answer→MCQ 271 Mark
If $x = 1$ is a common roots of the equations $ax^2+ ax + 3 = 0$ and $x^2+ x + b = 0$, then $ab =$
Answer$x = 1$ is the common roots given quadric equation are $ax^2+ ax + 3 = 0,$ and $x^2+ x + b = 0$
Then find the value of $q$.
Here, $ax^2+ ax + 3 = 0 ....(i)$
$x^2+ x + b = 0 ....(ii)$
Putting the value of $x = 1$ in equation $(i)$ we get
$a \times 1^2+ a \times 1 + 3 = 0$
$a + a + 3 = 0$
$2a = -3$
$\text{a}=-\frac{3}{2}$
Now, putting the value of $x = 1$ in equation $(ii)$ we get
$1^2+ 1 + b = 0$
$2 + b = 0$
$b = -2$
Then,
$\text{ab}=\frac{-3}{2}\times(-2)$
$=3$
Thus, the correct answer is $(a)$
View full question & answer→MCQ 281 Mark
If $x = 1$ is a common root of $ax^2+ ax + 2 = 0$ and $x^2+ x + b = 0,$ then $ab =$
Answer$x = 1$ is the common roots given quadric equation are $ax^2+ ax + 2 = 0$, and $x^2+ x + b = 0$
Then find the value of $ab.$
Here, $ax^2+ ax + 2 = 0 .....(i)$
$x^2+ x + b = 0 .....(ii)$
Putting the value of $x = 1$ in equation $(ii)$ we get
$1^2+ 1 + b = 0$
$2 + b = 0$
$b = -2$
Now, putting the value of $x = 1$ in equation $(i)$ we get
$a + a + 2 = 0$
$2a + 2 = 0$
$\text{a}=\frac{-2}{2}$
$a = -1$
$ab = (-1) \times (-2)$
Then, $ab = 2$
Thus, the correct answer is $(b)$
View full question & answer→MCQ 291 Mark
If the equation $x^2-b x+1=0 $ does not possess real roots, then:
- A
$-3 < b < 3$
- ✓
$-2 < b < 2$
- C
$b > 2$
- D
$b < -2$
AnswerCorrect option: B. $-2 < b < 2$
In the equation
$x^2-b x+1=0 $
$ \Rightarrow D=b^2-4 a c$
$ \Rightarrow D=(-b)^2-4 \times 1 \times 1 $
$ \Rightarrow D=b^2-4$
$\because$ The roots are not real
$\therefore D < 0$
$\Rightarrow b^2- 4 < 0$
$\Rightarrow b^2 < 4$
$\text{b}^2<(\pm2)^2$
$\therefore b < 2$ and $b > -2$ or $-2 < b$
$\therefore -2 < b < 2$
View full question & answer→MCQ 301 Mark
If $ax^2+ bx + c = 0$ has equal roots, then $c =$
- A
$\frac{-\text{b}}{2\text{a}}$
- B
$\frac{\text{b}}{2\text{a}}$
- C
$\frac{-\text{b}^2}{4\text{a}}$
- ✓
$\frac{-\text{b}^2}{4\text{a}}$
AnswerCorrect option: D. $\frac{-\text{b}^2}{4\text{a}}$
The given quadric equation is $ax^2+ bx + c = 0,$ and roots are equal
Then find the value of $c$.
Let $\alpha$ and $\beta$ be two roots of given equation $\alpha=\beta$
Then, as we know that sum of the roots
$\alpha+\beta=\frac{-\text{b}}{\text{a}}$
$\alpha+\alpha=\frac{-\text{b}}{\text{a}}$
$2\alpha=\frac{-\text{b}}{\text{a}}$
$\alpha=\frac{-\text{b}}{2\text{a}}$
And the product of the roots
$\alpha\cdot\beta=\frac{\text{c}}{\text{a}}$
$\alpha\alpha=\frac{\text{c}}{\text{a}}$
Putting the value of $\alpha$
$\frac{-\text{b}}{2\text{a}}\times\frac{-\text{b}}{2\text{a}}=\frac{\text{c}}{\text{a}}$
$\frac{\text{b}^2}{4\text{a}}=\text{c}$
Therefore, the value of $\text{c}=\frac{\text{b}^2}{4\text{a}}$
Thus, the correct answer is $(d)$
View full question & answer→MCQ 311 Mark
If $y = 1$ is a common root of the equations $ay^2+ ay + 3 = 0$ and $ y^2+ y + b = 0,$ then $ab$ equals:
- ✓
$3$
- B
$-\frac{1}{2}$
- C
$6$
- D
$-3$
Answer$\Rightarrow y = 1$
$\Rightarrow ay^2+ ay + 3 = 0$
$\therefore a \times (1)^2+ a.1 + 3 = 0$
$\Rightarrow a + a + 3 = 0$
$\Rightarrow 2a = -3$
$\Rightarrow\text{a}=\frac{-3}{2}$
and $y^2+ y + b = 0$
$\Rightarrow (1)^2+ (1) + b = 0$
$\Rightarrow 1 + 1 + b = 0$
$\Rightarrow 2 + b = 0$
$\therefore b = -2$
$\text{ab}=\frac{-3}{2}\times(-2)=3$
View full question & answer→MCQ 321 Mark
If $2$ is a root of the equation $x^2+ ax + 12 = 0$ and the quadratic equation $x^2+ ax + q = 0$ has equal roots, then $q =$
Answer$2$ is a root of equation $x^2+a x+12=0$
$ \Rightarrow(2)^2+\mathrm{a} \times 2+12=0 $
$ \Rightarrow 4+2 \mathrm{a}+12=0 $
$ \Rightarrow 2 \mathrm{a}=-(12+4) $
$ \Rightarrow 2 \mathrm{a}=-16 $
$ \Rightarrow \mathrm{a}=\frac{-16}{2} $
$ \Rightarrow \mathrm{a}=-8$
and in quadratic equation roots are equal $x^2+a x+q=0$
$ \Rightarrow b^2-4 a c=0 $
$ \Rightarrow a^2-4 q=0 $
$ \Rightarrow(-8)^2-4 q=0 $
$ \Rightarrow 64-4 q=0 $
$ \Rightarrow 4 q=64 $
$ \Rightarrow q=\frac{64}{4} $
$ \Rightarrow q=16 $
$ \therefore q=16$
View full question & answer→MCQ 331 Mark
If $x=0.2$ is a root of the equation $x^2-0.4 k=0$, then $k=$
View full question & answer→MCQ 341 Mark
The equation $\left(x^2+1\right)^2-x^2=0$ has
View full question & answer→MCQ 351 Mark
Which of the following equations has no real roots?
- ✓
$x^2-4 x+3 \sqrt{2}=0$
- B
$x^2+4 x-3 \sqrt{2}=0$
- C
$x^2-4 x-3 \sqrt{2}=0$
- D
$3 x^2+4 \sqrt{3} x+4=0$
AnswerCorrect option: A. $x^2-4 x+3 \sqrt{2}=0$
View full question & answer→MCQ 361 Mark
Which of the following equations has two distinct roots?
AnswerCorrect option: B. $x^2+x-5=0$
View full question & answer→MCQ 371 Mark
The quadratic equation $2 x^2-\sqrt{5} x+1=0$ has
View full question & answer→MCQ 381 Mark
Which of the following equations has the sum of its roots as 3 ?
AnswerCorrect option: B. $-x^2+3 x-3=0$
View full question & answer→MCQ 391 Mark
Which of the following equations has 2 as a root?
- A
$x^2-4 x+5=0$
- B
$x^2+3 x-12=0$
- ✓
$2 x^2-7 x+6=0$
- D
$3 x^2-6 x-2=0$
AnswerCorrect option: C. $2 x^2-7 x+6=0$
View full question & answer→MCQ 401 Mark
Which of the following is not a quadratic equation?
AnswerCorrect option: C. $(\sqrt{2} x+\sqrt{3})^2+x^2=3 x^2-5 x$
View full question & answer→MCQ 411 Mark
Which of the following is a quadratic equation?
AnswerCorrect option: D. $x^3-x^2=(x-1)^3$
View full question & answer→MCQ 421 Mark
The equation $9 x^2-6 x-2=0$ has
View full question & answer→MCQ 431 Mark
If $\alpha$ and $\beta$ are the zeroes of the polynomial $p(x)=k x^2-30 x+45 k$ and $\alpha+\beta=\alpha \beta$, then the value of $k$ is
- A
$-\frac{2}{3}$
- B
$-\frac{3}{2}$
- C
$\frac{3}{2}$
- ✓
$\frac{2}{3}$
AnswerCorrect option: D. $\frac{2}{3}$
(D)$\frac{2}{3}$
We have,
$
\begin{array}{ll}
& \alpha+\beta=-\frac{(-30)}{k} \text { and } \alpha \beta=\frac{45 k}{k} \\
\Rightarrow & \alpha+\beta=\frac{30}{k} \text { and } \alpha \beta=45 \\
\therefore \quad & \alpha+\beta=\alpha \beta \Rightarrow \frac{30}{k}=45 \Rightarrow k=\frac{2}{3}
\end{array}
$
View full question & answer→MCQ 441 Mark
If the discriminant of the quadratic equation $3 x^2-2 x+c=0$ is 16 , then the value of $c$ is
Answer(C)-1
It is given that: Discriminant $=16$.
$
\text { i.e. } \quad(-2)^2-4 \times 3 \times c=16 \Rightarrow 4-12 c=16 \Rightarrow 12 c=-12 \Rightarrow c=-1
$
View full question & answer→MCQ 451 Mark
The equation $x^2+x+1=0$ has
View full question & answer→MCQ 461 Mark
If one root of the quadratic equation $x^2-4 x+3$ is 1 , then the other root is
View full question & answer→MCQ 471 Mark
The ratio of the sum and product of the roots of the quadratic equation $5 x^2-6 x+21=0$ is
- A
$5: 21$
- ✓
$2: 7$
- C
$21: 5$
- D
$7: 2$
AnswerCorrect option: B. $2: 7$
View full question & answer→MCQ 481 Mark
The quadratic equation $x^2+x+1=0$ has _________ roots.
View full question & answer→MCQ 491 Mark
The value of $c$ for which the equation $a x^2+2 b x+c=0$ has equal roots is
- ✓
$\frac{b^2}{a}$
- B
$\frac{b^2}{4 a}$
- C
$\frac{a^2}{b}$
- D
$\frac{a^2}{4 b}$
AnswerCorrect option: A. $\frac{b^2}{a}$
View full question & answer→MCQ 501 Mark
24. If $a x^2+b x+c=0$ has equal roots, then $c=$
- A
$\frac{-b}{2 a}$
- B
$\frac{b}{2 a}$
- C
$\frac{-b^2}{4 a}$
- ✓
$\frac{b^2}{4 a}$
AnswerCorrect option: D. $\frac{b^2}{4 a}$
View full question & answer→MCQ 511 Mark
A quadratic equation whose roots are $2+\sqrt{3}$ and $2-\sqrt{3}$ is
- ✓
$x^2-4 x+1=0$
- B
$x^2+4 x+1=0$
- C
$4 x^2-3=0$
- D
$x^2-1=0$
AnswerCorrect option: A. $x^2-4 x+1=0$
(A)$x^2-4 x+1=0$
The quadratic equation whose roots are $\alpha=2+\sqrt{3}$ and $\beta=2-\sqrt{3}$ is $x^2-(\alpha+\beta) x+\alpha \beta=0$ or, $x^2-4 x+1=0$
View full question & answer→MCQ 521 Mark
The least positive value of $k$, for which the quadratic equation $2 x^2+k x-4=0$ has rational roots, is
- A
$\pm 2 \sqrt{2}$
- ✓
- C
$\pm 2$
- D
$\sqrt{2}$
Answer(B)2
Given quadratic equation will have rational roots, if its discriminant is a perfect square.
Let $D$ be the discriminant. Then, $D=k^2+32$.
Clearly, the least positive value of $k$ for which $D$ will be a perfect square, is 2 .
View full question & answer→MCQ 531 Mark
Which of the following quadratic equations has sum of its roots as 4 ?
AnswerCorrect option: B. $-x^2+4 x+4=0$
View full question & answer→MCQ 541 Mark
The roots of the equation $x^2+3 x-10=0$ are:
AnswerCorrect option: A. $2,-5$
View full question & answer→MCQ 551 Mark
The values of $k$ for which the quadratic equation $16 x^2+4 k x+9=0$ has real and equal roots are
AnswerCorrect option: C. $6,-6$
View full question & answer→MCQ 561 Mark
If $y=1$ is a common root of the equations $a y^2+a y+3=0$ and $y^2+y+b=0$, then $a b$ equals
View full question & answer→MCQ 571 Mark
The number of real roots of the equation $(x-1)^2+(x-2)^2+(x-3)^2=0$, is
Answer(D)none of these
For any real value of $x$, we find that $(x-1)^2+(x-2)^2+(x-3)^2 \neq 0$. Hence, the given equation has no real roots.
View full question & answer→MCQ 581 Mark
If the roots of the equation $a(b-c) x^2+b(c-a) x+c(a-b)=0$ are equal, then
- ✓
$b=\frac{2 a c}{a+c}$
- B
$b=\frac{a c}{a+c}$
- C
$c=\frac{2 a b}{a+b}$
- D
$a=\frac{2 b c}{b+c}$
AnswerCorrect option: A. $b=\frac{2 a c}{a+c}$
(A)$b=\frac{2 a c}{a+c}$
We observe that $x=1$ satisfies the given equation. So, it is a root of the equation. Thus, both the roots of the given equation are equal to 1.
$
\therefore \quad \text { Product of roots }=1 \times 1 \Rightarrow \frac{c(a-b)}{a(b-c)}=1 \Rightarrow c a-c b=a b-a c \Rightarrow 2 a c=a b+b c \Rightarrow b=\frac{2 a c}{a+c}
$
View full question & answer→MCQ 591 Mark
Quadratic equation whose roots are the reciprocal of the roots of the equation $a x^2+b x+c=0$, is
- A
$a x^2+c x+b=0$
- ✓
$c x^2+b x+a=0$
- C
$c x^2-b x+a=0$
- D
$c x^2+b x-a=0$
AnswerCorrect option: B. $c x^2+b x+a=0$
(B)$c x^2+b x+a=0$
Let $\alpha, \beta$ be the roots of the equation $a x^2+b x+c=0$. Then, $\alpha+\beta=-\frac{b}{a}$ and $\alpha \beta=\frac{c}{a}$. The equation whose roots are $\frac{1}{\alpha}$ and $\frac{1}{\beta}$ is
$
x^2-x\left(\frac{1}{\alpha}+\frac{1}{\beta}\right)+\frac{1}{\alpha \beta}=0 \text { or, } x^2-x\left(\frac{\alpha+\beta}{\alpha \beta}\right)+\frac{1}{\alpha \beta}=0 \text { or, } x^2+\frac{b x}{c}+\frac{a}{c}=0 \text { or, } c x^2+b x+a=0
$
View full question & answer→MCQ 601 Mark
If one root of the equation $a x^2+b x+c=0$ is three times the other, then
- A
$b^2=16 a c$
- B
$b^2=3 a c$
- ✓
$3 b^2=16 a c$
- D
$16 b^2=3 a c$
AnswerCorrect option: C. $3 b^2=16 a c$
(C)$3 b^2=16 a c$
Let the roots be $\alpha$ and $3 \alpha$. Then,
$\alpha+3 \alpha=-\frac{b}{a}$ and $\alpha \times 3 \alpha=\frac{c}{a} \Rightarrow \alpha=-\frac{b}{4 a}$ and $3 \alpha^2=\frac{c}{a} \Rightarrow 3 \times\left(-\frac{b}{4 a}\right)^2=\frac{c}{a} \Rightarrow 3 b^2=16 a c$
View full question & answer→MCQ 611 Mark
A quadratic equation whose one root is $1+\sqrt{2}$ and the sum of its roots is 2, is
- A
$x^2-2 x+1=0$
- ✓
$x^2-2 x-1=0$
- C
$x^2+2 x+1=0$
- D
$x^2+2 x-1=0$
AnswerCorrect option: B. $x^2-2 x-1=0$
(B) $x^2-2 x-1=0$
Let $\alpha, \beta$ be the roots of the desired equation and let $\alpha=1+\sqrt{2}$ and $\alpha+\beta=2$.
Then, $\alpha=\sqrt{2}+1$ and $\beta=1-\sqrt{2}$. So, the required equation is
$x^2-(\alpha+\beta) x+\alpha \beta=0$ or, $x^2-2 x-1=0$
View full question & answer→MCQ 621 Mark
If one root of the equation $3 x^2-8 x-(2 k+1)=0$ is seven times the other, then the value of $k$ is
- A
$\frac{7}{3}$
- B
$\frac{5}{3}$
- ✓
$-\frac{5}{3}$
- D
$-\frac{7}{3}$
AnswerCorrect option: C. $-\frac{5}{3}$
(C)$-\frac{5}{3}$
Let the roots of the given equation i.e. $3 x^2-8 x-(2 k+1)=0$ be $\alpha$ and $7 \alpha$. Then, $\alpha+7 \alpha=\frac{8}{3}$ and $\alpha \times 7 \alpha=-\frac{2 k+1}{3}$
$\Rightarrow \quad \alpha=\frac{1}{3}$ and $7 \alpha^2=-\frac{2 k+1}{3} \Rightarrow \frac{7}{9}=-\frac{2 k+1}{3} \Rightarrow 2 k+1=-\frac{7}{3} \Rightarrow k=-\frac{5}{3}$
View full question & answer→MCQ 631 Mark
If the sum and product of the roots of the equation $k x^2+6 x+4 k=0$ are equal, then the value of $k$ is
- ✓
$-\frac{3}{2}$
- B
$\frac{3}{2}$
- C
$\frac{2}{3}$
- D
$-\frac{2}{3}$
AnswerCorrect option: A. $-\frac{3}{2}$
(A)$-\frac{3}{2}$
Let $\alpha, \beta$ be the roots of the equation $k x^2+6 x+4 k=0$. Then, $\alpha+\beta=-\frac{6}{k}$ and $\alpha \beta=4$
It is given that
$
\alpha+\beta=\alpha \beta \Rightarrow-\frac{6}{k}=4 \Rightarrow k=-\frac{3}{2}
$
View full question & answer→MCQ 641 Mark
If $a, b, c$ are positive real numbers, then the number of real roots of the equation $a x^2+b|x|+c=0$, is
Answer(C)0
We find that for positive values of $a, b$ and $c ; a x^2+b|x|+c=a|x|^2+b|x|+c>0$ for all real values of $x$. Therefore, $a x^2+b|x|+c \neq 0$ for any real values of $x$.
Hence, $a x^2+b|x|+c=0$ has no real root.
View full question & answer→MCQ 651 Mark
If $c$ and $d$ are roots of the equation $(x-a)(x-b)-k=0$, then $a, b$ are roots of the equation
- A
$(x-c)(x-d)-k=0$
- ✓
$(x-c)(x-d)+k=0$
- C
$(x-a)(x-c)+k=0$
- D
$(x-b)(x-d)+k=0$
AnswerCorrect option: B. $(x-c)(x-d)+k=0$
(B)$(x-c)(x-d)+k=0$
If $c$ and $d$ are roots of the equation $(x-a)(x-b)-k=0$, then
$
\begin{array}{l}
(x-a)(x-b)-k=(x-c)(x-d) \\
\Rightarrow \quad(x-c)(x-d)+k=(x-a)(x-b) \Rightarrow a, b \text { are roots of the equation }(x-c)(x-d)+k=0
\end{array}
$
View full question & answer→MCQ 661 Mark
The number of real roots of the equation $x^2-3|x|+2=0$, is
Answer(A)4
We have, $x^2-3|x|+2=0$
$
\Rightarrow \quad|x|^2-3|x|+2=0 \quad\left[\because x^2=|x|^2\right]
$
$
\Rightarrow \quad|x|^2-2|x|-|x|+2=0 \Rightarrow(|x|-2)(|x|-1)=0 \Rightarrow|x|=1,|x|=2 \Rightarrow x= \pm 1, x= \pm 2
$
Hence, the given equation has four real roots.
View full question & answer→MCQ 671 Mark
The number of real roots of the equation $x^2+3|x|+2=0$, is
Answer(C)0
We find that $x^2+3|x|+2=|x|^2+3|x|+2 \neq 0$ for any real $x$. Hence, the given equation has no real root.
ALITER We find that $x^2+3|x|+2=|x|^2+2|x|+|x|+2=(|x|+2)(|x|+1) \neq 0$ for any $x$.
View full question & answer→MCQ 681 Mark
If $\alpha$ and $\beta$ are two roots of the quadratic equation $a x^2+b x+c=0$, then $a x^2+b x+c=$
- A
$a(x+\alpha)(x+\beta)$
- ✓
$a(x-\alpha)(x-\beta)$
- C
$b(x-\alpha)(x-\beta)$
- D
$c(x-\alpha)(x-\beta)$
AnswerCorrect option: B. $a(x-\alpha)(x-\beta)$
(B)$a(x-\alpha)(x-\beta)$
Since $\alpha$ and $\beta$ are roots of $a x^2+b x+c$.
$
\therefore \quad a x^2+b x+c=\lambda(x-\alpha)(x-\beta) \text { for some } \lambda \text { and all } x
$
$
\Rightarrow \quad a x^2+b x+c=i x^2-i x(\alpha+\beta)+\lambda \alpha \beta \text { for all } x \qquad [By comparing coefficients of x^2 ]$
$\Rightarrow \quad \lambda=a
$
Hence, $a x^2+b x+c=a(x-\alpha)(x-\beta)$.
View full question & answer→MCQ 691 Mark
Answer(C)exactly two roots
A quadratic equation has exactly two roots, say $\alpha$ and $\beta$, such that $\alpha+\beta=-\frac{b}{a}$ and $\alpha \beta=\frac{c}{a}$.
View full question & answer→MCQ 701 Mark
The equation $a x^2+b c+c=0$ is a quadratic equation for
Answer(B)all non-zero values of a
The equation $a x^2+b x+c=0, a \neq 0$ is defined as a quadratic equation for all values of $b$ and c. Hence, option (b) is correct.
View full question & answer→MCQ 711 Mark
If $a$ and $b$ are roots of the equation $x^2+a x+b=0$, then $a+b=$
View full question & answer→MCQ 721 Mark
If $\sin \alpha$ and $\cos \alpha$ are the roots of the equation $a x^2+b x+c=0$, then $b^2=$
- A
$a^2-2 a c$
- ✓
$a^2+2 a c$
- C
$a^2-a c$
- D
$a^2+a c$
AnswerCorrect option: B. $a^2+2 a c$
View full question & answer→MCQ 731 Mark
If $\left(a^2+b^2\right) x^2+2(a c+b d) x+c^2+d^2=0$ has no real roots, then
- A
$a d=b c$
- B
$a b=c d$
- C
$a c=b d$
- ✓
$a d \neq b c$
AnswerCorrect option: D. $a d \neq b c$
View full question & answer→MCQ 741 Mark
The number of quadratic equations having real roots and which do not change by squaring their roots is
MCQ 751 Mark
If $a$ and $b$ can take values $1,2,3,4$. Then the number of the equations of the form $a x^2+b x+1=0$ having real roots is
View full question & answer→MCQ 761 Mark
If the equation $x^2-b x+1=0$ does not possess real roots, then
View full question & answer→MCQ 771 Mark
If the roots of the equation $\left(a^2+b^2\right) x^2-2 b(a+c) x+\left(b^2+c^2\right)=0$ are equal, then
- A
$2 b=a+c$
- ✓
$b^2=a c$
- C
$b=\frac{2 a c}{a+c}$
- D
$b=a c$
AnswerCorrect option: B. $b^2=a c$
View full question & answer→MCQ 781 Mark
If the equation $\left(a^2+b^2\right) x^2-2(a c+b d) x+c^2+d^2=0$ has equal roots, then
- A
$a b=c d$
- ✓
$a d=b c$
- C
$a d=\sqrt{b c}$
- D
$a b=\sqrt{c d}$
AnswerCorrect option: B. $a d=b c$
View full question & answer→MCQ 791 Mark
The positive value of $k$ for which the equation $x^2+k x+64=0$ and $x^2-8 x+k=0$ will bot have real roots, is
View full question & answer→MCQ 801 Mark
If the equation $a x^2+2 x+a=0$ has two equal roots, if
- ✓
$a= \pm 1$
- B
$a=0$
- C
$a=0,1$
- D
$a=-1,0$
AnswerCorrect option: A. $a= \pm 1$
View full question & answer→MCQ 811 Mark
If the equation $9 x^2+6 k x+4=0$ has equal roots, then the roots are both equal to
- ✓
$\pm \frac{2}{3}$
- B
$\pm \frac{3}{2}$
- C
$0$
- D
$\pm 3$
AnswerCorrect option: A. $\pm \frac{2}{3}$
View full question & answer→MCQ 821 Mark
If the equation $x^2-a x+1=0$ has two distinct roots, then
- A
$|a|=2$
- B
$|a|<2$
- ✓
$|a|>2$
- D
AnswerCorrect option: C. $|a|>2$
View full question & answer→MCQ 831 Mark
If one root of the equation $4 x^2-2 x+(\lambda-4)=0$ be the reciprocal of the other, then $\lambda=$
View full question & answer→MCQ 841 Mark
If the sum of the roots of the equation $x^2-(k+6) x+2(2 k-1)=0$ is equal to half of the: product, then $k=$
View full question & answer→MCQ 851 Mark
If one root of the equation $a x^2+b x+c=0$ is three times the other, then $b^2: a c=$
- A
$3: 1$
- B
$3: 16$
- ✓
$16: 3$
- D
$16: 1$
AnswerCorrect option: C. $16: 3$
View full question & answer→MCQ 861 Mark
If $x^2+k(4 x+k-1)+2=0$ has equal roots, then $k=$
- A
$-\frac{2}{3}, 1$
- ✓
$\frac{2}{3},-1$
- C
$\frac{3}{2}, \frac{1}{3}$
- D
$-\frac{3}{2},-\frac{1}{3}$
AnswerCorrect option: B. $\frac{2}{3},-1$
View full question & answer→MCQ 871 Mark
If $p$ and $q$ are the roots of the equation $x^2+p x+q=0$, then
- ✓
$p=1, q=-2$
- B
$p=0, q=1$
- C
$p=-2, q=0$
- D
$p=-2, q=1$
AnswerCorrect option: A. $p=1, q=-2$
View full question & answer→MCQ 881 Mark
26. If 2 is a root of the equation $x^2+b x+12=0$ and the equation $x^2+b x+q=0$ has equal roots, then $q=$
View full question & answer→MCQ 891 Mark
The value of $\sqrt{6+\sqrt{6+\sqrt{6+}}} \ldots$ is
View full question & answer→MCQ 901 Mark
23. If the equation $x^2+4 x+k=0$ has real and distinct roots, then
- ✓
$k < 4$
- B
$k > 4$
- C
$k \geq 4$
- D
$k \leq 4$
AnswerCorrect option: A. $k < 4$
View full question & answer→MCQ 911 Mark
22. If $x=1$ is a common root of the equations $a x^2+a x+6=0$ and $x^2+x+b=0$, then $a b=$
View full question & answer→MCQ 921 Mark
21. If the sum of the roots of the equation $x^2-x=\lambda(2 x-1)$ is zero, then $\lambda=$
- A
- B
- ✓
$-\frac{1}{2}$
- D
$\frac{1}{2}$
AnswerCorrect option: C. $-\frac{1}{2}$
View full question & answer→MCQ 931 Mark
If the sum and product of the roots of the equation $k x^2+6 x+4 k=0$ are equal, then $k=$
- ✓
$-\frac{3}{2}$
- B
$\frac{3}{2}$
- C
$\frac{2}{3}$
- D
$-\frac{2}{3}$
AnswerCorrect option: A. $-\frac{3}{2}$
View full question & answer→MCQ 941 Mark
A quadratic equation whose one root is 2 and the sum of whose roots is zero, is
- A
$x^2+4=0$
- ✓
$x^2-4=0$
- C
$4 x^2-1=0$
- D
$x^2-2=0$
AnswerCorrect option: B. $x^2-4=0$
View full question & answer→MCQ 951 Mark
If one root of the equation $2 x^2+k x+4=0$ is 2 , then the other root is
View full question & answer→MCQ 961 Mark
If one root of the equation $x^2+a x+3=0$ is 1 , then its other root is
View full question & answer→MCQ 971 Mark
The discriminant of the quadratic equation $(x+2)^2=0$ is
View full question & answer→MCQ 981 Mark
A quadratic equation can have
MCQ 991 Mark
Which of the following equations has 3 as a root?
- ✓
$x^2-4 x+3=0$
- B
$x^2+4 x+3=0$
- C
$x^2+5 x+6=0$
- D
$x^2+7 x+12=0$
AnswerCorrect option: A. $x^2-4 x+3=0$
View full question & answer→MCQ 1001 Mark
If $-\frac{1}{2}$ is a root of the equation $x^2-k x-\frac{5}{4}=0$, then the value of $k$ is
- A
- ✓
- C
$\frac{1}{4}$
- D
$\frac{1}{2}$
View full question & answer→