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. $\frac{b^2}{4 a}$
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