MCQ 1011 Mark
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 1021 Mark
$2\text{x}^{2}+5\sqrt{3} \text{ x}+6 = 0$ have :
AnswerComparing the given equation to the below equation
$\text{ax}^{2} + \text{bx}+\text{c} = 0$
$\text{a} = 2,\text{b} = 5\sqrt{3},\text{c} = 6$
$\text{D} = \text{b}^{2}-\text{4ac}$
$\text{D} = (5\sqrt{3})^{2} - 4\times2\times6$
$\text{D} = 75 - 48$
$\text{D} = 27$
$\text{D}>0$
If $b^2-4 a c>0,$
then the equation has real and distinct roots.
View full question & answer→MCQ 1031 Mark
If the equation $4x^2- 3kx + 1 = 0$ has equal roots then $k = ?$
- A
$\pm\frac{2}{3}$
- B
$\pm\frac{1}{3}$
- C
$\pm\frac{3}{4}$
- ✓
$\pm\frac{4}{3}$
AnswerCorrect option: D. $\pm\frac{4}{3}$
Since the roots of the equation $4 {x}^2-3 {kx}+1=0$ are equal,
$D = 0$
$\Rightarrow b^2-4 a c=0$
$\Rightarrow(-3 k)^2-4 \times 4 \times 1=0$
$\Rightarrow 9 k^2-16=0$
$\Rightarrow 9 k^2=16$
$\Rightarrow\text{k}^2=\frac{16}{9}$
$\Rightarrow\text{k}=\pm\frac{4}{3}$
View full question & answer→MCQ 1041 Mark
The hypotenuse of a right triangle is 6m more than twice the shortest side. The third side is $2m$ less than the hypotenuse. The representation of the above situation in the form of a quadratic equation is :
- A
$(2 x+6)_2=x^2-(2 x+4)^2$
- B
- C
$(2 x+6)^2+x^2=(2 x+4)^2$
- ✓
$(2 x+6)^2=x^2+(2 x+4)^2$
AnswerCorrect option: D. $(2 x+6)^2=x^2+(2 x+4)^2$
Let the shortest side of a right angled triangle be $x$ meters.
Then according to question, its hypotenuse will be $(2x + 6)$ meters and,
The third side will be $(2x + 6 - 2) = (2x + 4)$ meters.
Now, using Pythagoras theorem, $\mathrm{(Hypotenuse)^2= (Base)^2+ (Perpendicular)^2}$
$\Rightarrow (2 x+6)^2=x^2+(2 x+4)^2$
View full question & answer→MCQ 1051 Mark
Which of the following is a quadratic equation ?
- A
$(\text{k}+1)\text{ x}^{2}+\frac{3}{2}\text{ x}-5 = 0, \text{k} = -1$
- ✓
$\text{x}^{3}-\text{x}^{2}=({\text{x}}-1)^{3}$
- C
$\text{x}^{2}+\text{2x}+1=(4-\text{x})^{2}+{3}$
- D
$-\text{2x}^{2} = (5-{\text{x}})(\text{2x}-\frac{2}{5})$
AnswerCorrect option: B. $\text{x}^{3}-\text{x}^{2}=({\text{x}}-1)^{3}$
In equation$ x^3-x^2=(x-1)^3 $
$ \Rightarrow x^3-x^2=x^3-1-3 x^2+3 x $
$ \Rightarrow-x^2+3 x^2-3 x+1=0 $
$ \Rightarrow 2 x^2-3 x+1=0 $
It is a quadratic equation as its degree is $2$.
View full question & answer→MCQ 1061 Mark
If $x = 2$ is a root of the quadratic equation $3x^2- px - 2 = 0,$ then the value of $p$ is :
AnswerGiven : $p(x)=3 x^2-p x-2=0$
$\therefore p(2)=3(2)^2-p(2)-2=0$
$\Rightarrow 12 - 2p - 2 = 0$
$\Rightarrow -2p = -10$
$\Rightarrow p = 5$
View full question & answer→MCQ 1071 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 1081 Mark
A quadratic equation whose one root is $3$ is :
- A
$ x^2-5 x-6=0 $
- B
$ x^2-6 x-6=0 $
- ✓
$ x^2-5 x+6=0 $
- D
$ x^2+6 x-5=0 $
AnswerCorrect option: C. $ x^2-5 x+6=0 $
since $3$ is the root of the equation $, x = 3$ must satisfy the equation.
Applying $x = 3$ in the equation $ x^2-5 x+6=0 $
gives, $(3)^2- 5(3) + 6 = 0$
$\Rightarrow 9 - 15 + 6 =0$
$\Rightarrow 15 - 15 = 0$
$\Rightarrow 0 = 0$
$\Rightarrow \text{L.H.S. = R.H.S}$
$ x^2-5 x+6=0 $ is a required equation which has $3$ as root.
View full question & answer→MCQ 1091 Mark
Directions : In the following questions, the Assertions $(A)$ and Reason $(s)\ (R)$ have been put forward. Read both the statements carefully and choose the correct alternative from the following :
Assertion : $3y^2+ 17y - 30 = 0$ have distinct roots.
Reason : The quadratic equation $ax^2+ bx + c = 0$ have distinct roots $($real roots$)$ if $D > 0$.
- ✓
If both assertion and reason are true and reason is the correct explanation of assertion.
- B
If both assertion and reason are true but reason is not the correct explanation of assertion.
- C
If assertion is true but reason is false.
- D
If both assertion and reason are false.
AnswerCorrect option: A. If both assertion and reason are true and reason is the correct explanation of assertion.
$3 y^2+17 y-30=0$
Therefore, $D=b^2-4 a c$
$\Rightarrow D=(17)^2-4(3)(-30)$
$\Rightarrow D = 289 + 360$
$\Rightarrow D = 649 > 0$
So, roots are real and distinct.
View full question & answer→MCQ 1101 Mark
The two numbers whose sum is $27$ and their product is $182$ are :
- A
$14$ and $15$
- B
$12$ and $13$
- ✓
$13$ and $14$
- D
$12$ and $15$
AnswerCorrect option: C. $13$ and $14$
Let the one number be $x.$
As the sum of numbers is $27,$ then the other number will be $(27 - x)$
According to question.
$x(27 - x) = 182$
$ \Rightarrow 27 x-x^2=182$
$ \Rightarrow x^2-27 x+182=0 $
$ \Rightarrow x^2-14 x-13 x+182=0 $
$\Rightarrow x(x - 14) -13(x - 14) = 0$
$\Rightarrow (x - 13) (x - 14) = 0$
$\Rightarrow x - 13 = 0$ and $x - 14 = 0$
$x = 13$ and $x = 14$
Now, the other number $= 27 - 13 = 14$ and $27 - 14 = 13$
$\therefore$ The required two numbers are $13$ and $14.$
View full question & answer→MCQ 1111 Mark
$x^2- 6x + 6 = 0$ have :
AnswerComparing the given equation to the below equation
$ a x^2+b x+c=0 $
$a=1, b=-6, c=6$
$ D=b^2-4 a c $
$ D=(-6)^2-4 \times 1 \times 6 $
$D = 36 - 24$
$D = 12$
$D > 0.$
If $b^2- 4ac > 0$, then the equation has real and distinct roots
Real and Distinct roots.
View full question & answer→MCQ 1121 Mark
The roots of the quadratic equation $9a^2b^2x^2- 16\text{ abcdx} - 25c^2d^2= 0$ are :
- A
$\frac{{-25}\text{cd}}{{9}\text{ab}}$ and $\frac{-\text{cd}}{\text{ab}}$
- B
$\frac{{-25}\text{cd}}{{9}\text{ab}}$ and $\frac{\text{cd}}{\text{ab}}$
- C
$\frac{{25}\text{cd}}{{9}\text{ab}}$ and $\frac{\text{cd}}{\text{ab}}$
- ✓
$\frac{{25}\text{cd}}{{9}\text{ab}}$ and $\frac{\text{-cd}}{\text{ab}}$
AnswerCorrect option: D. $\frac{{25}\text{cd}}{{9}\text{ab}}$ and $\frac{\text{-cd}}{\text{ab}}$
Using factorisation method $9 a^2 b^2 x^2-16 \text{ abcdx}-25 c^2 d^2=0$
$\Rightarrow 9 a^2 b^2 x^2-25 \text{ abcdx}+9 \text{ abcdx} -25 c^2 d^2=0$
$\Rightarrow abx(9abx - 25cd) + cd(9abx - 25cd) = 0$
$(abx + cd) (9abx - 25cd) = 0$
$abx + cd = 0$ and $9abx - 25cd = 0$
$\Rightarrow \text{x}=\frac{-\text{cd}}{\text{ab}}$ and $\text{x}=\frac{{25}\text{cd}}{{9}\text{ab}}$
View full question & answer→MCQ 1131 Mark
$9x^2+ 12x + 4 = 0$ have :
AnswerComparing the given equation to the below equation.
$ax^2+ bx + c = 0$
$a = 9, b = 12, c = 4$
$D = b^2- 4ac$
$D = 122 - 4 \times 9 \times 4$
$D = 144 - 144$
$D = 0$
If $b^2- 4ac = 0$ then equation have equal and real roots.
View full question & answer→MCQ 1141 Mark
The roots of the quadratic equation $x^2- 11x - 10 = 0$ are :
AnswerHere $, a = 1, b = -11, c = -10$
Then $,b^2-4 a c=(-11)^2-4 \times 1 \times(-10)$
$\Rightarrow 121+40=161$
Since $,b^2-4 a c>0$
$\therefore$ The roots of the quadratic equation $x^2-11 x-10=0$ is real and distinct.
View full question & answer→MCQ 1151 Mark
For what value of $p,$ the quadratic equation $x^2- 4x + p = 0$ has real roots ?
AnswerCorrect option: D. $\text{p} \underline{<} 4$
If the quadratic equation $x^2- 4x + p = 0$ has real roots,
then $\text{b}^{2} - 4\text{ac} \underline{>} 0$
$\Rightarrow (-4)^{2} - 4 \times 1 \times \text{ p}\underline{>} 0$
$\Rightarrow 16 - 4\text{p} \underline{>} 0$
$\Rightarrow4-\text{ p}\underline {>} 0$
$\Rightarrow\text{p}\underline{<}4$
View full question & answer→MCQ 1161 Mark
The discriminant of the equation $(2a + b) x = x^2+ 2ab$ is $......$
- A
$ \left(2 a+b^2\right)$
- ✓
$(2 a-b)^2 $
- C
$ (2 a+b)^2 $
- D
$\left(2 a-b^2\right)$
AnswerCorrect option: B. $(2 a-b)^2 $
$ (2 a+b) x=x^2+2 a b $
$ x^2-(2 a+b) x+2 a b=0 $
$ D=b^2-4 a c $
$ D=[-(2 a+b)]^2-4 \times 1 \times 2 a b $
$ D=4 a^2+b^2+4 a b-8 a b $
$ D=4 a^2+b^2-4 a b $
$ D=(2 a-b)^2 $
View full question & answer→MCQ 1171 Mark
Choose the correct answer from the given four options in the following questions : Which of the following is a quadratic equation?
- A
$\text{x}^2+2\text{x}+1=(4-\text{x})^2+3.$
- B
$-2\text{x}^2=(5-\text{x})\Big(2\text{x}-\frac{2}{5}\Big).$
- C
$(\text{k}+1)\text{x}^2+\frac{3}{2}\text{x}=7,\text{ where k}=-1.$
- ✓
$\text{x}^3-\text{x}^2=(\text{x}-1)^3.$
AnswerCorrect option: D. $\text{x}^3-\text{x}^2=(\text{x}-1)^3.$
$(a) $Given that, $x^2+2 x+1=(4-x)^2+3$
$\Rightarrow x^2+2 x+1=16+x^2-8 x+3$
$\Rightarrow 10x - 18 = 0$
Which is not of the form $\text{ax}^2+\text{bx}+\text{c},\text{ a}\neq0$.
Thus, the equation is not a quadratic equation.
$(b)$ Given that, $-2\text{x}^2=(5-\text{x})\Big(2\text{x}-\frac{2}{5}\Big)$
$\Rightarrow-2\text{x}^2=10\text{x}-2\text{x}^2-2+\frac{2\text{x}}{5}$
$\Rightarrow\ 50\text{x}+2\text{x}-10=0$
$\Rightarrow\ 52\text{x}-10=0$
which is also not a quadratic equation.
$(c)$ Given that, $\text{x}^2(\text{k} + 1)+\frac{3}{2}\text{x}=7$
Given, $\text{k}=-1$
$\Rightarrow\ \text{x}^2(-1+1)+\frac{3}{2}\text{x}=7$
$\Rightarrow\ 3\text{x}-14=0$
which is also not a quadratic equation.
$(d)$ Given that $, x^3-x^2=(x-1)^3 $
$ \Rightarrow x^3-x^2=x^3-3 x^2(1)+3 x(1)^2-(1)^3 $
$ {\left[\because(a-b)^3=a^3-b^3+3 a b^2-3 a^2 b\right]} $
$ \Rightarrow x^3-x^2=x^3-3 x^2+3 x-1 $
$ \Rightarrow-x^2+3 x^2-3 x+1=0 $
$ \Rightarrow 2 x^2-3 x+1=0 $
which represents a quadratic equation because it has the quadratic from $\text{ax}^2+\text{bx}+\text{c},\text{ a}\neq0$.
View full question & answer→MCQ 1181 Mark
The roots of $a x^2+b x+c=0, a \neq 0$ are real and unequal, if $\left(b^2-4 a c\right)$ is :
AnswerSince the roots of the equation $ax^2+ bx + c = 0,$
$\text{a}\neq0$ are real and unequal,
we must have $D > 0$
$\Rightarrow b^2- 4ac > 0$
View full question & answer→MCQ 1191 Mark
Choose the correct answer from the given four options in the following questions : If $\frac{1}{2}$ is a root of the equation $\text{x}^2+\text{kx}-\frac{5}{4}=0,$ then the value of $k$ is.
- ✓
$2$
- B
$-2$
- C
$\frac{1}{4}$
- D
$\frac{1}{2}$
AnswerSince, $\frac{1}{2}$ is a root of the quadratic equation $\text{x}^2+\text{kx}-\frac{5}{4}=0$.
Then, $\Big(\frac{1}{2}\Big)^2+\text{k}\Big(\frac{1}{2}\Big)-\frac{5}{4}=0$
$\Rightarrow\ \frac{1}{4}+\frac{\text{k}}{2}-\frac{5}{4}=0$
$\Rightarrow\ \frac{1+2\text{k}-5}{4}=0$
$\Rightarrow\ 2\text{k}-4=0$
$\Rightarrow\ 2\text{k}=4$
$\Rightarrow\ \text{k}=2.$
View full question & answer→MCQ 1201 Mark
A quadratic equation $ax^2+ bx + c = 0$ has real and distinct roots, if :
- A
- B
$ b^2-4 a c<0 $
- C
$ b^2-4 a c=0 $
- ✓
$ b^2-4 a c>0 $
AnswerCorrect option: D. $ b^2-4 a c>0 $
A quadratic equation $ax^2+ bx + c = 0$ has real and distinct roots, if $ b^2-4 a c>0 $
View full question & answer→MCQ 1211 Mark
$(x + 2)^3= 2x (x^2- 1)$ is a :
AnswerGiven : $(x+2)^3=2 x\left(x^2-1\right) $
$ \Rightarrow x^3+8+3 x x \times 2(x+2)=2 x^3-2 x $
$ \Rightarrow x^3+8+6 x^2+12 x=2 x^3-2 x $
$ \Rightarrow 2 x^3-x^3-6 x^2-12 x-2 x-8=0 $
$ \Rightarrow x^3-6 x^2-14 x-8=0 $
Here since the degree is $3,$
$\therefore$ it is a cubic equation.
View full question & answer→MCQ 1221 Mark
$4x^2- 2x - 3 = 0$ have :
Answer$ D=b^2-4 a c $
$ D=(-2)^2-4 \times 4 \times(-3) $
$D = 4 + 48$
$D = 52$
$D > 0$.
Real and Distinct roots.
View full question & answer→MCQ 1231 Mark
Directions : In the following questions, the Assertions $(A)$ and Reason $(s) \ (R)$ have been put forward. Read both the statements carefully and choose the correct alternative from the following :
Assertion $(A) : $ The equation $a^2+ 3x + 1 = (x - 2)^2$ is a quadratic equation.
Reason $(R) : $ Any equation of the form $ax^2+ bx + c = 0 \text{a}\neq0,$ is a quadratic equation.
- A
Both assertion $(A)$ and reason $(R)$ are true and reason $(R)$ isthe correct explanation of assertion $(A)$.
- B
Both assertion $(A)$ and reason $(R)$ are true but reason $(R)$ is not the correct explanation of assertion $(A)$.
- C
Assertion $(A)$ is true but reason $(R)$ is false.
- ✓
Assertion $(A)$ is false but reason $(R)$ is true
AnswerCorrect option: D. Assertion $(A)$ is false but reason $(R)$ is true
Assertion $(A)$ is false but reason $(R)$ is true
View full question & answer→MCQ 1241 Mark
$500$ bananas were divided equally among a certain number of students. If there were $25$ more students, each would have received one banana less. Then the number of students is :
AnswerLet the number of students be $x$
$\therefore$ Each student would get $= \frac{500}{\text{x}}$ bananas
$\therefore$ if there were $25$ more students, then each student would get $= \frac{500}{\text{x}+25}$ bananas
According to questions, $\frac{500}{\text{x}}-\frac{500}{\text{x}+25} = 1$
$\Rightarrow\frac{500\text{x}+12500-500\text{x}}{\text{x}(\text{x}+25)} = 1$
$\Rightarrow\frac{500}{{\text{x}}^{2}+25\text{x}} = 1$
$ \Rightarrow x^2+25 x-12500=0$
$\Rightarrow x^2+125 x-100 x-12500=0$
$\Rightarrow x(x + 125) -100(x + 125) = 0$
$\Rightarrow (x + 125) (x - 100) = 0$
$\Rightarrow x + 125 = 0$ and $x - 100 = 0$
$\Rightarrow x = -125$ and $x = 100 [x = -125$ is not possible$]$
$\therefore$ The number of student is $100$
View full question & answer→MCQ 1251 Mark
If the roots of $5 x^2-k x+1=0$ are real and distinct, then :
AnswerCorrect option: D. either $\text{k}>2\sqrt5$ or $\text{k}<-2\sqrt5$
Given, the roots of $5 x^2-k x+1=0$ are real and distinct.
$\Rightarrow D > 0$
$ \Rightarrow b^2-4 a c>0 $
$\Rightarrow(-k)^2-4 \times 5 \times 1>0 $
$ \Rightarrow k^2-20>0 $
$ \Rightarrow k^2>20$
$\Rightarrow\text{k}>\sqrt{20}$ or $\text{k}<-\sqrt{20}$
$\Rightarrow\text{k}>2\sqrt5$ or $\text{k}<-2\sqrt5$
View full question & answer→MCQ 1261 Mark
If $p = -7$ and $q = 12$ and $x^2+ px + q = 0,$ Then the value of $x$ is :
- ✓
$3$ and $4$
- B
$3$ and $-4$
- C
$-3$ and $-4$
- D
$-3$ and $4$
AnswerCorrect option: A. $3$ and $4$
Putting the values of $p$ and $q$ in given equation, we get
$ x^2+(-7) x+12=0 $
$ \Rightarrow x^2-7 x+12=0 $
$ \Rightarrow x^2-4 x-3 x+12=0 $
$\Rightarrow x(x - 4) -3(x - 4) = 0$
$\Rightarrow (x - 3) (x - 4) = 0$
$\Rightarrow x - 3 = 0 $ and $x - 4 = 0$
$\Rightarrow x = 3$ and $x = 4$
View full question & answer→MCQ 1271 Mark
$9x^2- 6x - 4 = 0$ have :
Answer$ D=b^2-4 a c $
$ D=(-6)^2-4 \times 9 \times(-4)$
$D = 36 + 144$
$D = 180$
$D > 0.$
Real and Distinct roots.
View full question & answer→MCQ 1281 Mark
Choose the correct answer from the given four options in the following questions:
Which of the following is not a quadratic equation?
- A
$2(\text{x} - 1)^2 = 4\text{x}^2 - 2\text{x} + 1.$
- B
$2\text{x} - \text{x}^2 = \text{x}^2 + 5.$
- ✓
$\big(\sqrt{2}\text{x}+\sqrt{3}\big)^2=3\text{x}^2-5\text{x}.$
- D
$(\text{x}^2 + 2\text{x})^2 = \text{x}^{4} + 3 + 4\text{x}^3.$
AnswerCorrect option: C. $\big(\sqrt{2}\text{x}+\sqrt{3}\big)^2=3\text{x}^2-5\text{x}.$
$\big(\sqrt{2}\text{x}+\sqrt{3}\big)^2=3\text{x}^2-5\text{x}.$
View full question & answer→MCQ 1291 Mark
In the equation $a x^2+b x+c=0$, it is given that $D=\left(b^2-4 a c\right)>0$. Then, the roots of the equation are :
AnswerFor equation $a x^2+b x+c=0,$ it is given that $D=\left(b^2-4 a c\right)>0$.
This means that the roots of the equation are real and unequal.
View full question & answer→MCQ 1301 Mark
If the roots of the equation $a x^2+b x+c=0$ are equal, then then $c = ?$
- A
$\frac{-\text{b}}{\text{2a}}$
- B
$\frac{\text{b}}{\text{2a}}$
- C
$\frac{-\text{b}^2}{\text{4a}}$
- ✓
$\frac{\text{b}^2}{\text{4a}}$
AnswerCorrect option: D. $\frac{\text{b}^2}{\text{4a}}$
Since roots of the equation $a x^2+b x+c=0$ are equal,
$D = 0$
$\Rightarrow b^2-4 a c=0$
$\Rightarrow b^2=4 a c$
$\Rightarrow\text{c}=\frac{\text{b}^2}{\text{4a}}$
View full question & answer→MCQ 1311 Mark
The perimeter of a rectangle is $82m$ and its area is $400m^2$. The breadth of the rectangle is :
AnswerPerimeter of a rectangle $= 82m$
Let the breadth of the rectangle be $x m.$
Then, length of the rectangle $=\frac{\text{Perimeter}}{2}-\text{Breadth}$
$=\frac{82}{2}-\text{x}=(41-\text{x})\text{m}$
Now Area $= 400m^2$
$\Rightarrow$ Length $\times$ Breadth $= 400$
$\Rightarrow x(41 - x) = 400$
$ \Rightarrow 41 x-x^2=400 $
$ \Rightarrow x^2-41 x+400=0 $
$ \Rightarrow x^2-25 x-16 x+400=0 $
$\Rightarrow x(x - 25) - 16(x - 25) = 0$
$\Rightarrow (x - 25)(x - 16) = 0$
$\Rightarrow x - 25 = 0$ or $x - 16 = 0$
$\Rightarrow x = 25$ or $x = 16$
Hence, the length is $25m$ and the breadth is $16m.$
View full question & answer→MCQ 1321 Mark
Choose the correct answer from the given four options in the following questions : Which of the following equations has two distinct real roots ?
- A
$2\text{x}^2-3\sqrt{2}\text{x}+\frac{9}{2}=0$
- ✓
$\text{x}^2+\text{x}-5=0$
- C
$\text{x}^2+3\text{x}+2\sqrt{2}=0$
- D
$5\text{x}^2-3\text{x}+1=0.$
AnswerCorrect option: B. $\text{x}^2+\text{x}-5=0$
$(a)$ Given equation is $2\text{x}^2-3\sqrt{2}\text{x}+\frac{9}{4}=0,$
On comparing with $ax^2+ bx + c = 0$
$\text{a}=2,\text{b}=-3\sqrt{2}$ and $\text{C}=\frac{9}{4}$
Now, $D = b^2- 4ac$
$=\big(-3\sqrt{2}\big)-4(2)\Big(\frac{9}{4}\Big)=18-18=0$
Thus, the equation has real and equal roots.
$(b)$ The given equation is $x^2+x-5=0$
On comparing with $a x^2+b x+c=0$, we get
$a = 1, b = 1$ and $c = -5$
The discriminant of $x^2+x-5=0$ is
$D=b^2-4 a c=(1)^2-4(1)(-5)$
$= 1 + 20 = 21$
$b^2-4 a c>0$
So, $x^2+x-5=0$ has two distinct real roots.
$(c)$ Given equation is $\text{x}^2+3\text{x}+2\sqrt{2}=0$
on comparing with $ax^2+ bx + c = 0$
$a = 1, b = 3$ and $\text{c}=2\sqrt{2}$
Now, $D = b^2- 4ac$
$=(3)^2-4(1)(2\sqrt{2})=9-8\sqrt{2}<0$
$\therefore$ Roots of the equation are not real.
$(d)$ Given equation is, $5x^2- 3x + 1 = 0$
On comparing with $ax^2+ bx + c = 0$
$a = 5, b = -3 c = 1$
Now, $D = b^2- 4ac$
$= (-3)^2- 4(5)(1) = 9 - 20 < 0$
Hence, roots of the equation are not real.
View full question & answer→MCQ 1331 Mark
if $x^2+ k(4x + k - 1) + 2 = 0$ has equal $r$ rots, 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 $x^2+ k(4x + k - 1) + 2 = 0,$ and roots are equal
Then find the value of $k$.
$x^2+ k(4x + k - 1) + 2 = 0$
$x^2+ 4kx + (k^2- k + 2) = 0$
Here $, a = 1, b = 4k$ and $c = k^2- k + 2$
As we know that $D = b^2- 4ac$
Putting the value of $a = 1, b = 4k$ 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 $D = 0$
$4\left(3 k^2+k-2\right)=0$
$3 k^2+k-2=0$
$3 k^2+3 k-2 k-2=0$
$3k(k + 1) - 2(k + 1) = 0$
$(k + 1)(3k - 2) = 0$
$(k + 1) = 0$ or $(3k - 2) = 0$
$k = -1$ or $\text{k}=\frac{2}{3}$
Therefore, the value of $\text{k}=\frac{2}{3},-1$
Thus, the correct answer is $(b)$
View full question & answer→MCQ 1341 Mark
The sum $S$ of first n even natural numbers is given by the relation $S = n (n + 1)$. If the sum is $420,$ then the value of $n$ is :
AnswerGiven : $n(n + 1) = 420$
$ \Rightarrow n^2+n=420 $
$ \Rightarrow n^2+n-420=0 $
$ \Rightarrow n^2+21 n-20 n-420=0 $
$\Rightarrow n(n + 21) -20(n + 21) = 0$
$\Rightarrow (n - 20) (n + 21) = 0$
$\Rightarrow n - 20 = 0, n + 21 = 0$
$\Rightarrow n = 20 $ and $n = -21 [n = -21$ is not possible$]$
$\therefore$ The value of $n$ is $20.$
View full question & answer→MCQ 1351 Mark
If the equation $x^2+ 6(k + 2)x + 9k = 0$ has equal roots then $k = ?$
- ✓
$1$ or $4$
- B
$-1$ or $4$
- C
$1$ or $-4$
- D
$-1$ or $-4$
AnswerCorrect option: A. $1$ or $4$
Since the roots of the equation $x^2+ 6(k + 2)x + 9k = 0$ are equal,
$D = 0$
$ \Rightarrow b^2-4 a c=0 $
$ \Rightarrow[2(k+2)]^2-4 \times 1 \times 9 k=0 $
$ \Rightarrow 4\left(k^2+4 k+4\right)-36 k=0 $
$ \Rightarrow 4 k^2+16 k+16-36 k=0 $
$ \Rightarrow 4 k^2-20 k+16=0 $
$ \Rightarrow k^2-5 k+4=0 $
$ \Rightarrow k^2-4 k-k+4=0 $
$\Rightarrow k(k - 4) -1(k - 4) = 0$
$\Rightarrow (k - 4)(k - 1) = 0$
$\Rightarrow k - 4 = 0$ or $k - 1 = 0$
$\Rightarrow k = 4$ or $k = 1$
View full question & answer→MCQ 1361 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 1371 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 1381 Mark
If the quadratic equations $\text{bx}^2 - 2\sqrt{\text{acx}}+\text{b} = 0$ has equal roots, then :
- A
$ 2 b^2= {ac} $
- ✓
$ {b}^2= {ac} $
- C
$ {b}^2=2 {ac}$
- D
$ {b}^2=-{ac}$
AnswerCorrect option: B. $ {b}^2= {ac} $
If the quadratic equation $\text{bx}^2 - 2\sqrt{\text{acx}}+\text{b} = 0$ has equal roots,
then $b^2- 4ac = 0$
$\Rightarrow(- 2\sqrt{\text{ac}})^{2}-4\times\text{b}\times\text{b} = 0$
$\Rightarrow4\text{ac} - 4\text{b}^2$
$\Rightarrow \text{b}^2 = \text{ac}$
View full question & answer→MCQ 1391 Mark
If one root of the equation $2x^2+ ax + 6 = 0$ is $2$ then $a = ?$
- A
$7$
- ✓
$-7$
- C
$\frac{7}{2}$
- D
$\frac{-7}{2}$
AnswerSince $x = 3$ is a solution of the equation $2x^2+ ax + 6 = 0,$ we have
$2(2)^2+ a(2) + 6 = 0$
$\Rightarrow 8 + 2a + 6 = 0$
$\Rightarrow 2a = -14$
$\Rightarrow a = -7$
View full question & answer→MCQ 1401 Mark
If $'a\ ’$ and $'b\ ’$ are the roots of the equation $x^2+a x+b=0$ then $a + b =$
AnswerSince Sum of the roots $ = \frac{-\text{b}}{\text{a}}$
$\therefore\text{a + b} = \frac{-\text{a}}{1} = -\text{a}$
View full question & answer→MCQ 1411 Mark
If the equation $9x^2+ 6kx + 4 = 0,$ 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
$9 x^2+6 k x+4=0$
$a = 9, b = 6k, c = 4$ 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 1421 Mark
If $x = 1$ is a common roots of the equations $a x^2+a x+3=0,$ and $x^2+x+b=0,$ then $ab =$
Answer$x = 1$ is the common roots given quadric equation are $a x^2+a x+3=0,$ and $x^2+x+b=0$
Then find the value of $q$.
Here, $a x^2+a x+3=0 ....(i)$
$x^2+x+b=0 ....(ii)$
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 1431 Mark
Choose the correct answer from the given four options in the following questions Which of the following equations has no real roots?
- ✓
$\text{x}^2-4\text{x}+3\sqrt{2}=0$
- B
$\text{x}^2+4\text{x}-3\sqrt{2}=0$
- C
$\text{x}^2-4\text{x}-3\sqrt{2}=0$
- D
$3\text{x}^2+4\sqrt{3}\text{x}+4=0$
AnswerCorrect option: A. $\text{x}^2-4\text{x}+3\sqrt{2}=0$
- The given equation $\text{x}^2-4\text{x}+3\sqrt{2}=0$
On comparing with $ax^2+ bx + c = 0$, we get
$a = 1, b = -4$ and $\text{c}=3\sqrt{2}$
The discriminant of $\text{x}^2-4\text{x}+3\sqrt{2}=0$ is
$D = b^2- 4ac$
$=(-4)^2-4(1)(3\sqrt{2})=16-12\sqrt{2}$
$=16-12\times(1.=41)$
$=16-16.92=-0.92$
$\Rightarrow\ \text{b}^2-4\text{ac}<0$
- The given equation is $\text{x}^2+4\text{x}-3\sqrt{2}=0$
On comparing the equation with $ax^2+ bx + c = 0$, we get
$a = 1, b = 4$ and $\text{c}=-3\sqrt{2}$
Then, $\text{D}=\text{b}^2-4\text{ac}=(-4)^2-4(1)(-3\sqrt{2})$
$=16+12\sqrt{2}>0$
Hence, the eqaution has real roots.
- Given equation is $\text{x}^2-4\text{x}-3\sqrt{2}=0$
Om comparing the equation with $ax^2+ bx + c = 0$, we get
$a = 1, b = -4 $ and $\text{c}=3\sqrt{2}$
Then, $D= b^2- 4ac$
$=(-4)^2-4(1)(-3\sqrt{2})$
$=16 + 12\sqrt{2}>0$
Hence, the equation has real roots.
- Given equation is $3\text{x}^2+4\sqrt{3}\text{x}+4=0$
On comparing the equation with $ax^2+ bx + c = 0$, we get
$a = 3, \text{b}=4\sqrt{3}$ and $c = 4$
Then, $D = b^2- 4ac$
$=4(\sqrt{3})^2-4(3)(4)=48-48=0$
Hence, the equation has real roots.
Hence, $\text{x}^2-4\text{x}+3\sqrt{2}=0$ has no real roots. View full question & answer→MCQ 1441 Mark
If $x = 1$ is a common root of $a x^2+a x+2=0$, and $x^2+x+b=0,$ then $ab =$
Answer$x = 1$ is the common roots given quadric equation are $a x^2+a x+2=0,$ and $x^2+x+b=0$
Then find the value of $ab$.
Here, $a x^2+a x+2=0 .....(i)$
$x^2+x+b=0 ....(ii)$
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 1451 Mark
Which of the following equations has $2$ as a root ?
- ✓
$ 2 x^2-7 x+6=0 $
- B
$ x^2-4 x+5=0 $
- C
$ 3 x^2-6 x-2=0 $
- D
$ x^2+3 x-12=0 $
AnswerCorrect option: A. $ 2 x^2-7 x+6=0 $
Given $, 2 x^2-7 x+6=0 $
If $2$ satisfies the above equation then $2$ is a root.
Now $,2(2)^2- 7(2) + 6 = 0$
$\therefore 2$ is a root of this equation
View full question & answer→MCQ 1461 Mark
If the equation $x^2- bx + 1 = 0$ 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 1471 Mark
$3x^2+ 4x + 5 = 0$ have :
Answer$D = b^2- 4ac$
$D = 42 - 4 \times 3 \times 5$
$D = 16 - 60$
$D = - 44$
$D < 0.$
No Real roots.
View full question & answer→MCQ 1481 Mark
The product of two successive integral multiples of $5$ is $1050$. Then the numbers are :
- A
$35$ and $40$
- B
$25$ and $30$
- C
$25$ and $35$
- ✓
$30$ and $35$
AnswerCorrect option: D. $30$ and $35$
Let one multiple of $5$ be $x$ then the next consecutive multiple will be $(x + 5)$
According to question,
$x(x + 5) = 1050$
$ \Rightarrow x^2+5 x-1050=0 $
$ \Rightarrow x^2+35 x-30 x-1050=0 $
$\Rightarrow x(x + 35) -30(x + 35) = 0$
$\Rightarrow (x - 30) (x + 35) = 0$
$\Rightarrow x - 30 = 0$ and $x + 35 = 0$
$\Rightarrow x = 30$ and $x = -35$
$x = -35$ is not possible therefore $x = 30$
Then the other multiple of $5$ is
$= x + 5$
$= 30 + 5 = 35$
Then the number are $30$ and $35$.
View full question & answer→MCQ 1491 Mark
$4x^2- 20x + 25 = 0$ have :
Answer$D = b^2- 4ac$
$D = (-20)^2- 4 \times 4 \times 25$
$D = 400 - 400$
$D = 0$.
Real and equal roots.
View full question & answer→MCQ 1501 Mark
If the sum of the roots of a quadratic equation is $6$ and their product is $6,$ the equation is :
- A
$ x^2-6 x-6=0 $
- B
$x^2+6 x-6=0 $
- ✓
$ x^2-6 x+6=0 $
- D
$ x^2+6 x+6=0 $
AnswerCorrect option: C. $ x^2-6 x+6=0 $
Required equation is $ x^2-6 x+6=0 $
View full question & answer→