MCQ

MCQ 11 Mark

The value(s) of k for which the quadratic equation $2x^2– kx + k = 0$ has equal roots is (are)

A
0 only
B
4
C
8
✓
$0, 8$

Answer

Correct option: D.

$0, 8$

$2x^2– kx + k = 0$
a = 2, b = -k, c = k
$∴ b^2- 4ac$
$= (-k)^2- 4 x 2 x 4$
$= k^2- 8k$
∴ Roots are equal.
$∴ b^2- 4ac = 0$
$k^2- 8k = 0$
⇒ k(k - 8) = 0
Either k = 0
or
k - 8 = 0,
then k = 8
k = 0, 8.

View full question & answer→

MCQ 21 Mark

If the equation $2 x^2-5 x+(k+3)=0$ has equal roots then the value of $k$ is

A
$\frac{9}{8}$
B
$-\frac{9}{8}$
✓
$\frac{1}{8}$
D
$-\frac{1}{8}$

Answer

Correct option: C.

$\frac{1}{8}$

$\begin{aligned} & 2 x^2-5 x+(k+3)=0 \\ & a=2, b=-5, c=k+3 \\ & \therefore b^2-4 a c \\ & =(-5)^2-4 \times 2 x(k+3) \\ & =25-8(k+3) \\ & \therefore \text { Roots are equal. } \\ & \therefore b^2-4 a c=0 \\ & \therefore 25-8(k+3)=0 \\ & 25-8 k-24=0 \\ & 1-8 k=0 \\ & \Rightarrow 8 k=1 \\ & \therefore k=\frac{1}{8} .\end{aligned}$

View full question & answer→

MCQ 31 Mark

If one root of a quadratic equation with rational coefficients is $\frac{3-\sqrt{5}}{2}$, then the other

A
$\frac{-3-\sqrt{5}}{2}$
B
$\frac{-3+\sqrt{5}}{2}$
✓
$\frac{3+\sqrt{5}}{2}$
D
$\frac{\sqrt{3}+5}{2}$

Answer

Correct option: C.

$\frac{3+\sqrt{5}}{2}$

One root of a quadratic equation is $\frac{3-\sqrt{5}}{2}$
then other root will be $\frac{3+\sqrt{5}}{2}$.

View full question & answer→

MCQ 41 Mark

The roots of the equation $x^2-3 x-10=0$ are

A
2,- 5
✓
$-2,5$
C
2, 5
D
$-2,-5$

Answer

Correct option: B.

$-2,5$

$
\begin{aligned}
& x^2-3 x-10=0 \\
& x=\frac{-(-3) \pm \sqrt{(-3)^2-4 \times 1 \times(-10)}}{2 \times 1} \\
& =\frac{3 \pm \sqrt{9+40}}{2} \\
& =\frac{3 \pm \sqrt{49}}{2} \\
& =\frac{3+7}{2} \\
& \therefore x=\frac{3+7}{2}=5
\end{aligned}
$
and
$
\begin{aligned}
& x=\frac{3-7}{2}=\frac{-4}{2}=-2 \\
& x=5,-2 \text { or }-2,5 .
\end{aligned}
$

View full question & answer→

MCQ 51 Mark

If $\frac{1}{2}$ is a root of the quadratic equation $4 x^2-4 k x+k+5=0$, then the value of $k$ is

A
$-6$
B
$-3$
C
$3$
✓
$6$

Answer

Correct option: D.

$6$

$\frac{1}{2}$ is a root of the equation
$
4 x^2-4 k x+k+5=0
$
Substituting the value of $x=\frac{1}{2}$ in the equation
$
\begin{aligned}
& 4\left(\frac{1}{2}\right)^2-4 \times k \times \frac{1}{2}+k+5=0 \\
& 1-2 k+k+5=0 \\
& \Rightarrow-k+6=0 \\
& k=6
\end{aligned}
$

View full question & answer→

MCQ 61 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

✓
2
B
$-2$
C
$\frac{1}{4}$
D
$\frac{1}{2}$

Answer

Correct option: A.

$\frac{1}{2}$ is a root of the equation
$
x ^2+ kx -\frac{5}{4}=0
$
Substituting the value of $x=\frac{1}{2}$ in the equation
$
\begin{aligned}
& \left(\frac{1}{2}\right)^2+k \times \frac{1}{2}-\frac{5}{4}=0 \\
& \Rightarrow \frac{1}{4}+\frac{k}{2}-\frac{5}{4}=0 \\
& \Rightarrow \frac{k}{2}-1=0 \\
& \Rightarrow k =1 \times 2=2 \\
& \therefore k =2 .
\end{aligned}
$

View full question & answer→

MCQ 71 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
$3 x^2-6 x-2=0$

Answer

Correct option: C.

$2 x^2-7 x+6=0$

$2 x^2-4 x+5=0$
$\Rightarrow(2)^2-4 x^2+5=0$
$\Rightarrow 4-8+5=0$
$\Rightarrow 9-8 \neq 0$
2 is not its root.
$x 2+3 x-12=0$
$\Rightarrow(2)^2-3 \times 2-12=0$
$\Rightarrow 4-6-12$
$=4-18$
$=-14$
$\therefore 2$ is not its roots.
$2 x^2-7 x+6=0$
$\Rightarrow 2(2)^2-7 \times 2+6=0$
$\Rightarrow 8-14+6=0$
$\Rightarrow 0=0$
$\therefore 2$ is its roots.
$3 x^2-6 x-2=0$
$\Rightarrow 3(2)^2-6 \times 2-2=0$
$\Rightarrow 12-12-2=0$
$\Rightarrow 12-14=0$
$\therefore 2$ is not its roots.

View full question & answer→

MCQ 81 Mark

Which of the following is a quadratic equation?

A
$(x-2)(x+1)=(x-1)(x-3)$
B
$(x+2)^3=2 x\left(x^2-1\right)$
C
$x^2+3 x+1=(x-2)^2$
✓
$8(x-2)^3=(2 x-1)^3+3$

Answer

Correct option: D.

$8(x-2)^3=(2 x-1)^3+3$

$(x-2)(x+1)=(x-1)(x-3)$
$\Rightarrow x^2+x-2 x-2=x^2-3 x-x+3$
$\Rightarrow 3 x+x-2 x+x=3+2$
$\Rightarrow 3 x=5$
It is not a quadratic equation
$(x+2)^3=2 x\left(x^2-1\right)$
$x^3+6 x^2+12 x+8=2 x^3-2 x$
$x^3+6 x^2+12 x+8-2 x^3+2 x=0$
$-x^3+6 x^2+14 x+8=0$
It is not a quadratic equation.
$x^2+3 x+1=(x-2)^2$
$x^2+3 x+1=x^2-4 x+4$
$3 x+1+4 x-4=0$
$\Rightarrow 7 x-3=0$
It is not a quadratic equation.
$8(x-2)^3=(2 x-1)^3+3$
$8\left(x^3-6 x^2+12 x-8\right)=8 x^3-12 x^2+6 x-1+3$
$8 x^3-48 x^2+96 x-64-8 x^3+12 x^2-6 x+1-3=0$
$-36 x^2+90 x-66=0$
It is a quadratic equation.

View full question & answer→

MCQ 91 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
$3 x^2+4 \sqrt{3} x+4=0$

Answer

Correct option: A.

$x^2-4 x+3 \sqrt{2}=0$

$x^2-4 x+3 \sqrt{2}=0$
Explanation:
(A) The given equation is $x^2-4 x+3 \sqrt{2}=0$
On comparing with $ax^2 + bx + c = 0,$ we get
$a = 1, b = – 4$ and $c =3 \sqrt{2}$
The discriminant of $x^2-4 x+3 \sqrt{2}=0$ is
$D=b^2-4 a c $
$=(-4)^2-4(1)(3 \sqrt{2}) $
$ =16-12 \sqrt{2} $
$ =16-12 \times(1.41) $
$ =16-16.92 $
$ =-0.92$
$b^2 – 4ac < 0$
$(B)$ The given equation is$x^2+4 x-3 \sqrt{2}=0$
On comparing the equation with $ax^2 + bx + c = 0$, we get
$a = 1, b = 4$ and $c =-3 \sqrt{2}$
Then, $D = b^2 – 4ac$
$=(-4)^2-4(1)(-3 \sqrt{2}) $
$ =16+12 \sqrt{2}>0$
Hence, the equation has real roots.
(C) Given equation is $x^2-4 x-3 \sqrt{2}=0$
On comparing the equation with $ax^2 + bx + c = 0,$ we get
$a = 1, b = – 4$ and $c =-3 \sqrt{2}$
Then, $D = b^2 – 4ac$
$=(-4)^2-4(1)(-3 \sqrt{2}) $
$ =16+12 \sqrt{2}>0$
Hence, the equation has real roots.
$(D)$ Given equation is $3 x^2+4 \sqrt{3} x+4=0$
On comparing the equation with $ax^2 + bx + c = 0,$ we get
$a = 3, b =4 \sqrt{3}$ 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, $x^2-4 x+3 \sqrt{2}=0$ has no real roots.

View full question & answer→

MCQ 101 Mark

Which of the following equations has two distinct real roots?

A
$2 x^2-3 \sqrt{2} x+\frac{9}{4}=0$
B
$x^2+x-5=0$
C
$x^2+3 x+2 \sqrt{2}=0$
D
$5 x^2-3 x+1=0$

Answer

$
\begin{aligned}
& 2 x^2-3 \sqrt{2} x+\frac{9}{4}=0 \\
& b^2-4 a c \\
& =(-3 \sqrt{2})^2-4 \times 2 \times \frac{9}{4} \\
& =18-18 \\
& =0 \\
& \therefore \text { Roots are real and equal. } \\
& x^2+x-5=0 \\
& b^2-4 a c \\
& =(1)^2-4 \times 1 \times(-5) \\
& =1+20 \\
& =\sqrt{21}>0
\end{aligned}
$
$\therefore$ Roots are real and distinct.

View full question & answer→

MCQ 111 Mark

The quadratic equation $2 x^2-\sqrt{5} x+1=0$ has

A
two distinct real roots
B
two equal real roots
✓
no real roots
D
more than two real roots

Answer

Correct option: C.

no real roots

$C$
The quadratic equation $2 x^2-\sqrt{ 5} x+1=0$ has no real roots.
Explanation:
Given equation is $2 x^2-\sqrt{5} x+1=0$
On comapring with $ax^2 + bx + c = 0$, we get
$a=2, b=-\sqrt{5}$ and $c=1$
$\therefore$ Discriminant, $D = b^2 – 4ac$
$=(-\sqrt{5})^2-4 \times(2) \times(1)$
$ =5-8 $
$ =-3<0$
Since, discrimant is negative,
Therefore quadratic equation$2 x^2-\sqrt{5} x+1=0$has no real roots
i.e., imaginary roots.

View full question & answer→

MCQ 121 Mark

If the equation $2 x^2-6 x+p=0$ has real and different roots, then the values of $p$ are given by

✓
$p < \frac{9}{2}$
B
$p \leq \frac{9}{2}$
C
$p>\frac{9}{2}$
D
$p \geq \frac{9}{2}$

Answer

Correct option: A.

$p < \frac{9}{2}$

$ 2 x^2-6 x+p=0 $
$ \text { Here, } a=2, b=-6, c=p $
$ b^2-4 a c $
$ =(-6)^2-4 \times 2 \times p $
$ =36-8 p$
$\because$ Roots are real and unequal,
$ \therefore b^2-4 a c>0 $
$ \Rightarrow 36-8 p>0 $
$ \Rightarrow 36>8 p $
$ \Rightarrow \frac{36}{8}>p $
$ \Rightarrow p<\frac{36}{8} $
$ \Rightarrow p<\frac{9}{2} .$

View full question & answer→

MCQ 131 Mark

If the equation $\{k+1) x^2-2(k-1) x+1=0$ has equal roots, then the values of $k$ are

A
1, 3
✓
$0, 3$
C
$0, 1$
D
$0, 1$

Answer

Correct option: B.

$0, 3$

$(k+1) x^2-2(k-1) x+1=0$
Here, $a=k+1, b=-2(k-1), c=1$
$\therefore b^2-4 a c$
$=[-2(k-1)]^2-4(k+1)(1)$
$=4\left(k^2-2 k+1\right)-4 k-4$
$=4 k^2-8 k+4-4 k-4$
$=4 k^2-12 k$
$\because$ Roots are equal.
$\therefore b ^2-4 ac =0$
$\therefore 4 k ^2-12 k =0$
$4 k ( k -3)=0$
$\Rightarrow 4 k ( k -3)=0$
$\Rightarrow k ( k -3)=0$
Either $k=0$
or
$k-3=0$
then $k=3$
$k =0,3$

View full question & answer→

MCQ 141 Mark

If the equation $3 x^2-k x+2 k=0$ roots, then the the value $(s)$ of $k$ is $($are$)$

A
$6$
B
$0 $
✓
$24 $
D
$5$

Answer

Correct option: C.

$24 $

$3 x^2-k x+2 k=0$
Here, $a=3, b=-k, c=2 k$
$b^2-4 a c$
$=(-k)^2-4 \times 3 \times 2 k$
$=k^2-24 k$
$\therefore $ Roots are equal.
$\therefore b^2-4 a c=0$
$\therefore k^2-24 k=0$
$\Rightarrow k(k-24)=0$
Either $ k=0$ or $k-24=0$
then $ k=24$
$\therefore k=0,24$

View full question & answer→

MCQ 151 Mark

Which of the following is not a quadratic equation?

A
$(x+2)^2=2(x+3)$
B
$x^2+3 x=(-1)(1-3 x)$
C
$(x+2)(x-1)=x^2-2 x-3$
D
$x^3-x^2+2 x+1=(x+1)^3$

Answer

$(x+2)^2=2(x+3)$
$\Rightarrow x^2+4 x+4=2 x+6$
$\Rightarrow x^2+4 x-2 x+4-6=0$
$\Rightarrow x^2+2 x-2$
It is a quadratic equation.
$x^2+3 x=(-1)(1-3 x)$
$\Rightarrow x^2+3 x=-1+3 x$
$\Rightarrow x^2+1=0$
it is also quadratic equation.
$(x+2)(x-1)=x^2-2 x-3$
$x^2-x+2 x-2=x^2-2 x-3$
$x^2-x^2+x+2 x-2+3=0$
$\Rightarrow 3 x+1=0$
It is not a quadratic equation.
$x^3-x^2+2 x+1=(x+1)^3$
$=x^3+3 x^2+3 x+1$
$x^3-x^2+2 x+1$
$3 x^2+x^2-2 x-1+3 x+1=0$
$\Rightarrow 4 x^2+x=0$
It is a quadratic equation.

View full question & answer→

Mathematics MCQ

Test yourself on this topic