M.C.Q (1 Marks)

MCQ 11 Mark

Choose the correct answer from the given four options in the following questions: $\left(x^2+1\right)^2-x^2=0$ has:

A
Four real roots.
B
Two real roots.
✓
No real roots.
D
One real root.

Answer

Correct option: C.

No real roots.

Given equation is $\left(x^2+1\right)^2-x^2=0$
$\Rightarrow x^4+1+2 x-x^2=0$
${\left[\because(a+b)^2=a^2+b^2+2 a b\right]}$
$\Rightarrow x^4+x^2+1=0$
$\text { Let } x^2=y$
$\therefore\left(x^2\right)^2+x^2+1=0$
$y^2+y+1=0$
On comparing with $a y^2+b y+c=0$,
we get $a=1, b=1$ and $c=1$
Discrinimant, $D=b^2-4 a c$
$=(1)^2-4(1)(1)$
$=1-4=-3$
SInce, $D <0$
$\therefore y^2+y+1=0 \text { i.e., } x^4+x^2+1=0 \text { or }$
$\left(x^2+1\right)^2-x^2=0$ has no real roots.

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MCQ 21 Mark

Choose the correct answer from the given four options in the following questions:
Which constant must be added and subtracted to solve the quadratic equation $9\text{x}^2+\frac{3}{4}\text{x}-\sqrt{2}=0$ by the method of completing the square?

A
$\frac{1}{8}$
✓
$\frac{1}{64}$
C
$\frac{1}{4}$
D
$\frac{9}{64}$

Answer

Correct option: B.

$\frac{1}{64}$

Given equation is $9\text{x}^2+\frac{3}{4}\text{x}-\sqrt{2}=0.$$(3\text{x})^2+\frac{1}{4}(3\text{x})-\sqrt{2}=0$
On putting 3x = y, we have $\text{y}^2+\frac{1}{4}\text{y}-\sqrt{2}=0$
$\text{y}^2+\frac{1}{4}\text{y}+\Big(\frac{1}{8}\Big)^2-\Big(\frac{1}{8}\Big)^2-\sqrt{2}=0$
$\Big(\text{y}+\frac{1}{8}\Big)^2=\frac{1}{64}+\sqrt{2}$
$\Big(\text{y}+\frac{1}{8}\Big)^2=\frac{1+64\sqrt{2}}{64}$
Thus, $\frac{1}{64}$ must be added and subtracted to solve the given equation.

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MCQ 31 Mark

Choose the correct answer from the given four options in the following questions: Values of $k$ for which the quadratic equation $2x^2 - kx + k = 0$ has equal roots is:

A
$0$ only.
B
$4.$
C
$8$ only.
✓
$0, 8.$

Answer

Correct option: D.

$0, 8.$

Given equation is $2 x^2-k x+k=0$
On comparing with $ax ^2+ bx + c =0$,
we get $a =2, b=- k$ and $c = k$
For equal roots, the discriminant must be zero.
i.e., $D=b^2-4 a c=0$
$\Rightarrow(- k )^2-4(2) k =0$
$\Rightarrow k ^2-8 k =0$
$\Rightarrow k ( k -8)=0$
$\therefore k =0,8$
Hence, the required values of $k$ are $0$ and $8 .$

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MCQ 41 Mark

Choose the correct answer from the given four options in the following questions:
The quadratic equation $2\text{x}^2-\sqrt{5}\text{x}+1=0$ has.

A
Two distinct real roots.
B
Two equal real roots.
✓
No real roots.
D
More than 2 real roots.

Answer

Correct option: C.

No real roots.

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

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MCQ 51 Mark

Choose the correct answer from the given four options in the following questions:
Which of the following equations has the sum of its roots as $3?$

A
$2\text{x}^2-3\text{x}+6=0.$
✓
$-\text{x}^2+3\text{x}-3=0.$
C
$\sqrt{2}\text{x}^2-\frac{3}{\sqrt{2}}\text{x}+1=0.$
D
$3\text{x}^2-3\text{x}+3=0$

Answer

Correct option: B.

$-\text{x}^2+3\text{x}-3=0.$

$a.$ Given that, $2 x^2-3 x+6=0$
On comparing with $a x^2+b x+c=0$, we get $a=2, b=-3$ and $c=6$
$\therefore$ Sum of the roots $=\frac{- b }{ a }=\frac{-(-3)}{2}=\frac{3}{2}$
So, sum of the roots of the quadratic equation $2 x^2-3 x+6=0$ is not $3$ ; so it is not the answer.
$b.$ Given that, $-x^2+3 x-3=0$
On compare with $a x^2+b x+c=0$, we get $a=-1, b=3$ and $c=-3$
$\therefore$ Sum of the roots $=\frac{- b }{ a }=\frac{-(3)}{-1}=3$
So, sum of the roots of the quadratic equation $-x^2+3 x-3=0$ is $3$ , so it is the answer.
$c.$ Given that, $\sqrt{2} x ^2-\frac{3}{\sqrt{2}} x +1=0$
$\Rightarrow 2 x^2-3 x+\sqrt{2}=0$
On comparing with $a x^2+b x+c=0$, we get $a =2, b=-3$ and $c =\sqrt{2}$
$\therefore$ Sum of the roots $=\frac{- b }{ a }=\frac{-(-3)}{2}=\frac{3}{2}$
So, sum of the roots of the quadratic equation $\sqrt{2} x ^2-\frac{3}{\sqrt{2}} x +1=0$ is not $3$ , so it is not the answer,
$d.$ Given that, $3 x^2-3 x+3=0$
$\Rightarrow x^2-x+1=0$
On comparing with $a x^2+b x+c=0$, we get $a=1, b=-1$ and $c=1$
$\therefore$ Sum of the roots $=\frac{- b }{ a }=\frac{-(-1)}{1}=1$
So, sum of the roots of the quadratic equation $3 x^2-3 x+3=0$ is not $3$ , so it is not the answer.

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MCQ 61 Mark

Choose the correct answer from the given four options in the following questions:Which of the following is not a quadratic equation?

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.

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

$a.$ Substituting $x=2$ in $x^2-4 x+5$, we get
$(2)^2-4(2)+5$
$=4-8+5=1 \neq 0$
So, $x=2$ is not a root of $x^2-4 x+5=0$.
$b.$ Substituting $x=2$ in $x^2+3 x-12$, we get
$(2)^2+3(2)-12$
$=4+6-12=-2 \neq 0$
So, $x=2$ is not a root $x^2-3 x-12=0$.
$c.$ Substituting $x=2$ in $2 x^2-7 x+6$, we get
$2(2)^2-7(2)+6=2(4)-14+6$
$=8-14+6=14-14=0$
So, $x=2$ is root of the equation $2 x^2-7 x+6=0$.
$d.$ Substituting $x=2$ in $3 x^2-6 x-2$, we get
$3(2)^2-6(2)-2$
$=12-12-2=-2 \neq 0$
So, $x=2$ is not a root of $3 x^2-6 x-2=0$.

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MCQ 71 Mark

Choose the correct answer from the given four options in the following questions:
Which of the following is a quadratic equation?

✓
$\text{x}^2+2\text{x}+1=(4-\text{x})^2+3.$
B
$-2\text{x}^2=(5-\text{x})\Big(2\text{x}-\frac{2}{5}\Big).$
C
$(\text{k}+1)\text{x}^2+\frac{3}{2}\text{x}=7,\text{ where k}=-1.$
D
$\text{x}^3-\text{x}^2=(\text{x}-1)^3.$

Answer

Correct option: A.

$\text{x}^2+2\text{x}+1=(4-\text{x})^2+3.$

$a.$ Given that, $x^2+2 x+1=(4-x)^2+3$
$\Rightarrow x^2+2 x+1=16+x^2-8 x+3$
$\Rightarrow 10 x-18=0$
Which is not of the form $ax ^2+ bx + c , a \neq 0$.
Thus, the equation is not a quadratic equation.
$b.$ Given that, $-2 x ^2=(5- x )\left(2 x -\frac{2}{5}\right)$
$\Rightarrow-2 x^2=10 x-2 x^2-2+\frac{2 x}{5}$
$\Rightarrow 50 x+2 x-10=0$
$\Rightarrow 52 x-10=0$
which is also not a quadratic equation.
$c.$ Given that, $x ^2( k +1)+\frac{3}{2} x =7$
Given, $k =-1$
$\Rightarrow x^2(-1+1)+\frac{3}{2} x=7$
$\Rightarrow 3 x-14=0$
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 $ax ^2+ bx + c , a \neq 0$.

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MCQ 81 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
$\frac{1}{2}$

Answer

Correct option: A.

$2$

Since, $\frac{1}{2}$ is a root of the quadratic equation $\text{x}^2+\text{kx}-\frac{5}{4}=0$.Then, $\Big(\frac{1}{2}\Big)^2+\text{k}\Big(\frac{1}{2}\Big)-\frac{5}{4}=0$
$\Rightarrow\ \frac{1}{4}+\frac{\text{k}}{2}-\frac{5}{4}=0\Rightarrow\ \frac{1+2\text{k}-5}{4}=0$
$\Rightarrow\ 2\text{k}-4=0$
$\Rightarrow\ 2\text{k}=4\Rightarrow\ \text{k}=2.$

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MCQ 91 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
$(\text{x}^2 + 2\text{x})^2 = \text{x}^{4} + 3 + 4\text{x}^3.$

Answer

Correct option: C.

$\big(\sqrt{2}\text{x}+\sqrt{3}\big)^2=3\text{x}^2-5\text{x}.$

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MCQ 101 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
$5\text{x}^2-3\text{x}+1=0.$

Answer

Correct option: B.

$\text{x}^2+\text{x}-5=0$

Given equation is $2 x ^2-3 \sqrt{2} x +\frac{9}{4}=0$,

On comparing with $ax ^2+ bx + c =0$
$a =2, b=-3 \sqrt{2}$ and $C =\frac{9}{4}$
Now, $D=b^2-4 a c$
$=(-3 \sqrt{2})-4(2)\left(\frac{9}{4}\right)=18-18=0$
Thus, the equation has real and equal roots.
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
$D=b^2-4 a c=(1)^2-4(1)(-5)$
$=1+20=21$
$\Rightarrow b^2-4 a c > 0$
So, $x^2+x-5=0$ has two distinct real roots.
Given equation is $x ^2+3 x +2 \sqrt{2}=0$
on comparing with $ax ^2+ bx + c =0$
$a=1, b=3 \text { and } c=2 \sqrt{2}$
Now, $D=b^2-4 a c$
$=(3)^2-4(1)(2 \sqrt{2})=9-8 \sqrt{2} < 0$
$\therefore$ Roots of the equation are not real.
Given equation is, $5 x^2-3 x+1=0$
On comparing with $ax ^2+ bx + c =0$
$a=5, b=-3 c=1$
Now, $D=b^2-4 a c$
$=(-3)^2-4(5)(1)=9-20 < 0$
Hence, roots of the equation are not real.

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MCQ 111 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
$\text{5x}^2+4\text{x}-3\sqrt{2}=0$
C
$\text{2x}^2-4\text{x}-3\sqrt{2}=0$
D
$3\text{x}^2+4\sqrt{3}\text{x}+4=0$

Answer

Correct 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
$D=b^2-4 a c$
$=(-4)^2-4(1)(3 \sqrt{2})=16-12 \sqrt{2}$
$=16-12 \times(1 .=41)$
$=16-16.92=-0.92$
$\Rightarrow b^2-4 a c < 0$
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, $D = b ^2-4 ac =(-4)^2-4(1)(-3 \sqrt{2})$
$=16+12 \sqrt{2}>0$
Hence, the eqaution has real roots.
Given equation is $x ^2-4 x -3 \sqrt{2}=0$
Om comparing the equation with $a x^2+b x+c=0$,
we get $a =1, b=-4$ and $c =3 \sqrt{2}$
Then, $D=b^2-4 a c$
$=(-4)^2-4(1)(-3 \sqrt{2})$
$=16+12 \sqrt{2}>0$
Hence, the equation has real roots.
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, $D=b^2-4 a c$
$=4(\sqrt{3})^2-4(3)(4)=48-48=0$
Hence, the equation has real roots.

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Maths M.C.Q (1 Marks)

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