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CBSE Questions for Class 10 Maths Quadratic Equations Quiz 2 - MCQExams.com

The mentioned equation is in which form?

$\dfrac{q^{2}- 4}{q^{2}}\, =\, -3$

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Linear
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Quadratic
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Cubic
0%
none

Explanation

Given equation is

$\dfrac{q^2 - 4}{q^2} = -3$

$\Rightarrow q^2 - 4 = -3q^2$

$\Rightarrow 4q^2 = 4$

$\Rightarrow q^2 = 1$
The highest power of

$q$ is

$2$ . Hence, it is a quadratic equation.

Is the following equation quadratic?

$\displaystyle -\frac{5}{3}\, x^{2}\, =\, 2x\, +\, 9$

0%
Yes
0%
No
0%
Ambiguous
0%
Data insufficient

Explanation

$Answer=1$
The equation,

$\displaystyle -\frac{5}{3}\, x^{2}\, =\, 2x\, +\, 9$ has the highest power as 2.
Thus, the equation is quadratic.

Is the following equation quadratic?

$13\, =\, -5y^{2}\, -\, y^{3}$

0%
Yes
0%
No
0%
Ambiguous
0%
Data insufficient

Explanation

Answer=0
The equation

$13 = -5y^2 - y^3$ has the highest power as 3. Thus, it is not quadratic equation.

The condition for a general quadratic equation such that its both roots are equal, is

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$b^{2}+4ac= 0$
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$b^{2}-4ac= 0$
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$b^{2}+ac= 0$
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$b^{2}+2ac= 0$

Explanation

General quadratic equation is

$ax^2+bx+c=0$
For both roots to be equal discriminant of this quadratic should be zero

$\Rightarrow b^2-4ac=0$

Find discriminant of

$5x^2+2x+1=0$

0%
$-13$
0%
$12$
0%
$-16$
0%
$14$

Explanation

The discriminant

$D={ b }^{ 2 }- 4ac = 4 - 4(5)(1) = -16$

Choose the best possible option.

$\displaystyle { x }^{ 2 }+\frac { 1 }{ 4{ x }^{ 2 } } -8=0$ is a quadratic equation.

0%
Yes
0%
No
0%
Can't predict
0%
None

Explanation

$\displaystyle { x }^{ 2 }+\frac { 1 }{ 4{ x }^{ 2 } } -8=0$

$\displaystyle \frac { 4{ x }^{ 4 }+1-8\times 4{ x }^{ 2 } }{ 4{ x }^{ 2 } } =0$

$\displaystyle 4{ x }^{ 4 }+1-32{ x }^{ 2 }=0$
The degree of the equation is 4

$\displaystyle \therefore \quad { x }^{ 2 }+\frac { 1 }{ 4{ x }^{ 2 } } -8=0$ is not a quadratic equation

The discriminant of

$x^2 - 3x + k = 0$ is

$1$ then the value of

$k = .............$

0%
$-2$
0%
$4$
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$-4$
0%
$2$

Explanation

Comparing the equation

$x^2 - 3x + k = 0$ with

$ax^2 + bx + c = 0$ , we get

$a = 1, b = - 3, c = k$

Here, discriminant

$= 1$

$\therefore b^2 - 4ac = 1$ ......

$[\because D=b^2 - 4ac]$

$\therefore (-3)^2 - 4(1)(k) = 1$

$\therefore 9 - 4k = 1$

$\therefore -4k = - 8$

$\therefore k = 2$

Find a quadratic equation, from the equations given below, having same discriminant as

$5x^2+2x+1=0$

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$-x^2+4x-8$
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$x^2+3x-7$
0%
$4x^2-5$
0%
$-2x^2-3x+6$

Explanation

Given: set of quadratic equations

To find: the equation having same discriminant as

$5x^2+2x+1=0$

Sol: The equation

$5x^2+2x+1=0$ is of form

$ax^2+bx+c=0$

$\therefore a=5, b=2, c=1$

The discriminant of this equation becomes

$b^2-4ac=2^2-4(5)(1)=4-20=-16$

(i)

$-x^2+4x-8$ this is also of the form

$ax^2+bx+c$

$\therefore a=-1, b=4, c=-8$

The discriminant of this equation becomes

$b^2-4ac=4^2-4(-1)(-8)=16-32=-16$

(ii)

$x^2+3x-7$ this is also of the form

$ax^2+bx+c$

$\therefore a=1, b=3, c=-7$

The discriminant of this equation becomes

$b^2-4ac=3^2-4(1)(-7)=9+28=37$

(iii)

$4x^2-5\implies 4x^2-0x-5$ this is also of the form

$ax^2+bx+c$

$\therefore a=4, b=0, c=-5$

The discriminant of this equation becomes

$b^2-4ac=0^2-4(4)(-5)=0+80=80$

(iv)

$-2x^2-3x+6$ this is also of the form

$ax^2+bx+c$

$\therefore a=-2, b=-3, c=6$

The discriminant of this equation becomes

$b^2-4ac=(-3)^2-4(-2)(6)=9+48=57$

State true or false:

If

$(x\, -\, 3)^{2}\, =\, 25$ , then

$x$ is

$8$ or

$-2$ .

0%
True
0%
False

Explanation

$(x-3)^2 = 25$

$x^2 - 6x +9= 25$

$x^2 - 6x - 16 = 0$

$x^2 -8x +2x -16 =0$

$(x-8)(x +2) =0$

$x = -2,8$

Find the roots of the quadratic equation by applying the quadratic formula

$\displaystyle 2x^2 - 7x + 3 = 0$

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$3,\displaystyle -\frac{1}{2}$
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$-3,\displaystyle \frac{1}{2}$
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$3,\displaystyle \frac{1}{2}$
0%
None of these

Explanation

Given equation is

$2x^{2}-7x+3=0$
Hence,

$a=2,b=-7,c=3$
Therefore,

$x=\dfrac{-b\pm\sqrt{b^{2}-4ac}}{2a}$

$=\dfrac{7\pm\sqrt{25}}{4}$

$=\dfrac{7\pm5}{4}$

Thus,

$x=3$ and

$x=\dfrac{1}{2}$

Is the following equation a quadratic equation?

$\displaystyle 3x + \frac{1}{x} - 8 = 0$

0%
Yes
0%
No
0%
Ambiguous
0%
Data insufficient

Explanation

Given,

$3x + \dfrac{1}{x} - 8 = 0$

$\Rightarrow \dfrac{3x^2+1-8x}{x}=0$

$\Rightarrow 3x^2+1-8x=0$
Comparing it with the standard form of quadratic equation

$x^2+bx+c=0, a\neq 0$
Here,

$a=3$
Thus, it is a quadratic equation.

If

$\cfrac{1}{x}+\cfrac{1}{y}=a$ and

$xy=\cfrac{1}{b}$ and

$\cfrac{1}{{x}^{2}}+\cfrac{1}{{y}^{2}}$

0%
${a}^{2}-2b$
0%
$\cfrac{1}{{a}^{2}}-2b$
0%
${a}^{2}-\cfrac{2}{b}$
0%
$\cfrac{1}{{a}^{2}}-\cfrac{2}{b}$

Explanation

Given,

$\frac { 1 }{ x } +\frac { 1 }{ y } =a$ ..................eq (1)

$xy=\frac { 1 }{ b }$ .........................................eq.(2)

$\Rightarrow \frac { 1 }{ x } =b$
Squaring both sides in eq (1)

${ \left( \frac { 1 }{ x } +\frac { 1 }{ y } \right) }^{ 2 }={ a }^{ 2 }$

$\Rightarrow { \left( { \frac { 1 }{ x } } \right) }^{ 2 }+2\cdot \frac { 1 }{ x } \cdot \frac { 1 }{ y } +{ \left( \frac { 1 }{ y } \right) }^{ 2 }={ a }^{ 2 }$ \\

$\Rightarrow { \left( { \frac { 1 }{ { x }^{ 2 } } } \right) }+\frac { 2 }{ xy } +{ \left( \frac { 1 }{ y^{ 2 } } \right) }={ a }^{ 2 }$

$\Rightarrow { \left( { \frac { 1 }{ { x }^{ 2 } } } \right) }+2b+{ \left( \frac { 1 }{ y^{ 2 } } \right) }={ a }^{ 2 }$ putting

$\Rightarrow \frac { 1 }{ x } =b$

$\Rightarrow \frac { 1 }{ { x }^{ 2 } } +\frac { 1 }{ { y }^{ 2 } } ={ a }^{ 2 }-2b$

Is the following equation a quadratic equation?

$(x + 2)^3 = x^3 - 4$

0%
Yes
0%
No
0%
Ambiguous
0%
Data insufficient

Explanation

Given equation is

$(x + 2)^3 = x^3 - 4$

$\rightarrow x^3+2^3+3\times x\times 2(x+2)=x^3-4$

$\Rightarrow x^3+8+6x(x+2)=x^3-4$

$\Rightarrow x^3+8+6x^2+12x-x^3+4=0$

$\Rightarrow 6x^2+12x+12=0$
Comparing it with the standard form of quadratic equation

$x^2+bx+c=0, a\neq 0$ .
Here,

$a=6$
Thus, it is a quadratic equation.

Check whether the following is a quadratic equation.

$(x - 3) (2x + 1) = x (x + 5)$

0%
Yes
0%
No
0%
Ambiguous
0%
Data insufficient

Explanation

Given,

$(x-3)(2x+1)=(x)(x+5)$

$\Rightarrow 2x^{2}+x-6x-3=x^{2}+5x$

$\Rightarrow x^{2}-10x-3=0$
Using quadratic formula, we get

$x=\dfrac{10\pm\sqrt{100-12}}{2}$

$=\dfrac{10\pm7\sqrt{2}}{2}$

Is the following equation a quadratic equation?

$\displaystyle \frac{3x}{4} - \frac{5x^2}{8} = \frac{7}{8}$

0%
Yes
0%
No
0%
Ambiguous
0%
Data insufficient

Explanation

Given,

$\dfrac{3x}{4}-\dfrac{5x^{2}}{8}=\dfrac{7}{8}$
Hence,

$6x-5x^{2}=7$

$\Rightarrow 5x^{2}-6x+7=0$

$\Rightarrow ax^{2}+bx+c=0$
Hence, it is a quadratic equation.

Is the following equation a quadratic equation?

$16x^2 - 3 = (2x + 5) (5x - 3)$

0%
Yes
0%
No
0%
Ambiguous
0%
Data insufficient

Explanation

$16x^2 - 3 = (2x + 5) (5x - 3)$

$\Rightarrow 16x^2 - 3=10x^2-6x+25x-15$

$\Rightarrow 16x^2-10x^2+6x-25x+15-3=0$

$\Rightarrow 6x^2-19x+12=0$
Comparing it with the standard form of quadratic equation

$x^2+bx+c=0, a\neq 0$ .
Here

$a=6$
Thus, it is a quadratic equation.

$\displaystyle a^2x^2 - 3abx + 2b^2$ = 0 then

$x = \cfrac{a}{b}\ and\ x = \cfrac{b}{a}$ are the roots of the equation.

0%
True
0%
False
0%
Ambiguos
0%
Can't say

Explanation

$a^{2}x^{2}-3abx+2b^{2}=0$
Therefore

$2b^{2}.a^{2}=2b^{2}a^{2}$

$=2ab.ab$
Hence

$a^{2}x^{2}-abx-2abx+2b^{2}=0$

$ax(ax-b)-2b(ax-b)=0$

$(ax-2b)(ax-b)=0$
Thus, the roots are

$\dfrac{2b}{a},\dfrac{b}{a}$

Hence, the given statement is false.

Discriminant of the equation

$-3x^2 + 2x -8 = 0$ is

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$- 92$
0%
$- 29$
0%
$39$
0%
$49$

Explanation

The quadratic equation is

$-3x^2 + 2x -8 = 0$
Comparing it with

$ax^2+bx+c=0$ we get,

$a=-3, b=2, c=-8$
Therefore,

$D=b^2-4ac$

$2^2-4\times (-3)\times (-8)$

$=4-96$

$=-92$

Choose the best possible option.

$\displaystyle { x }^{ 3 }-5x+2{ x }^{ 2 }+1=0$ is quadratic equation.

0%
Yes
0%
No
0%
Can't predict
0%
None

Explanation

The degree of equation is 3.

$\displaystyle \therefore \quad { x }^{ 3 }-5x+2{ x }^{ 2 }+1=0$ is not quadratic.
It is cubic equation.

Determine whether the equation

$\displaystyle 5{ x }^{ 2 }=5x$ is quadratic or not.

0%
Yes
0%
No
0%
Complex equation
0%
None

Explanation

The degree of the equation is 2

$\displaystyle \therefore \quad 5{ x }^{ 2 }=5x$

$\displaystyle =\quad 5{ x }^{ 2 }-5x=0$ is quadratic equation.

Solve

$(b -c)x^2 + (c -a)x + (a -b) =0$

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$\dfrac {a+b}{b-c}, 1$
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$\dfrac {a-b}{b+c}, 1$
0%
$\dfrac {a-b}{b-c}, -1$
0%
$\dfrac {a-b}{b-c}, 1$

Explanation

$x=\frac {-(c-a)\pm \sqrt {(c-a)^2-4(b-c)(a-b)}}{2(b-c)}$

$=\frac {-(c-a)\pm \sqrt {(a+c-2b)^2}}{2(b-c)}=\frac {-(c-a)\pm (a+c-2b)}{2(b-c)}$

$=\frac {-c + a + a + c-2b}{2(b-c)}$ or

$\frac {-c + a -a -c + 2b}{2(b-c)}$

$=\frac {2(a-b)}{2(b-c)}$ or

$\frac {2(b-c)}{2(b-c)}$
Hence, the roots are

$\frac {a-b}{b-c}$ and 1.

Choose best possible option.

$\displaystyle \left( x+\frac { 1 }{ 2 } \right) \left( \frac { 3x }{ 2 } +1 \right) =\frac { 6 }{ 2 } \left( x-1 \right) \left( x-2 \right)$ is quadratic.

0%
Yes
0%
No
0%
Complex equation
0%
None

Explanation

$\displaystyle \left( x+\frac { 1 }{ 2 } \right) \left( \frac { 3x }{ 2 } +1 \right) =\frac { 6 }{ 2 } \left( x-1 \right) \left( x-2 \right)$

$\displaystyle \left( 2x+2 \right) \left( 3x+2 \right) =12\left( x-1 \right) \left( x-2 \right)$

$\displaystyle 12{ x }^{ 2 }-36x+24-6{ x }^{ 2 }-7x-2=0$

$6x^2$

$-43x$ +

$22$ =0
The degree of equation is 2
Therefore, it is a quadratic equation.

Choose the best possible answer

$\displaystyle 32{ x }^{ 2 }-6=\left( 4x+10 \right) \left( 10x-6 \right)$ is quadratic equation

0%
Yes
0%
No
0%
Complex
0%
None

Explanation

$\displaystyle 32{ x }^{ 2 }-6=\left( 4x+10 \right) \left( 10x-6 \right)$

$\displaystyle 32{ x }^{ 2 }-6=40{ x }^{ 2 }-24x+100x-60$

$\displaystyle 8{ x }^{ 2 }+76x-54=0$
The degree of the equation is 2

$\displaystyle \therefore$ It is a quadratic equation

Calculate the zeroes of the quadratic equation

$(x+3)^2=49$ .

0%
$x=-10, x=4$
0%
$x=-10, x=10$
0%
$x=-4, x=10$
0%
$x=3\pm 2\sqrt{3}$

Explanation

$(x+3)^2=49$ , given equation

$\Rightarrow x+3=\pm \sqrt{49}=\pm 7$

$\Rightarrow x=\pm 7-3$

$\Rightarrow x =-7-3, 7-3$

Therefore, the values are

$x =-10,4$

Which of the following is not a quadratic equation?

0%
$x^{2}+ 6y + 2$
0%
$(x - 2) + (x + 2)^{2} + 3$
0%
$(3x - 2)^{3} + \dfrac{1}{2}x - 4$
0%
$3x^{2} - 6x + \dfrac{1}{2}$

Explanation

A quadratic equation can be expressed in the form

$ax^{2} + bx + c = 0$ .
Here, the only equation that has a third degree variable is

$(3x - 2)^{3} + \dfrac{1}{2}x - 4$ and all other equations have two degree variables.
So,

$(3x - 2)^{3} + \dfrac{1}{2}x - 4$ is not a quadratic equation.

Discriminant of a quadratic equation

$\displaystyle p{ x }^{ 2 }+qx+r=0$ is given by.

0%
$\displaystyle { \left( { { q }^{ 2 }-4pr } \right) }$
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$\displaystyle \sqrt { { q }^{ 2 }+4qr }$
0%
$\displaystyle \sqrt { { q }^{ 2 }-4pr }$
0%
$\displaystyle { q }^{ 2 }+4{ q }^{ 2 }$

Explanation

Given equation is

$px^2+qx+r=0$

Here

$a = p$ ,

$b=q$ and

$c=r$

Discriminant

$D\displaystyle = {{ b }^{ 2 }-4ac}$

$=\displaystyle {{ q }^{ 2 }-4pr}$

Hence, option A is correct.

If

$x^{2} + 10 x = 24,$ where

$x>0$ , then the value of

$x + 5$ is

0%
$7$
0%
$5$
0%
$-12$
0%
$2$

Explanation

$x^2+10x-24 = 0$

$x^2+12x-2x-24=0$

On factoring we get,

$(x – 2)(x + 12) = 0$

$x = 2, x = -12$

Condition: $x > 0$ , then the value of $x + 5$ is $2 + 5 = 7$ .

Identify the standard format of quadratic equation.

0%
$ax^{2}+bx + c = 0$
0%
$ax^{3}+ bx^{2} + 2$
0%
$ax^{3}+ bx^{3}+13$
0%
$a(x - 1)^{2}+ bx^{3}+ c$

Explanation

A quadratic equation can be expressed in the standard form is

$ax^{2} + bx + c = 0$ and all other equations have third degree variables.

Hence, option A is correct.

The roots of the quadratic equations

$(x-1)^2 = \frac{9}{4}$ , are

0%
$x=-\frac{5}{3}, x=\frac{5}{3}$
0%
$x=-\frac{1}{2}, x=\frac{5}{2}$
0%
$x=\frac{5}{9}, x=\frac{13}{9}$
0%
$x=1\pm \sqrt{\frac{2}{3}}$

Explanation

Given,

$(x-1)^2=\dfrac{9}{4}=\left(\dfrac{3}{2}\right)^2$

$\Rightarrow x-1=\pm \dfrac{3}{2}$

$\Rightarrow x=\pm \dfrac{3}{2}+1$

$\Rightarrow x=-\dfrac{1}{2}, \dfrac{5}{2}$

Choose the best possible option.

$\displaystyle (x+5)(x-8)=0$ is quadratic equation.

0%
Yes
0%
No
0%
Complex equation
0%
None

Explanation

$\displaystyle \left( x+5 \right) \left( x-8 \right) =0$

$\displaystyle { x }^{ 2 }-8x+5x-40=0$

$\displaystyle { x }^{ 2 }-3x-40=0$ is quadratic , because the degree of equation is 2

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