MCQ Questions for Class 10 Maths Quadratic Equations Quiz 1

Solve the following equations.

$x^4\,-\,3x^2\,+\,2\,=\,0$ , roots are

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$x\,=\,\pm\,\sqrt3,\,\pm\,1$
0%
$x\,=\,\pm\,\sqrt2,\,\pm\,1$
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$x\,=\,\pm\,\sqrt2,\,\pm\,3$
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none of the above

Explanation

Given equation is

$x^4 - 3x^2 +2 = 0$

$x^4 - 2x^2 -x^2 + 2 =0$

$(x^2 - 1)(x^2 - 2) = 0$

$\therefore x^2 = 2 \Rightarrow x = \pm \sqrt{2}$

$\therefore x^2 = 1 \Rightarrow x = \pm 1$

Which of the following methods is used to derive the Standard Quadratic Formula for the Quadratic Equation

$ax^2+bx+c=0$ ?

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Factorisation Method
0%
Completing Square Method
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Hit and Trial Method
0%
All the above.

Explanation

We have learnt that in order to derive the standard quadratic formula from the standard form of a quadratic equation

$ax^2+bx+c=0$ ,we follow the steps involved in the Completing Square Method . Hence

$b$ is the correct option.

Let

$\alpha(a)$ and

$\beta(a)$ be the roots of the equation

$\\ \left( \sqrt [ 3 ]{ 1+a } -1 \right) x^{ 2 }+\left( \sqrt { 1+a } -1 \right) x+\left( \sqrt [ 6 ]{ 1+a } -1 \right) =0$

$\displaystyle \lim _{ a\rightarrow 0 }{ \alpha } \left( a \right)$ and

$\displaystyle \lim _{ a\rightarrow 0 }{ \beta } \left( a \right)$ are

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$-\dfrac{5}{2}$ and $1$
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$-\dfrac{11}{2}$ and $-1$
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$-\dfrac{7}{2}$ and $2$
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$-\dfrac{9}{2}$ and $3$

The roots of the equation

$\sqrt{3y + 1} = \sqrt{y - 1}$ are?

0%
$- 1$
0%
$2, 3$
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$2, 1$
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None of these

Explanation

Given: equation

$\sqrt {3y+1}=\sqrt {y-1}$

To find the roots of the equation

Sol:

$\sqrt {3y+1}=\sqrt {y-1}$

Take square on both sides, we get

${3y+1}= {y-1}\\\implies 3y-y=-1-1\\\implies 2y=-2$

or,

$y=-1$

But

$y\ne-1$ as

$\sqrt{y-1} = \sqrt{-2}$ which is not possible. Hence, none of the given options is the answer.

A quadratic equation in

$x$ is

$ax^2 + bx + c = 0$ , where

$a, b, c$ are real numbers and the other condition is

0%
$a \neq 0$
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$b \neq 0$
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$c \neq 0$
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$b = 0$

Explanation

$a, b$ and

$c$ are constants in the equation.
As

$x$ is raised to power

$2$ it is a quadratic equation.
If

$a=0$ , then it will nullify

${ x }^{ 2 }$ , making it a linear equation.
So only if

$a\neq 0$ , the equation remains quadratic.
Option A is correct.

Find the roots of following quadratic equation

$x^2\,+\,3x\,-\,2\,=\,0$

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$\displaystyle\,x\,=\,\frac{-3\,\pm\,\sqrt{15}}{2}$
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$\displaystyle\,x\,=\,\frac{-3\,\pm\,\sqrt{17}}{2}$
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$\displaystyle\,x\,=\,\frac{-3\,\pm\,\sqrt{19}}{2}$
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$\displaystyle\,x\,=\,\frac{-3\,\pm\,\sqrt{21}}{2}$

Explanation

Gven equation is

$x^2 + 3x - 2 = 0$
Using quadratic formula,

$a=1, b=3, c=-2$

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

$x = \frac{-3 \pm \sqrt{9 - 4(1)(-2)}}{2\times1}$

$\therefore x = \frac{-3 \pm \sqrt{17}}{2}$

$\therefore$ the roots of the given equation are

$\displaystyle \frac {-3\pm \sqrt{17}}{2}$

Find the discriminant for the given quadratic equation:

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

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$26$
0%
$32$
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$38$
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$44$

Explanation

Given equation is,

$\sqrt{3}x^2\,+\,2\sqrt{2}x\,-\,2\sqrt{3}\,=\,0$
We know,

$D = b^2 - 4ac$

$a=\sqrt 3, b=2 \sqrt 2, c=-2\sqrt3$

$D = (2\sqrt{2})^2 - 4\times\sqrt{3}\times(-2\sqrt{3})$

$D = 8 + 24$

$\therefore D = 32$

Find the discriminant for the given equation:

$3x^2\,+\,2x\,-\,1\,=\,0$

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$11$
0%
$13$
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$15$
0%
$16$

Explanation

Given equation is,

$3x^2 + 2x- 1 = 0$
We know,

$D = b^2 - 4c$

$a=3,b=2,c=-1$

$D = 2^2 - 4\times3\times(-1)$

$D = 4 + 12$

$\therefore D = 16$

Find the discriminant for the given quadratic equation:

$x^2\,+\,4x\,+\,k\,=\,0$

0%
$6\,-\,4k$
0%
$16\,-\,4k$
0%
$16\,-\,3k$
0%
$13\,-\,2k$

Explanation

Given equation is,

$x^2 + 4x + k = 0$
We know,

$D = b^2 - 4ac$

$a=1, b=4, c=k$

$D = (4)^2 - 4(1)(k)$

$\therefore D = 16 - 4k$

Find the value of discriminant for the following equation.

$x^{2}\, +\, 4x\, +\, k\, =\, 0$

0%
$\Delta\, =\, 16\, + 4k$
0%
$\Delta\, =\, 16\, -4k$
0%
$\Delta\, =\, 16\, - k$
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$\Delta\, =\, 16+k$

Explanation

Given equation is:

$x^2+4x+k=0$

Discriminant =

$b^2 - 4ac$

$a=1, b=4, c=k$

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

$= 16 - 4k$

$\therefore \Delta =16-4k$

Say true or false.

A quadratic equation cannot be solved by the method of completing the square.

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True
0%
False

Explanation

Given a quadratic equation is

$ax^2+bx+c=0$

There are

$3$ methods to solve the quadratic equations :

$(1)$ Factorisation(Find factors of the equation)

$(2)$ Using quadratic formula i.e

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

$(3)$ Completing the square

Thus, completing the square is one of the methods to solve quadratic equations.

Hence, the given statement is false.

The discriminant of

$ax^{2}-(a+b)x+b=0$ is:

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$a-b$
0%
$a+b$
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$(a+b)^{2}$
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$(a-b)^{2}$

Explanation

(D)

$b^{2}-4ac=\left [ (a+b) \right ]^{2}-4ab$

$=(a+b)^{2}-4ab$

$= a^2+b^2-2ab$

$=(a-b)^{2}$

If a given expression is a complete square, then which of the following formulae we use to factorize it?

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

Explanation

If the given expression is a complete square, then we use the following formulae to factorize it.

$a^2+2ab+b^2=(a+b)^2$ and

$a^2-2ab+b^2=(a-b)^2$

For the expression

$ax^2 + 7x + 2$ to be quadratic, the necessary condition is

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$a=0$
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$a \neq 0$
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$a> \cfrac{7}{2}$
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$a<-1$

Explanation

For the expression

$a^2+7x+2$ to be quadratic the possible values of

$a$ must be non zero real numbers because we know that the expression

$ax^2+bx+c$ where

$a, b, c$ are real numbers is quadratic if

$a\neq 0$

Find the value of discriminant for the following equation.

$2x^{2}\, +\, x\, +\, 1\, =\, 0$

0%
$7$
0%
$-7$
0%
$9$
0%
$-9$

Explanation

The value of discriminant of

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

$D = b^2 - 4ac$

$a=2, b=1, c=1$

$\therefore D = (1)^2 - 4\times2\times1$

$= 1 - 8$

$= -7$

The discriminant (D) of

$\sqrt{x^{2}+x+1}=2$ is:

0%
-3
0%
13
0%
11
0%
12

Explanation

Squaring on the both sides of the given equation, we get

$x^{2}+x+1=4$

$x^{2}+x-3=0$

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

$=1-4(1)(-3)$

$=1+12$

$=13$

If the discriminant of

$3x^{2}-14x+k=0$ is

$100$ , then

$k=$

0%
$8$
0%
$32$
0%
$16$
0%
$24$

Explanation

Given that, the discriminant of

$3x^2-14x+k=0$ is

$100$ .

To find out: The value of

$k$ .

We know that, the discriminant of a quadratic equation of the form

$ax^2+bx+c=0$ is given by,

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

Here,

$a=3, b=-14, c=k$

So,

$D=(-14)^{2}-4(3)k$

$\Rightarrow (-14)^{2}-4(3)k=100$

$\Rightarrow 196-12k=100$

$\Rightarrow 12k=96$

$\Rightarrow k=8$

Hence, the value of

$k$ is

$8$ .

Which of the following is a quadratic polynomial in one variable?

0%
$\sqrt{2x^{3}}+5$
0%
$2x^2 + 2x^{-2}$
0%
$x^2$
0%
$2x^2+y^2$

Explanation

Polynomials in one variable are algebraic expressions that consist of terms in the form

$ax^n$

Here

$n$ is a non-negative (i.e. positive or zero) integer and

$a$ is a real number and is called the coefficient of the term.

The degree of a polynomial in one variable is the largest exponent in the polynomial.

And for quadratic polynomial

$n=2.$

Hence the only quadratic polynomial in one variable is

$x^2$

The mentioned equation is in which form?

$3y^{2}\,-\, 7\, =\, \sqrt{3}\,y$

0%
linear
0%
Quadratic
0%
Cubic
0%
None

Which of the following is a quadratic equation ?

0%
$x^{\frac{1}{2}}+2x+3=0$
0%
$(x^2-1)(x+4)=x^{2}+1$
0%
$x^{2}-3x+5=0$
0%
$(2x^2+1)(3x-4)=6x^{2}+3$

Explanation

Standard form of a quadratic equation is

$ax^{2}+bx+c=0$

The quadratic equation is an equation having

$2$ highest degree of the variable, which is possible only in the equation

$x^2-3x+5=0$ out of the four given equations.

Hence,

$Op-C$ is correct.

The condition for

$px^{2}+qx+r=0$ to be pure quadratic is:

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$p=0$
0%
$q=0$
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$r=0$
0%
$p=q=0$

Explanation

A quadratic equation in which the term containing x raised to the power of 1 is not present is called a pure quadratic equation. In other words, ax^2 + c = 0 is a pure quadratic equation.

$\therefore\ q\ = 0$

The mentioned equation is in which form?

$\cfrac {3}{4}y^{2}\, =\, 2y\, +\,7$

0%
cubic
0%
quadratic
0%
linear
0%
none of these

Explanation

Given equation is

$\dfrac {3}{4}y^2=2y+7$

The highest power of

$x$ is

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

The mentioned equation is in which form?

$(y\, -\, 2)\, (y\, +\, 2)\, =\, 0$

0%
cubic
0%
quadratic
0%
linear
0%
none of these

Explanation

Given,

$(y -2)(y +2) =0$

$\Rightarrow y^2 - 4 = 0$
The highest power of

$y$ is

$2$ .

Thus, it is a quadratic equation.

The mentioned equation is in which form?

$m^{3}\, +\, m\, +\, 2\, =\, 4m$

0%
Linear
0%
Quadratic
0%
Quartic
0%
None

The mentioned equation is in which form?

$y^{2}\, -\, 4\, =\, 11y$

0%
Quadratic
0%
linear
0%
Cubic
0%
None

Explanation

Given equation is

$y^2-4=11y$

The highest power of

$y$ is

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

The mentioned equation is in which form?

$z\, -\, \cfrac{7}{z}\, =\, 4z\, +\, 5$

0%
Linear
0%
Quadratic
0%
cubic
0%
None of these

Explanation

Given,

$z - \dfrac{7}{z} = 4z + 5$

$\Rightarrow \dfrac { { z }^{ 2 }-7 }{ z } =4z+5$

$\Rightarrow z^2 - 7 = 4z^2 + 5z$

$\Rightarrow 3z^2 + 5z + 7 = 0$

The highest power of z is 2. Hence, it is a quadratic equation.

The mentioned equation is in which form?

$n\, -\, 3\, =\, 4n$

0%
linear
0%
Quadratic
0%
constant
0%
None

STATEMENT - 1 :

$(x-2)(x+1)$

$=$

$(x-1)(x+3)$ is a quadratic equation.
STATEMENT - 2 : If

$p(x)$ is a quadratic polynomial, then

$p(x)$

$=$

$0$ is called a quadratic equation.

0%
Statement - 1 is True, Statement - 2 is True, Statement - 2 is a correct explanation for Statement - 1
0%
Statement - 1 is True, Statement - 2 is True : Statement 2 is NOT a correct explanation for Statement - 1
0%
Statement - 1 is True, Statement - 2 is False
0%
Statement - 1 is False, Statement - 2 is True

Explanation

Given equation

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

To find whether the given equation is a quadratic equation or not

Sol:

$(x-2)(x+1)=(x-1)(x+3)\\\implies x^2+x-2x-2=x^2+3x-x-3\\\implies x^2-x^2+x-2x-3x+x-2+3=0\\\implies -3x+1=0$

This is not a quadratic equation as an equation of degree two is called a quadratic equation.

And, a polynomial when equated to zero or some value becomes an equation.

Is the following equation quadratic?

$(x\, +\, 3) (x\, -\, 4)\, =\, 0$

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

Explanation

$Answer=1$
The equation

$(x\, +\, 3) (x\, -\, 4)\, =\, 0$ can be formed as

$x^2 - x -12 =0$ has the highest power as 2. Thus, it is quadratic equation.

Is the following equation quadratic?

$n^{3}\, -\, n\, +\, 4\, =\, n^{3}$

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

Explanation

A quadratic equation is a second-order polynomial equation in a single variable

$x$ .

$ax^2+bx +c=0$ where

$a\neq 0$ .

The equation

$n^3 - n + 4 = n^3$ can be framed as

$-n + 4 =0$ by subtracting

${ n }^{ 3 }$ on both sides,
Thus, it is not a quadratic equation as it is not an equation of degree

$2.$

CBSE Questions for Class 10 Maths Quadratic Equations Quiz 1 - MCQExams.com

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Practice Class 10 Maths Quiz Questions and Answers

CBSE Questions for Class 10 Maths Quadratic Equations Quiz 1 - MCQExams.com

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Practice Class 10 Maths Quiz Questions and Answers

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