MCQ Questions for Class 10 Maths Quadratic Equations Quiz 3

Solution of the equation

$2x^{2} = 4x + 3$ is

0%
$\dfrac {2\pm \sqrt {10}}{2}$
0%
$\dfrac {2\pm 5\sqrt {2}}{2}$
0%
$1\pm \sqrt {5}$
0%
$1\pm \sqrt {10}$

Explanation

We know the formula,

$\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}$
Here the equation,

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

$a = 2, b = -4, c = -3$
Substituting the values, we get

$=$

$\dfrac{4\pm\sqrt{(-4)^2-4\times 2\times -3}}{2\times 2}$

$=$

$\dfrac{4\pm\sqrt{16+24}}{4}$

$=$

$\dfrac{4\pm\sqrt{40}}{4}$
Take

$2$ as common, we get

$=$

$\dfrac{2\pm\sqrt{10}}{2}$

$\therefore$ solution of the equation

$2x^2-4x-3$ is

$\dfrac{2\pm\sqrt{10}}{2}$ .

The discrimination of

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

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

Explanation

$D = (b^{1})^{2}-4a^{1}c^{1}$

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

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

$\Rightarrow \boxed{D = (a-b)^{2}}$

1097679_517000_ans_0ab9cca3d9724db596cdaa817fbe4348.jpg

Which of the following is a quadratic equation?

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

Explanation

i) As highest Power is '1'

$\therefore$ not quadratic

ii)

$(x-1)(x+4) = x^{2}+1\Rightarrow x^{2}+3x-4 = x^{2}+1\Rightarrow \boxed{3x-5 = 0}$

$\therefore$ not quadratic (higher degree 1)

iii) Highest delrle is '2'

$\therefore$ quadratic

iv)

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

$\Rightarrow 5x+7 = 0$ (

$\therefore$ not quadratic)

Solve the given equation by the method of Completion of Squares:

${x}^{2}+4x+4=0$ ?

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

Explanation

Given that:

$p(x)=x^2+4x+4=0$

Solution:

Zeroes of the polynomial

$p(x)=x^2+4x+4=0$

$\Rightarrow x^2 + 2 \times 2 \times x + (2)^2 = (x + 2)^2$

or,

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

or,

$x=-2$ or

$x=-2$

Therefore, the correct answer is

$-2$

The discriminant (

$D$ ) of the equation

$5x-6+\cfrac { 1 }{ x } =0$ is ............

0%
$4$
0%
$\sqrt { 56 }$
0%
$16$
0%
$56$

Explanation

Given:

$5x-6+\cfrac { 1 }{ x } =0$

$\Rightarrow 5{ x }^{ 2 }-6x+1=0$

Comparing above equation with standard quadratic equation

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

$a=5,b=-6,c=1$

Now, Discriminant

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

$=36-20$

$=16$

The roots of quadratic equation

$\dfrac{x}{k} = \dfrac{k}{x}$ are ...........

0%
$k, -k$
0%
$-k, -k$
0%
$k, k$
0%
$k^2, - k^2$

Explanation

Given:

$\dfrac{x}{k} = \dfrac{k}{x}$

$\Rightarrow x^2 = k^2$

$\Rightarrow x = \sqrt{k^2}$

$\Rightarrow x = \pm k$

$\Rightarrow x = k, -k$

Hence, the roots are

$k,-k$

The following equation is a quadratic equation.

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

0%
True
0%
False

Explanation

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

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

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

$6x^2+12x+12=0$

$6(x^2+2x+2)=0$

$\Rightarrow$

$x^2+2x+2=0$

$\Rightarrow$ A quadratic equation is an equation of the second degree, meaning it contains at least one term that is squared.

$\Rightarrow$ Comparing equation

$x^2+2x+2=0$ with standard form of quadratic equation

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

$\Rightarrow$ We get,

$a=1,\,b=2$ and

$c=2$

$\therefore$ The equation

$(x+2)^3=x^3-4$ is quadratic equation is true.

__________ is true for the discriminant of a quadratic equation

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

0%
$D=0$
0%
$D<0$
0%
$D>0$
0%
$D$ is a perfect square

Explanation

For

$x^2+x+1=0$

$a=b=c=1$ { by comparing with the general equation

$ax^2+bx+c$ }

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

$D=1^{2}-4\times 1\times 1$

$D=1-4=-3$

Therefore D is negative

The number of triples

$(x, y, z)$ of real numbers satisfying the equation

$x^{4} + y^{4} + z^{4} + 1 = 4xyz$ is

0%
$0$
0%
$4$
0%
$8$
0%
More than $8$

Explanation

Applying

$AM \geq GM$

We get

$\dfrac{ x^{4} + y^{4} + z^{4} + 1}{4} \geq \sqrt[4]{x^4y^4z^4}$

$\Longrightarrow x^{4} + y^{4} + z^{4} + 1 \geq 4xyz$

$\textrm {Thus, min value of } x^{4} + y^{4} + z^{4} + 1 \textrm { must be } 4xyz$

Clearly our function is increasing for x,y,z>0 and decreasing for x,y,z<0.

Thus min value must occur for integral values like 1,0,-1.

For 0, it doesn't satisfy the equation

$x^{4} + y^{4} + z^{4} + 1 = 4xyz$

But triplets of 1 and -1 do.

Thus, we can check that the triplets

$(1,-1,-1); (-1,1,-1); (-1,-1,1) and (1,1,1)$ satisfy the equation.

We don't need to look for further values as we won't get min value of

$x^{4} + y^{4} + z^{4} + 1$ by them.

Thus, these are the only solutions.

The following equation is a qudratic equation.

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

0%
True
0%
False

Explanation

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

$\Rightarrow$

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

$\Rightarrow$

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

$\Rightarrow$

$6x^2-19x+12=0$

$\Rightarrow$ A quadratic equation is an equation of the second degree, meaning it contains at least one term that is squared.

$\Rightarrow$ Comparing given equation

$6x^2-19x+12=0$ with standard form of quadratic equation

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

$\Rightarrow$ We get,

$a=6,\,b=-19$ and

$c=12$

$\therefore$ The equation

$16x^2-3=(2x+5)(5x-3)$ is quadratic equation is true.

Check whether the given equation is a quadratic equation or not.

$2{ x }^{ 2 }-7x=0\quad$

0%
True
0%
False

Explanation

$2x^2-7x=0$

A quadratic equation is an equation of the second degree, meaning it contains at least one term that is squared.

On comparing equation

$2x^2-7x=0$ with standard form of quadratic equation

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

We get,

$a=2,\,b=-7,\,c=0$

Hence, the equation

$2x^2-7x=0$ is quadratic equation.

Which of the following methods is/are guaranteed to solve any Quadratic Equation?

0%
Factorisation Method
0%
Completing Square Method
0%
Standard Quadratic Formula
0%
Both option b and option c

Explanation

We have learnt that the Factorisation Method might not be helpful in solving different quadratic equations. But the Completing Square Method and the Standard Quadratic Formula are guaranteed to solve any given Quadratic Equation. Hence both options

$b$ and

$c$ are correct. Thus option

$d$ is the correct answer.

Identify which of the following equations is a Quadratic Equation.

0%
$5x + 8 = 0$
0%
$-2x^2+x=21$
0%
$ax + by + c = 0$
0%
None of these.

The Completing Square Method converts the quadratic equation $ax^2+bx+c=0$ into which of the following forms?

0%
$(px+q)(rx+s)=0$
0%
$px+q=0$
0%
$(x+p)^2=q$
0%
All of the above.

Explanation

The Completing Square Method involves converting a quadratic equation of the form

$ax^2+bx+c=0$ into the form

$(x+p)^2=q$ so that this equation can be solved easily as it now contains only one term of

$x$ unlike the original quadratic equation which has two terms in

$x$ ,

$ax^2$ and bx thus making it difficult to solve the equation. Hence option c is the correct answer.

For a quadratic equation in the standard form $ax^2+bx+c=0$ , the nature of roots is dependent on the value of ?

0%
$b^2+4ac$
0%
$b^2 \pm 4ac$
0%
$\sqrt{b^2}-4ac$
0%
$b^2-4ac$

Explanation

Nature of roots is dependent on the value of determinant and discriminant

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

Check whether the given equation is a quadratic equation or not.

${ x }^{ 2 }+2\sqrt { x } -3$

0%
True
0%
False

Explanation

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

$\Rightarrow$ A quadratic equation is an equation of the second degree, meaning it contains at least one term that is squared.

$\Rightarrow$ Comparing equation

$x^2+2\sqrt{x}-3=0$ with standard form of quadratic equation

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

$\Rightarrow$ So, here we do not get

$b$ term

$\therefore$ The equation

$x^2+2\sqrt{x}-3=0$ is not quadratic equation.

The discriminant of the quadratic equation

${ x }^{ 2 }-4x+a=0$ is

$16-4a$ .

0%
True
0%
False

Explanation

Given:

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

$\Rightarrow$ Here,

$a=1,\,b=-4,\,c=a$ on comparing with the standard quadratic equation.

$\Rightarrow$

$Discriminant=b^2-4ac$

$=(-4)^2-4(1)(a)$

$=16-4a$

$\therefore$ Discriminant

$=16-4a$

Which of the following is a Quadratic Equation?

0%
$5x + 8 = 0$
0%
$6x^2 + 7x =19$
0%
$x + 1$
0%
None of these

Explanation

A Quadratic Equation is an equation with one variable and degree

$2$ . In options a, and c we have equations in one variable i.e. x . but the degree of the variable

$x$ is

$1$ . While option b has one variable and the degree is

$2$ , hence

$b$ is the correct option.

Assertion (A): The equation

$x-\displaystyle \cfrac{2}{x-1}=1-\cfrac{2}{x-1}$ has no root.
Reason (R):

$x-1\neq 0$ , then only above equation is defined.

0%
Both A and R are true and R is the correct explanation of A.
0%
Both A and R are true and R is not the correct explanation of A.
0%
A is true but R is false.
0%
A is false but R is true

Explanation

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

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

$x=1$
However at

$x=1$ the expression

$\dfrac{2}{x-1}$ is not defined.
Hence the above equation has no real roots.

Find roots of equation

$x^2-11x+28=0$ by quadratic formula

0%
$4,7$
0%
$-4,-7$
0%
$4,-7$
0%
$-4,7$

Explanation

$x^2-11x+28=0$

We have a formula for solving quadratic equation

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

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

$x=\dfrac{11\pm\sqrt{121-112}}{2}\\$

$x=\dfrac{11\pm\sqrt{9}}{2}\\$

$x=\dfrac{11\pm3}{2}\\$

$x=\dfrac{11-3}{2};x=\dfrac{11+3}{2}\\$

$x=\dfrac82;x=\dfrac{14}2\\$

$x=4,7$

If the roots of a quadratic equation

$ax^2+bx+c=0$ are all real then which one of the following is true?

0%
$b^2-4ac=0$
0%
$b^2-4ac>0$
0%
$b^2-4ac\ge 0$
0%
$b^2-4ac<0$

Explanation

Given, the roots of a quadratic equation

$ax^2+bx+c=0$ ........(1) are all real.

Then the discreminant of the equation (1) is

$\ge 0$ .

The discreminant of the equation (1) is

$b^2-4ac$ .

$b^2-4ac\ge 0$ .

$px^2 +qx+r=0$ has equal roots, then

$q^2 =$

0%
$4p$
0%
$4pr$
0%
$8pr$
0%
$pr$

Explanation

$px^2 +qx+r=0$

The standard quadratic equation is

$ax^2+bx+c=0$

For roots to be equal

$\begin{array}{l}{b^2} - 4ac = 0\\or,{b^2} = 4ac\\so,{(q)^2} = 4 \times p \times r\\{q^2} = 4pr \end{array}$

If the roots of the equation

$ax^2+bx+c=0$ are all real equal then which one of the following is true?

0%
$b^2-4ac=0$
0%
$b^2-4ac\ge0$
0%
$b^2-4ac<0$
0%
$b^2-4ac>0$

Explanation

Given, the roots of a quadratic equation

$ax^2+bx+c=0$ are all real and equal.

Then the discriminant of this equation is

$= 0$ .

The discriminant of this equation is

$b^2-4ac$ .

$b^2-4ac= 0$ .

If the roots of a quadratic equation are equal, than discriminate is

0%
$1$
0%
$0$
0%
greater than $0$
0%
less than $0$

Explanation

We know that to find the roots we have

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

If roots are equal

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

$(Discriminate )$

Which is a quadratic equation?

0%
$x+\dfrac{1}{x}=2$
0%
$x^{2}+1=(x+3)^{2}$
0%
$x(x+2)$
0%
$x+\dfrac{1}{x}$

The number of integral values of

$'x'$ for which

$x^{2}+19x+92$ is a perfect square is:

0%
$0$
0%
$1$
0%
$2$
0%
$3$

Explanation

The above quadratic equation can be expressed as :

$x^2 + 19x + 92 = \left(x + \dfrac{19}{2} \right)^2 + 92 - \dfrac{361}{4}$

$= \left(x + \dfrac{19}{2} \right)^2 + \dfrac{7}{4}$

From the above expression, we can infer that no integer value of x will make it a perfect square because we have a fraction in the expression.

The solution of the equation

$x+\cfrac{1}{x}=2$ will be

0%
$2,-1$
0%
$0,-1,-\cfrac{1}{5}$
0%
$-1,-\cfrac{1}{5}$
0%
None of these

Explanation

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

$x^{2}-2x+1=0$

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

$x=1$

$81$ is discriminant of

$2{ x }^{ 2 }+5x-k=0$ then the value of

$k$ is

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

Explanation

the given equation

$2x^2+5x-k=0$ is in the form of

$ax^2+bx+c$

By comparing

$a=2, b=5, c=-k$

Given, Descriminant

$D=81$

$(D=b^2-4ac)$

By equating,

$\Rightarrow b^2-4ac=81$

$\Rightarrow 25+8k=81$

$\Rightarrow 8k=81-25$

$\Rightarrow 8k=56$

$\Rightarrow k=7$

The value of

$\left( \frac { x - 3 } { x ^ { 2 } - x - 6 } + \frac { 2 x - 1 } { 2 x ^ { 2 } + 5 x - 3 } - \frac { 2 x + 5 } { x ^ { 2 } + 5 x + 6 } \right)$ is:

0%
$\frac { x - 3 } { x - 2 }$
0%
$- 1$
0%
$1$
0%
$0$

Explanation

$\quad \frac { x-3 }{ { x }^{ 2 }-x-6 } +\frac { 2x-1 }{ { 2x }^{ 2 }+5x-3 } -\frac { 2x+5 }{ { x }^{ 2 }+5x+6 } \\ =\frac { x-3 }{ \left( x-3 \right) \left( x+2 \right) } +\frac { 2x-1 }{ \left( x+3 \right) \left( 2x-1 \right) } -\frac { 2x+5 }{ \left( x+2 \right) \left( x+3 \right) } \quad \left( x\neq 3,-3,-2,\frac { 1 }{ 2 } \right) \\ =\frac { 1 }{ x+2 } +\frac { 1 }{ x+3 } -\frac { \left( x+2 \right) +\left( x+3 \right) }{ \left( x+2 \right) \left( x+3 \right) } \\ =\frac { 1 }{ x+2 } +\frac { 1 }{ x+3 } -\frac { 1 }{ x+3 } -\frac { 1 }{ x+2 } \\ =0$

so option D is correct

Use the quadratic formula to solve

$x^2-4x-8=0$ .

0%
$2 \pm 2\sqrt 3$
0%
$1 \pm \sqrt 3$
0%
$3-3\sqrt 3$
0%
None of these

Explanation

Given quadratic equation is

$x^2-4x-8=0$

We know that, for a quadratic equation of the form

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

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

Here,

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

$\therefore \ x=\dfrac{-(-4) \pm \sqrt{(-4)^2-4(1)(-8)}}{2(1)}$

$\Rightarrow x=\dfrac{4 \pm \sqrt{16+32}}{2}$

$\Rightarrow x=\dfrac{4 \pm \ 4 \sqrt{3}}{2}$

$\therefore \ x=2 \pm 2\sqrt 3$

Hence, option A is correct.

CBSE Questions for Class 10 Maths Quadratic Equations Quiz 3 - MCQExams.com

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Practice Class 10 Maths Quiz Questions and Answers

CBSE Questions for Class 10 Maths Quadratic Equations Quiz 3 - MCQExams.com

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Practice Class 10 Maths Quiz Questions and Answers

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