MCQ Questions for Class 10 Maths Quadratic Equations Quiz 9

Solve the quadratic equation by using quadratic formula

$9x^2-3( a+b) x+ ab=0$

0%
$x=\dfrac {a}{3}, \dfrac {b}{3}$
0%
$x=\dfrac {-a}{3}, \dfrac {-b}{3}$
0%
$x=\dfrac {-a}{3}, \dfrac {b}{3}$
0%
$x=\dfrac {a}{3}, \dfrac {-b}{3}$

Explanation

$\Rightarrow$ Here,

$a=9,\,b=-3(a+b),\,c=ab$

$\Rightarrow$

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

$=\dfrac{-[-3(a+b)]\pm\sqrt{[-3(a+b)]^2-4(9)(ab)}}{2(9)}$

$=\dfrac{3a+3b\pm\sqrt{9(a^2+b^2+2ab)-36ab}}{18}$

$=\dfrac{3a+3b\pm\sqrt{9a^2+9b^2+18ab-36ab}}{18}$

$=\dfrac{3a+3b\pm\sqrt{9a^2+9b^2-18ab}}{18}$

$\dfrac{3a+3b\pm\sqrt{9(a^2+b^2-2ab)}}{18}$

$=\dfrac{3a+3b\pm\sqrt{9(a-b)^2}}{18}$

$=\dfrac{3a+3b\pm3(a-b)}{18}$

$=\dfrac{3a+3b\pm 3a-3b}{18}$

$x=\dfrac{3a+3b+3a-3b}{18}$ and

$x=\dfrac{3a+3b-3a+3b}{18}$

$x=\dfrac{6a}{18}$ and

$x=\dfrac{6b}{18}$

$\therefore$

$x=\dfrac{a}{3},\,\dfrac{b}{3}$

The solution for

$2x^2-9x+7=0$ is

0%
$2,3$
0%
$1,\dfrac 94$
0%
$\dfrac 14,\dfrac {13}4$
0%
$1,\dfrac {13}4$

Explanation

Given Quadratic equation is

$2x^2-9x+7=0$

Comparing it with the general form of quadratic equation

$ax^2+bx+c=0$

we write,

$a=2,b=-9,c=7$

Quadratic formula is

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

$\implies x=\dfrac{-(-9)\pm\sqrt {(-9)^2-4(2)(7)}}{2(2)}$

$\implies x=\dfrac{9\pm\sqrt {81-56}}{4}$

$\implies x=\dfrac{9\pm\sqrt {25}}{4}$

$\implies x=\dfrac{9\pm 5}{4}$

$\implies x=\dfrac{9 +5}{4} \quad x=\dfrac{9-5}4$

$\implies x=1,\dfrac{7}2$

Find the roots of the following equations, if they exist, by the completing the square:

0%
$2x^{2}-7x+3=0$
0%
$2x^{2}+x-4=0$
0%
$4x^{2}+\sqrt4{3x}+3=0$
0%
$2x^{2}+x+4=0$

Explanation

$so\quad the\quad root\quad of\quad quadratic\quad equation\quad are\quad$

$X=3\quad and\quad X=\dfrac { 1 }{ 2 }$

The root of

$x^{2}+kx+k=0$ are real and equal , find

$k$ .

0%
$0$
0%
$4$
0%
$0$ or $4$
0%
$2$

Explanation

The roots of

$x^2 + kx + k = 0$ are real and equal

i.e.,

$b^2 - 4ac = 0$

$\implies k^2 - 4(1)k = 0$

$\implies k^2 - 4k = 0$

$\implies k(k - 4) = 0$

$\therefore k = 0$ or

$4$

Find the root of the following quadratic equation, if they exist, by the method of completing the square.

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

0%
$\dfrac{-1-{\sqrt 33}}{4}$
0%
$\dfrac{-1-{\sqrt 33}}{44}$
0%
$\dfrac{-1+{\sqrt 33}}{44}$
0%
$\dfrac{1-{\sqrt 33}}{4}$

If both roots of quadratic equation

$(\alpha +1)x^{2}-2(1+3\alpha )x+1+8\alpha =0$ are real and distinict, the

$\alpha$ be-

0%
-2
0%
1
0%
2
0%
3

Explanation

$\left( \alpha + 1 \right) {x}^{2} - 2 \left( 1 + 3 \alpha \right) x + 1 + 8 \alpha = 0$

Here,

$a = \alpha + 1$

$b = - 2 \left( 1 + 3 \alpha \right)$

$c = 1 + 8 \alpha$

If both roots are real and distinct,

$D > 0$

$\Rightarrow {b}^{2} - 4ac > 0$

$\Rightarrow {\left( - 2 \left( 1 + 3 \alpha \right) \right)}^{2} - 4 \left( 1 + \alpha \right) \left( 1 + 8 \alpha \right) > 0$

$\Rightarrow 4 \left( 1 + 9 {\alpha}^{2} + 6 \alpha \right) - 4 \left( 1 + 9 \alpha + 8 {\alpha}^{2} \right) > 0$

$\Rightarrow 4 \left( 1 + 9 {\alpha}^{2} + 6 \alpha - 1 - 9 \alpha - 8 {\alpha}^{2} \right) > 0$

$\Rightarrow {\alpha}^{2} - 3 \alpha > 0$

$\Rightarrow \alpha \left( \alpha - 3 \right) > 0$

$\Rightarrow \alpha \in \left( - \infty, 0 \right) \cup \left( 3, \infty \right)$

Hence among the given values,

$\alpha$ will be

$-2$ .

Hence the correct answer is

$\left( A \right) -2$ .

Identify the standard form of the given equation

$(4x-5)(4x+5)=0$ .

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$16x^2-25=0$
0%
$16x^3-25=0$
0%
$16x^2-2=0$
0%
$6x^2-25=0$

Explanation

$(4x+5)(4x-5) = 0$

$4x(4x-5)+5(4x-5) = 0$

$16x^2-20x+20x-25 = 0$

$16x^2-25 =0$ which is in standard form of quadratic equation

$x^{2}-5x+1=0$ , then

$\dfrac {x^{10}+1}{x^{5}}$ has the value

0%
$2524$
0%
$2525$
0%
$2424$
0%
$2010$

Explanation

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

$\therefore x^2+1=5x$

$\therefore x+\dfrac{1}{x}=5$

$\therefore x^2+\dfrac{1}{x^2}=25-2=23$

$\therefore x^3+\dfrac{1}{x^3}=\left(x+\dfrac{1}{x}\right)^3-3\left(x+\dfrac{1}{x}\right)$

$\therefore x^3+\dfrac{1}{x^3}=125-15$

$\therefore x^3+\dfrac{1}{x^3}=110$

Now

$\left(x^3+\dfrac{1}{x^2}\right)\left(x^3+\dfrac{1}{x^3}\right)=56\times 23$

$\therefore x^5+x+\dfrac{1}{x}+\dfrac{1}{x^5}=110\times 23$

$\therefore x^5+\dfrac{1}{x^5}=2530-5$

$\therefore x^5+\dfrac{1}{x^5}=2525$ .

$a,b,c,x$ are real numbers and

$\left( { a }^{ 2 }+{ b }^{ 2 } \right) { x }^{ 2 }-2b\left( a+c \right) x+\left( { b }^{ 2 }+{ c }^{ 2 } \right) =0$ has real & equal roots, then

$a,b,c$ are in

0%
$A.P$
0%
$G.P$
0%
$H.P$
0%
$None of these$

Explanation

Since

$(a^2+b^2)x^2-2b(a+c)x+(b^2+c^2)=0$ has real & equal roots.

$\therefore$

$4b^2(a^2+c^2+2ac)-4(a^2+b^2)(b^2+c^2)=0$

$4b^2a^2+4b^2c^2+8acb^2-4a^2b^2-4a^2c^2-4b^2-4b^2c^2=0$

$8acb^2-4a^2c^2+4b^4=0$

$(2ac-2b^2)^2=0$

$\boxed{ac=b^2}$

The discriminant of the quadratic equation

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

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

Explanation

The given quadratic equation is

$5x^2-6x+1=0$ is in the form of

$ax^2+bx+c$

by comparing

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

Now, Discriminant

$D=b^2-4ac$

$D=36-4\times 5 \times 1$

$D=36-20=16$

Let

${ \lambda }_{ 1 }$ and

${\lambda}_{2}$ be two values of

${\lambda}$ for which the expression

${x}^{2}+(2-\lambda)x+\lambda-{3}$ becomes a perfect square. The value of

$(\lambda_{1}^{2}+\lambda_{2}^{2})$ equals

0%
$32$
0%
$25$
0%
$50$
0%
$100$

$\begin{array} { l } { \text { If } \alpha , \beta \text { are the roots of } a x ^ { 2 } + b x + c \text { and } \alpha + h } , { \beta + h \text { are the roots of } p x ^ { 2 } + q x + r = 0 } \\{ \text { and } D _ { 1 } \text { , } D _ { 2 } \text { are the respective } } { \text { discriminants of these equations, } } \\ { \text { then } D _ { 1 } : D _ { 2 } = } \end{array}$

0%
$\cfrac{a ^ { 2 } }{ p ^ { 2 }}$
0%
$\cfrac{b ^ { 2 }} { q ^ { 2 }}$
0%
$\cfrac{c ^ { 2 } }{r ^ { 2 }}$
0%
$1$

Explanation

Let

$A=\alpha+h$ and

$B=\beta+h$

Then,

$A-B=\left(\alpha+h\right)-\left(\beta+h\right)$

$\Rightarrow\,A-B=\alpha-\beta$

$\Rightarrow\,{\left(A-B\right)}^{2}={\left(\alpha-\beta\right)}^{2}$

$\Rightarrow\,{\left(A+B\right)}^{2}-4AB={\left(\alpha+\beta\right)}^{2}-4\alpha\beta$

$\Rightarrow\,\dfrac{{b}^{2}}{{a}^{2}}-\dfrac{4c}{a}=\dfrac{{q}^{2}}{{p}^{2}}-\dfrac{4r}{p}$

$\Rightarrow\,\dfrac{{b}^{2}-4ac}{{a}^{2}}=\dfrac{{q}^{2}-4rp}{{p}^{2}}$

$\Rightarrow\,\dfrac{{D}_{1}}{{a}^{2}}=\dfrac{{D}_{2}}{{p}^{2}}$

$\Rightarrow\,\dfrac{{D}_{1}}{{D}_{2}}=\dfrac{{a}^{2}}{{p}^{2}}$

Hence

${D}_{1}:{D}_{2}={a}^{2}:{p}^{2}=\dfrac{{a}^{2}}{{p}^{2}}$

Find the root of the quadratic equation

${x^2} + 2\sqrt {2x} + 6 = 0$ by using the quadratic formula

0%
$x = \sqrt 2 \pm 2i$
0%
$x = - \sqrt 2 \pm 2i$
0%
$x = - \sqrt 4 \pm 2i$
0%
$x = - \sqrt 2 \pm 4i$

Explanation

According to the given quadratic equation,

$a=1,b=2\sqrt2$ and

$c=6$

So, by quadratic formula,

$x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}=\dfrac{-2\sqrt2\pm\sqrt{(2\sqrt2)^2-4\times1\times6}}{2\times1}$

$=\dfrac{-2\sqrt2\pm\sqrt{8-24}}{2}=\dfrac{-2\sqrt2\pm\sqrt{-16}}{2}=\dfrac{-2\sqrt2\pm4i}{2}$

Thus,

$x=-\sqrt2\pm2i$

Let f (x) =

$x^{2}+\lambda x+\mu cosx,\lambda$ is +ve integer

$\mu$ is a real number. The number of ordered pairs (

$\lambda ,\mu$ ) for which f (x) = 0 and f (f(c)) = 0 have same set of real roots.

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

$\dfrac{\sqrt{1296}}{x}=\dfrac{x}{2.25}$ then value

$x$ is:

0%
$0.9$
0%
$0.09$
0%
$9$
0%
$1.9$

Explanation

we know that,

$\sqrt {1296} = 36$

$\dfrac {36}{x} = \dfrac{x}{2.25}$

$x ^ 2 = 36 \times 2.25$

$x ^ 2 = 81$

$x = \pm \sqrt {81}$

$x =\pm 9$

Find the value of

$x:$

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

0%
$2$
0%
$1$
0%
$0$
0%
None of the above

Explanation

Given equation is:

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

It can also be written as:

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

Above equation is of the form of following identity:

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

Hence, the equation

$(1)$ can be written as follows:

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

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

$x=2,2$

$k> 0$ and the product of roots of the equations

$x^{2}-3kx+2e^{\log k}-1=0$ is

$7$ , then find the sum of the roots-

0%
$6$
0%
$4$
0%
$3$
0%
$-12$

Explanation

Given:

${ x^{ 2 } }-3kx+2.{ e^{ 2\log k } }-1=0$

$\Rightarrow { x^{ 2 } }-3kx+2{ e^{ \log { k^{ 2 } } } }-1=0$

$\Rightarrow { x^{ 2 } }-3kx+\left( { 2{ k^{ 2 } }-1 } \right) =0$

Let

$\alpha$ and

$\beta$ be the roots

Hence

$\alpha +\beta =3k$

and

$\alpha \beta =2{ k^{ 2 } }-1=7$

$Rightarrow 2{ k^{ 2 } }=8=4$

$\Rightarrow k=\pm 2$

$\Rightarrow k=2$ So,

$k>0$

Now,

$\alpha +\beta =3k$

$=3\left( 2 \right) =6$

$x ^ { 2 } - 5 x + 1 = 0$ then

$\frac { x ^ { 10 } + 1 } { x ^ { 5 } }$ is equal to:-

0%
2424
0%
3232
0%
2525
0%
None of these

$(x+1)^2+(y-1)^2 +(x-5)^2+(y-1)^2$ has the value equal to

0%
$16$
0%
$25$
0%
$36$
0%
$49$

The ratio of the roots of the equation

${ ax }^{ 2 }+bx+c=0$ is same as the ratio of the roots of equation

${ px }^{ 2 }+qx+r=0$ . If

${ D }_{ 1 }$ and

${ D }_{ 2 }$ are the discriminants of

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

${ px }^{ 2 }+qx+r=0$ respectively, then

${ D }_{ 1 }:{ D }_{ 2 }=$

0%
$\dfrac { { b }^{ 2 } }{ { q }^{ 2 } }$
0%
$\dfrac { b }{ q }$
0%
$\dfrac { b }{ { q }^{ 2 } }$
0%
none of these

Explanation

$\textbf{Step 1: Using quadratic formula,}$

$\boldsymbol{x=\dfrac{-b\pm\sqrt{D}}{2a}}$

$ax^2+bx+c=0$

$x=\dfrac{-b\pm\sqrt{D_1}}{2a}$

$[\text{ Using Quadratic Formula }]$

$px^2+qx+r=0$

$x=\dfrac{-q\pm\sqrt{D_2}}{2a}$

$[\text{ Using Quadratic Formula }]$

$\text{As per given question, We have }$

$\dfrac{-b+\sqrt{D_1}}{-b-\sqrt{D_1}}=\dfrac{-q+\sqrt{D_2}}{-q-\sqrt{D_2}}$

$\textbf{Step 2: Using Componendo and Dividendo, we get}$

$\dfrac{-\sqrt{D_1}}{b}=\dfrac{-\sqrt{D_2}}{q}$

$\text{On squaring both side}$

$\dfrac{D_1}{b^2}=\dfrac{D_2}{q^2}$

$\Rightarrow \dfrac{D_1}{D_2}=\dfrac{b^2}{q^2}$

Which of the following is a quadratic equation ?

0%
$6x^2 = 20 -x^3$
0%
$x^2-\left ( \dfrac{1}{x^2} \right )$ = $\dfrac{7}{2}$
0%
$\dfrac{3}{x}$ $= 4$ $x^2$
0%
$5x^2 + 7 = 3x$

Explanation

For

$(A)$

$6x^2=20-x^3$

$\Rightarrow x^3+6x^2-20=0$

In this, the maximum power of

$x$ is

$3$ .

So, it is not a quadratic equation.

For

$(B)$

$x^2-\left(\dfrac{1}{x^2}\right)=\dfrac{7}{2}$

$\Rightarrow 2x^4-2=7x^2$

$\Rightarrow 2x^4-7x^2-2=0$

In this, the maximum power of

$x$ is

$4$ .

So, it is not a quadratic equation.

For

$(C)$

$\dfrac{3}{x}=4x^2$

$\Rightarrow 3=4x^3$

In this, the maximum power of

$x$ is

$3$ .

So, it is not a quadratic equation.

For

$(D)$

$5x^2+7=3x$

$\Rightarrow 5x^2-3x+7=0$

In this, the maximum power of

$x$ is

$2$ .

So, it is a quadratic equation.

Hence,

$Op-D$ is correct.

Solve the quadratic equation by completing the square method:

${x}^{2}+8x-9=0$

0%
$-9,-1$
0%
$2,1$
0%
$-9,1$
0%
$9,1$

Explanation

$x^2+8x-9=0$

$\Rightarrow x^2+2\times \left( 4\right) \times x +16-25=0$

$\left( x+4\right)^2=25$

$\left( x+4\right)^2=5^2$

$x+4=\pm 5$

$\Rightarrow x=1$ and

$-9$

Hence, the answer is

$1,-9.$

Discriminant is ................. for the equation

$5x-6=-\frac{1}{x}$

0%
-56
0%
16
0%
-16
0%
0

$(1+m^2)x^2-2(1+3m)x+(1+8m)=0$ . Then numbers of value(s) of m for which this equation has no solution.

0%
$3$
0%
infinitely many
0%
$2$
0%
$1$

Explanation

We know that

$D=b^2-4ac$

$D=4(1+3m)^2-4(1+m^2)(1+8m)$

$=4(1+9m^2+6m-(1+8m+m^2+8m^3))$

$=4(8m^2-2m-8m^3)$

$=-8(4m^3-4m^2+m)$

$=-8m(4m^2-4m+1)$

$=-8m(2m-1)^2 < 0$ .........

$\because (2m-1)^2 \, is \, always +ve$

$=m>0$

Hence infinitely many values.

Quadratic equation among the following is:

0%
${ x }^{ 2 }+1={ x }^{ 2 }+2x$
0%
$x\left( x+1 \right) =x\left( x-3 \right)$
0%
${ x }^{ 2 }-2x=3\left( 5x+2 \right)$
0%
${ x }^{ 2 }+2x+3={ x }^{ 2 }+5x$

For what value of

$c$ , the root of

$(c-2) x^{2}+2(c-2) x+2=0$ are not real

0%
$( 1,2)$
0%
$( 2,3)$
0%
$( 3,4)$
0%
$(2,4)$

Explanation

Given the quadratic equation is

$(c-2) x^{2}+2(c-2) x+2=0$ .

Now the roots of the equation will be not real if the discriminant of the equation is less than

$0$ .

Then,

$\{2(c-2)\}^2-4.2.(c-2)<0$

or,

$4\{(c-2)(c-4)\}<0$

or,

$(c-2)(c-4)<0$

or,

$2<c<4$ .

$c\in (2,4)$ .

Which constant must be added and subtracted to solve the quadratic equation

$9x^2 + \dfrac{3}{4}x - \sqrt{2} = 0$ by the method of completing the square ?

0%
$\dfrac{1}{8}$
0%
$\dfrac{1}{64}$
0%
$\dfrac{1}{4}$
0%
$\dfrac{9}{64}$

Explanation

Given equation is

$9x^2 + \dfrac{3}{4}x - \sqrt{2} = 0$

$\Rightarrow (3x)^2 + 2\left(\dfrac{1}{8}\right)(3x) - \sqrt{2} = 0$

Hence, to make it a perfect square we must add and subtract

$\left(\dfrac{1}{8}\right)^2=\dfrac1{64}$

$\alpha$ and

$\beta$ are roots of the equation

$x^2 + x\sin \theta - 2 \sin \theta = 0$ then

$\dfrac{\alpha^{12} + \beta^{12}}{(\alpha^{-12} + \beta^{-12})(\alpha - \beta)^{24}}$ is equal to

0%
$\dfrac{2^{24}}{(8 + \sin \theta)^{12}}$
0%
$\dfrac{2^{12}}{(8 + \sin \theta)^{12}}$
0%
$\dfrac{2^{12}}{(8 - \sin \theta)^{12}}$
0%
$\dfrac{1}{2^{24}}$

Explanation

$x^2 + x\sin \theta - 2 \sin \theta = 0$ , has roots

$\alpha$ and

$\beta$

$\dfrac{\alpha^{12}+\beta^{12}}{(\alpha^{-12}+\beta^{-12})(\alpha - \beta)^{24}} = \dfrac{\alpha^{12} . \beta^{12}}{(\alpha - \beta)^{24}} = \dfrac{(\alpha \beta)^{12}}{(\alpha - \beta)^{24}}$

$= \dfrac{(-2\sin \theta)^{12}}{(\sqrt{\sin^2 \theta + 8 \sin \theta})^{24}} = \dfrac{2^{12}}{(8 + \sin \theta)^{12}}$

If the discriminant of

$3 x ^ { 2 } - 2 x + K = 0$ is zero then

$\mathrm { K } =$ _____

0%
$3$
0%
$\dfrac { 1 } { 3 }$
0%
$-3$
0%
$\dfrac { -1 } { 3 }$

Which constant must be added and subtracted to solve the Quadratic equation

$9 x ^ { 2 } + \dfrac { 3 } { 4 } x - \sqrt { 2 }$ by the method of completion the square.

0%
$\dfrac { 1 } { 8 }$
0%
$\dfrac { 1 } { 4 }$
0%
$\dfrac { 1 } { 64 }$
0%
$\dfrac { 9 } { 64 }$

CBSE Questions for Class 10 Maths Quadratic Equations Quiz 9 - MCQExams.com

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Practice Class 10 Maths Quiz Questions and Answers

CBSE Questions for Class 10 Maths Quadratic Equations Quiz 9 - MCQExams.com

0:0:2

Practice Class 10 Maths Quiz Questions and Answers

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