MCQ Questions for Class 10 Maths Quadratic Equations Quiz 7

Value(s) of

$x$ which satisfies the equation

$x^{2} + 2kx = \dfrac {j}{3}$ , where

$j$ and

$k$ are constants, is/are

0%
$x = -k\pm \dfrac {\sqrt {3(3k^{2} + j)}}{3}$
0%
$x = -6k\pm \dfrac {\sqrt {3(3k^{2} + j)}}{3}$
0%
$x = -k\pm \dfrac {\sqrt {3(3k^{2} + j)}}{6}$
0%
$x = -6k\pm \left (k + \dfrac {\sqrt {3j}}{3}\right )$

Explanation

We know the quadratic equation formula,

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

$x^2+2kx=\dfrac{j}{3}$
On cross multiplying, we get

$3x^2+6kx-j=0$
Here

$a = 3, b = 6k, c = -j$
On substituting the values, we get

$=$

$\dfrac{-6k\pm\sqrt{(6k)^2-4\times 3\times -j}}{2\times 3}$

$=$

$\dfrac{-6k\pm\sqrt{36k^2+12 j}}{6}$

$=$

$\dfrac{-6k\pm\sqrt{12(3k^2+j)}}{6}$

$=$

$\dfrac{-6k\pm4\sqrt{3(3k^2+j)}}{6}$
Take

$2$ as common, we get

$=$

$-3k\pm\dfrac{\sqrt{3(3k^2+j)}}{3}$
Therefore,

$x =-k\pm \dfrac{\sqrt{3(3k^2+j)}}{3}$

What are the real factors of

$x^2 + 4$ ?

0%
$(x^2+2)$ and $(x^2-2)$
0%
$(x+2)$ and $(x-2)n$
0%
Does not exist
0%
$(x^2 + 2)$ and $(x + 2)$

Explanation

Real factors of the equation does not exist as it doesn't have real roots

$x^{2}=-4$

$\Rightarrow x=+2i$ or

$- 2i$

Which of the following is an approximate of a zero of the equation

$x^2 - 3x = 7$ ?

0%
-4.54
0%
-1.54
0%
1.54
0%
3.54
0%
5.54

Explanation

$x^2-3x-7=0$

$x=\dfrac{3\pm \sqrt{9+28}}{2}$ (using

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

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

$=\dfrac{3\pm 6.08}{2}$

$x=4.54$ &

$x=-1.54$

$x=4.54$ not in options

$\therefore x=-1.54$ (Approximate zero).

If A and B are whole numbers such that

$9A^{2} = 12A + 96$ and

$B^{2} = 2B + 3$ , find the value of

$5A + 7B$ .

0%
$31$
0%
$37$
0%
$41$
0%
$43$

Explanation

A,B are whole numbers

$9{ A }^{ 2 }=12A+96$ &

${ B }^{ 2 }=2B+3$

$9{ A }^{ 2 }-12A-96=0$ &

${ B }^{ 2 }-2B-3=0$

$3{ A }^{ 2 }-4A-32=0$

$3{ A }^{ 2 }-12A+8A-32=0$ &

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

$(3A+8)(A-4)=0$ &

$(B-3)(B+1)=0$

$\therefore A=4,B=3$ (

$\because A,B$ are whole numbers

$\because (A\neq \cfrac { -8 }{ 3 })$ &

$(B\neq -1$ )

$5A+7B=5(4)+7(3)$

$=41$

Given that '

$x$ ' is real then the solution set of the equation

$\sqrt { x-1 } +\sqrt { x+1 } =1$ .

0%
No solution
0%
Unique solution
0%
$2$ solutions
0%
None of these

Explanation

$\sqrt{x-1}+\sqrt{x+1}=1$ As

$x > 1$

$\because x-1$ can't be negative.

$\Rightarrow \sqrt{x+1} > 1$

$0\sqrt{x-1}$ is possitve

$\therefore$ there sum is

$> 1$ always . hence no solution.

Let

$a$ be the solution of the equation

$4x^2-12x+9=16$ , then the value of

$10a-15$ is

0%
20
0%
25
0%
30
0%
35

Explanation

$4x^2 - 12x + 9 = 16$

$\Rightarrow 4x^2 - 12x - 7 = 0$

$\Rightarrow x = \cfrac{12 \pm \sqrt{144 + 112}}{8}$

$\Rightarrow x = \cfrac{12 \pm 16}{8}$

$\Rightarrow x = -0.5, 3.5$

Substituting the value of a in

$10a - 15$ , we get

$35 - 15 = 20$ and

$5 - 15 = -10$

$x = 7 + 4\sqrt {3}$ and

$xy = 1$ , what is the value of

$\dfrac {1}{x^{2}} + \dfrac {1}{y^{2}}$ ?

0%
$64$
0%
$134$
0%
$194$
0%
$\dfrac {1}{49}$

Explanation

$x=7+4\sqrt{3}$ &

$xy=1$

$\Rightarrow y=\dfrac{1}{7+4\sqrt{3}}=\dfrac{7-4\sqrt{3}}{49-48}=7-4\sqrt{3}$

$\dfrac{1}{x^{2}}+\dfrac{1}{y^{2}}=(\dfrac{1}{x}+\dfrac{1}{y})^{2}-2 \Rightarrow \dfrac{1}{x^{2}}+\dfrac{1}{y^{2}}=(\dfrac{1}{7+4\sqrt{3}}+\dfrac{1}{7-4\sqrt{3}})^{2}-2$

$=(\dfrac{14}{49-48})^{2}-2=196-2$

$=\boxed{194}$

The discriminant (D) of

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

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

Explanation

Discrimant (D) is applied only to quadratic equations, but the equation

$\sqrt{x^{2}+x+1} = 2$ is not an quadratic equation,

Hence, first convert the equation into quadratic by squaring it on both sides,

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

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

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

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

$D = 1-4\times 1\times (-3)$

$\boxed{D = 13}$

$f(x)=x^2+4x+a$ is a perfect square function then calculate the value of

$a$ .

0%
$3$
0%
$4$
0%
$2$
0%
$6$

Explanation

Since f(x) is perfect square, then both the roots are equal.

$\therefore$ D=0

$16-4a=0$

$a=4$

Identify which of the following is/are a quadratic polynomial function:

0%
$f(x)=(x+1)^3-(x+2)^3$
0%

$g(x)=\dfrac{x^4}{x^2}$
0%
$h(x)=(x+1)^2-(x+2)^2$
0%
All of these

Explanation

Option 1)

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

$=x^3+1+3x(x+1)-{x^3+8+6x(x+2)}$

$=x^3+1+3x^2+3x-x^3-8-6x^2-12x$

The power of

$x^3$ cancels out and results in to quadratic equation.

Option 2) g(x) =

$x^2$

Option 3)

$x^2+1+2x-x^2-4-8x$

The power of

$x^2$ cancels out which doesn't result in to quadratic equation.

The roots of the quadratic equation

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

0%
$(a+b+c)$ and $(a-b-c)$
0%
$\dfrac 12$ and $a-2b+c$
0%
$a-2b+c$ and $\dfrac 1{(a+b-2c)}$
0%
None of the above.

Explanation

Clearly we see that x=1 satisfies the equation

$\Rightarrow$ x=1 is a root of the that equation

Let

$\alpha$ be other root

Product of roots =

$1×\alpha =\dfrac{a-2b+c}{a+b-2c}$

other root

$=\alpha =\dfrac {a-2b+c}{a+b-2c}$

Sum of the roots of the equation

$(x+3)^2-4|x+3|+3=0$ is-

0%
$4$
0%
$12$
0%
$-12$
0%
$-4$

Explanation

${ (x+3) }^{ 2 }-4|x+3|+3=0\\ let\quad |x+3|=t\\ \therefore t\ge 0\\ { t }^{ 2 }-4t+3=0\\ (t-1)(t-3)=0\\ \therefore t=1\ or\ 3\\ when\quad t=1\\ |x+3|=1\\ x+3=1\quad or\quad x+3=-1\\ x=-2\quad or\quad x=-4\\ when\quad t=3\\ |x+3|=3\quad or\quad x+3=-3\\ x=0\quad or\quad x=-6\\ sum\quad of\quad all\quad roots\quad =\quad (-2)+(-4)+(0)+(-6)\\ =-12$

The roots of the equation ,

$(x^2+1)^2=x(3x^2+4x+3)$ , are given by-

0%
$2-\sqrt 3$
0%
$\dfrac {-1+i\sqrt 3}2$
0%
$2+\sqrt 3$
0%
$\dfrac {-1-i\sqrt 3}2$

Explanation

${ ({ x }^{ 2 }+1) }^{ 2 }=x(3{ x }^{ 2 }+4x+3)\\ { x }^{ 4 }+2{ x }^{ 2 }+1=3{ x }^{ 3 }+4{ x }^{ 2 }+3x\\ { x }^{ 4 }-{ 3x }^{ 3 }-2{ x }^{ 2 }-3x+1=0\\ \therefore ({ x }^{ 2 }-4x+1=0)({ x }^{ 2 }+x+1)=0\\ { x }^{ 2 }-4x+1=0\quad or\quad { x }^{ 2 }+x+1=0\\ \therefore x=\cfrac { 4\pm \sqrt { 16-4 } }{ 2 } \quad \\ x=\cfrac { 4\pm \sqrt [ 2 ]{ 3 } }{ 2 } \\ x=2\pm \sqrt { 3 } \\ \therefore x=(2+\sqrt { 3 } )\ or\ (2-\sqrt { 3 } )\\ x=\cfrac { -1\pm \sqrt { 1-4 } }{ 2 } \\ x=\cfrac { -1\pm i\sqrt { 3 } }{ 2 } \\ x=\cfrac { -1+i\sqrt { 3 } }{ 2 }\ or\ \cfrac { -1-i\sqrt { 3 } }{ 2 }$

$\alpha,\beta$ are roots of the equation

$2x^2-35x+2=0$ then the value of

$(2\alpha-35)^3(2\beta-35)^3$ is:

0%
$1$
0%
$8$
0%
$4$
0%
$64$

Explanation

We will substitute

$\alpha, \beta$ in the above equation

$\Rightarrow 2\alpha - 35+\dfrac {2}{\alpha} =0$

$\Rightarrow 2\alpha - 35=\dfrac {-2}{\alpha}$

Similarly for beta

$\Rightarrow (2\alpha - 35)^{3}(2\beta - 35)^{3}=\dfrac {-8}{\alpha ^{3}}×\dfrac {-8}{\beta^{3}}=\dfrac {64}{1}=64$

$A$ and

$B$ solve an equation

$x^2+px+q=0$ . In solving

$A$ commits a mistake in reading

$p$ and finds the root

$2$ and

$6$ and

$B$ commits a mistake in reading

$q$ and finds the roots

$2$ and

$-9$ . Find the correct roots.

0%
$2$
0%
$-3$
0%
$-4$
0%
$-2$

Explanation

${ x }^{ 2 }+px+q=0$

As A copied q correct so he will get product of roots correct

As B copied P correct so he will get sum of roots correct

$sum\quad =\quad 2-9=-7\\ product\quad =\quad 2\times6=12\\ \therefore equation\\ { x }^{ 2 }+7x+12=0\\ (x+4)(x+3)=0\\ \therefore x=-3\quad or\quad -4$

The minimum value of the expression

$4x^2+2x+1$ is-

0%
$1$
0%
$\dfrac{1}{3}$
0%
$\dfrac{1}{2}$
0%
$\dfrac{3}{4}$

Explanation

For a general quadratic equation

${ ax }^{ 2 }+bx+c=0,\quad where\quad a>0\quad the\quad minimum\quad value\quad of\quad expression\quad occurs\quad at\quad \cfrac { -D }{ 4a } \\ \therefore { 4x }^{ 2 }+2x+1\\ minimum\quad value\quad =\quad \cfrac { -({ 2 }^{ 2 }-4.4.1) }{ 4.4 } \\ =\cfrac { 12 }{ 16 } =\cfrac { 3 }{ 4 }$

The number of values of

$a$ for which

$(a^2-3a+2)x^2+(a^2-5a+6)x+a^2-4=0$ is an identity in

$x$ is-

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

Explanation

Equation is a identity if all its coefficients are equal to zero simultaneously

$\Rightarrow a^{2}-3a+2=0$

$\Rightarrow a=2,1$

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

$\Rightarrow a^{2}-4=0$

$\Rightarrow a=\pm 2$

$\Rightarrow$ coefficients are equal to zero simultaneously when

$a=2$

Therefore only one value of 'a' the equation is identity in

$x$

$b\in F'$ then the roots of the equation

$\left( 2+b \right) { x }^{ 2 }+(3+b)x+(4+b)=0\quad$ is

0%
real and imaginary
0%
real and equal
0%
imagenary
0%
cannot predicted

Explanation

$(2+b)x^{2}+(3+b)x+(4+b)=0$

$D=(3+b)^{2}-4(2+b)(4+b)=b^{2}+6{b}+9-4{b^{2}}-24{b}-32=-3{b^{2}}-18{b}-23=-3(b^{2}+6{b}+9)+4$

$=-3(b+3)^{2}+9<0$

So roots are imaginary

The number of integral value of k such that the given quadratic equation has imaginary roots are?

0%
$0$
0%
$\sqrt{2/3}$
0%
$2$
0%
$3$

$a, b,c$ are distinct and the roots of

$(b-c)x^{2} +(c-a)x +(a-b)=0$ are equal, then

$a,b,c$ are in:

0%
Arithmetic Progression
0%
Geometric progression
0%
Harmonic progression
0%
Arithmetico-Geometric progression

Explanation

$f\left(x\right)=\left(b-c\right)x^{2}+\left(c-a\right)x+\left(a-b\right)=0$

Given that the roots are equal.

$\Rightarrow\Delta=0$

Now,

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

$\Rightarrow\left(c-a\right)^{2}-4\left(b-c\right)\left(a-b\right)=0$

$\Rightarrow{c}^{2}-2ac+a^{2}+4b^{2}+4ac-4bc-4ab=0$

$\Rightarrow\left(a-2b+c\right)^{2}=0$

$\Rightarrow\left(a-2b+c\right)=0$

$\Rightarrow2b=a+c$

$\Rightarrow{a},b, c$ are in A.P.

The discriminant value of equation

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

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

Explanation

The given equation is

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

$(i)$

General form of quadratic equation is given by

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

$(ii)$

Comparing equation

$(i)$ with

$(ii)$ we have,

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

The discriminant

$D$ is given by

$D={b}^{2}-4ac={(-6)}^{2}-4(5)(1)=36-20=16$

$\therefore$ Discriminant value is

$D=16$

............... is true for discriminate of quadratic equation

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

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

Explanation

The polynomial is

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

Comparing with

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

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

$D = b^2 - 4ac = 1 - 4(1) (1) = -3 < 0$

Solve the following quadratic equation by completing the square:

$\dfrac{5x+7}{x-1}=3x+2$

0%
$\left \{ -1, 3 \right \}$
0%
$\left \{ 1, 3 \right \}$
0%
$\left \{ -1, -3 \right \}$
0%
None of these

Explanation

Given

$\dfrac{5x+7}{x-1}=3x+2$

$\Rightarrow$

$5x+7=(3x+2)(x-1)$

$\Rightarrow$

$5x+7=3x^2-3x+2x-2$

$\Rightarrow$

$3x^2-6x-9=0$

$\Rightarrow$

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

$\Rightarrow$ Here,

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

$\Rightarrow$

$x^2-2x=3$

Now,

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

Adding

$1$ on both sides,

$\Rightarrow$

$x^2-2x+1=3+1$

$\Rightarrow$

$(x-1)^2=4$

Taking square root on both sides

$\Rightarrow$

$x-1=\pm2$

$\Rightarrow$

$x-1=2$ and

$x-1=-2$

$\therefore$

$x=3$ and

$x=-1$

Let

$\alpha$ be the root of the equation

$25\cos^{2}\theta + 5\cos \theta - 12 = C$ , where

$\dfrac {\pi}{2} < \alpha < \pi$ .

What is

$\tan \alpha$ equal to?

0%
$-\dfrac {3}{4}$
0%
$\dfrac {3}{4}$
0%
$-\dfrac {4}{3}$
0%
$-\dfrac {4}{5}$

Explanation

Given equation is

$25\cos^2\theta+5\cos\theta-12=0$

$\Rightarrow 25\cos^2\theta+20\cos\theta-15\cos\theta-12=0$

$\Rightarrow 5\cos\theta(5\cos\theta+4)-3(5\cos\theta+4)=0$

$\Rightarrow (5\cos\theta-3)(5\cos\theta+4)=0$

So,

$\cos\theta=\dfrac{3}{5}$ or

$\cos\theta=\dfrac{-4}{5}$

Since

$\alpha$ is the root of the equation and it lies between

$\dfrac{\pi}{2}<\alpha<\pi\Rightarrow \cos \alpha\$ is '-ve'.

Thus

$\alpha=\cos^{-1}\left (\dfrac{-4}{5}\right)$

Since it lies in the range

$\dfrac{\pi}{2}<\alpha<\pi$ , so

$\tan \alpha$ is '-ve'

$\tan\alpha=\tan\left [\cos^{-1}\left (\dfrac{-4}{5}\right)\right]$

$=\tan \left [\tan^{-1}\left (-\dfrac{3}{4}\right)\right]$

$=-\dfrac{3}{4}$

If the roots of the equation

$x^{2} + px + c = 0$ are

$(2, -2)$ and the roots of the equation

$x^{2} + bx + q = 0$ are

$(-1, -2)$ , then the roots of the equation

$x^{2} + bx + c = 0$ are

0%
$-3, -2$
0%
$-3, 2$
0%
$1, -4$
0%
$-5, 1$

Explanation

Given, roots of the equation

$x^{2} + px + c = 0$ are

$2$ and

$-2$ .

$\therefore -p = 2 - 2 \Rightarrow p = 0$

and

$(2)\cdot (-2) = c\Rightarrow c = -4$

Again, roots of the equation

$x^{2} + bx + q = 0$ are

$-1$ and

$-2$ .

$\therefore -b = -1 -1 \Rightarrow b = 3$

and

$(1-) \cdot (-2) = q\Rightarrow q = 2$

$\therefore x^{2} + bc + c \equiv x^{3} + 3x - 4 =0$

$\Rightarrow (x - 1)(x + 4) = 0$

So, the required roots are

$1$ and

$-4$ .

Sum of the roots of the equation

${ \left| x-3 \right| }^{ 2 }+\left| x-3 \right| -2=0$ is

0%
$2$
0%
$4$
0%
$6$
0%
$16$
0%
$-2$

Explanation

Given equation is

${ \left| x-3 \right| }^{ 2 }+\left| x-3 \right| -2=0$

$\Rightarrow { \left| x-3 \right| }^{ 2 }+2\left| x-3 \right| -\left| x-3 \right| -2=0$

$\Rightarrow \left( \left| x-3 \right| +2 \right) \left( \left| x-3 \right| -1 \right) =0$

Therefore,

$\left| x-3 \right| =-2,1$

But

$\left| x-3 \right| \neq -2$

Hence,

$\left| x-3 \right| =1$

$\Rightarrow x-3=\pm 1$

$\Rightarrow x=3\pm 1$

$\Rightarrow x=4, 2$

Now, sum of the roots

$=4+2=6$ .

The equation

${ e }^{ \sin { x } }-{ e }^{ -\sin { x } }-4=0$ has

0%
No solution
0%
Two solutions
0%
Three solutions
0%
None of these

Explanation

Given,

${ e }^{ \sin { x } }-{ e }^{ -\sin { x } }-4=0$

multiply throughout by

$e^{sinx}$

$\Rightarrow { e }^{ 2\sin { x } }-4{ e }^{ \sin { x } }-1=0$

this is quadratc equation in

$e^{sinx}$

converting from exponential to logarithmic form.

${ e }^{ \sin { x } }=\dfrac { 4\pm \sqrt { 16+4 } }{ 2 } =2\pm \sqrt { 5 }$

$\Rightarrow \sin { x } =\log { \left( 2+\sqrt { 5 } \right) }$ [

$\because \log { \left( 2-\sqrt { 5 } \right) }$ is not defined ]

Since,

$2+\sqrt { 5 } > e \Rightarrow \log { \left( 2+\sqrt { 5 } \right) } > 1$

$\Rightarrow \sin { x } > 1$ , which is not possible.
Hence, no solution exist.

The root of the equation

$2(1+i)x^2-4(2-i)x-5-3i=0$ , where

$i=\sqrt{-1}$ , which has eater modulus is

0%
$(3-5i)/2$
0%
$(5-3i)/2$
0%
$(3+i)/2$
0%
$(1+3i)/2$

Which of the following equations, is not a quadratic equation?

0%
$4x^{2} - 7x + 3 = 0$
0%
$3x^{2} - 4x + 1 = 0$
0%
$2x - 7 = 0$
0%
$4x^{2} - 3 = 0$

Explanation

Alternate

$(A) : 4x^{2} - 7x + 3 = 0$
The maximum exponent of variable

$x$ is

$2$ . So it is a quadratic equation.
Alternate

$(B) : 3x^{2} - 4x + 1 = 0$
The maximum exponent of variable

$x$ is

$2$ . So it is a quadratic equation.
Alternate

$(C) : 2x - 7 = 0$
The maximum exponent of variable

$x$ is

$1$ . So it is not a quadratic equation.

The discriminant of quadratic equation

$3x^{2} - 4x - 1 = 0$ is _______.

0%
$0$
0%
$4$
0%
$12$
0%
$28$

Explanation

Quadratic equation:

$3x^{2} - 4x - 1 = 0$

Comparing the given equation with

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

we have

$a = 3, b = -4, c = -1$

Now, discriminant

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

$= (-4)^{2} - 4(3) (-1)$

$= 16 + 12$

$= 28$ .

CBSE Questions for Class 10 Maths Quadratic Equations Quiz 7 - MCQExams.com

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Practice Class 10 Maths Quiz Questions and Answers

CBSE Questions for Class 10 Maths Quadratic Equations Quiz 7 - MCQExams.com

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Practice Class 10 Maths Quiz Questions and Answers

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