MCQ Questions for Class 10 Maths Quadratic Equations Quiz 10

Which one of the following is the quadratic equation?

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

The integral values of $$ m $$ for which the roots of the equation $$ m x^{2}+(2 m-1) x+(m-2)=0 $$ are rational are given by the expression [where $$ n$$ is integer ]

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$${n}^{2}$$
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$$n(n+2)$$
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$$n(n+1)$$
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none of these

Which among the below are quadratic equations?

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$$\frac{5} {x} - 3 = x^{2}$$
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$$x \left(x + 5\right) = 2$$
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$$n - 1 = 2n$$
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$$\dfrac{1} {x^{2}} \left(x + 2\right) = 2$$

If $$ax^2 + bx + c = 0$$ then quadratic formula is

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

Discriminant of quadratic equation $$3\sqrt 3x^2+10x+\sqrt 3=0$$

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$$10$$
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$$64$$
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$$46$$
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$$30$$

Let $$a, b, c$$ be the lengths of sides of a scalene triangle. If the roots of the equation $$\displaystyle x^{2}+2(a+b+c)x+3\lambda (ab+bc+ca)=0$$ $$(\lambda \in R)$$ are real then

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$$\displaystyle \lambda <\dfrac{4}{3}$$
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$$\displaystyle \lambda >\frac{5}{3}$$
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$$\displaystyle \lambda \in \left ( \frac{1}{3}, \frac{5}{3} \right )$$
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$$\displaystyle \lambda \in \left ( \frac{4}{3}, \frac{5}{3} \right )$$

The equation $$\sqrt {2x-2x^2-5}=x^2-2x-3$$, where $$x$$ is real has

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No real solution
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Exactly one real solution
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Exactly two real solutions
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Exactly four real solutions

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Both Assertion and Reason are correct and Reason is the correct explanation for Assertion.
0%
Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion.
0%
Assertion is correct but Reason is incorrect.
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Assertion is incorrect but Reason is correct.

All solutions of the equation $$\displaystyle 4x^2 - 40x + 51 = 0$$ lie in the interval

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$$\displaystyle \left(\frac {23}{10}, \frac {83}{10}\right)$$
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$$\displaystyle \left(\frac {23}{10}, \frac {15}{2}\right)$$
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$$\displaystyle \left(7, \frac {83}{10}\right)$$
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none of these

The values of $$a$$ for which both the roots of the equation $$(a-6)x^2=a(x-3)$$ are positive and are given by

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$$(0,6)$$
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$$\displaystyle \left( 0,\frac { 72 }{ 11 } \right) $$
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$$\displaystyle \left( 6,\frac { 72 }{ 11 } \right) $$
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$$\displaystyle \left( \frac { 72 }{ 13 } ,6 \right) $$

The roots of $$(x-a)(x-c)+k(x-b)(x-d)=0$$ are real and distinct for all real $$k$$ if

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$$a < b < c < d$$
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$$a > b < c < d$$
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$$a < b > c < d$$
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$$a < b = c < d$$

The quadratic equation $$\displaystyle x^{2}-2x-\lambda=0,\lambda\neq 0,$$

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cannot have a real root if $$\lambda < -1$$
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can have a rational root if $$\lambda$$ is a perfect square
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cannot have an integral root if $$n^{2}-1< \lambda< n^{2}+2n$$ where $$n=0,1,2,3,.....$$
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none of these

If $$1$$ is a root of the equations $$ay^2$$ + $$ay + 3 = 0$$ and $$y^2$$+ $$y + b = 0$$, then find the value of $$ab$$

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$$3$$
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$$4$$
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$$5$$
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$$6$$

If $${ x }^{ 2 }-\left( a+b+c \right) x+\left( ab+bc+ca \right) =0$$ has non real roots, where $$a,b,c\in { R }^{ + }$$, then $$\sqrt { a } ,\sqrt { b } ,\sqrt { c } $$

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Can be the sides of a triangle
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Cannot be the sides of triangle
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Nothing can be said
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None of these

The value of $$x^2-6x+13$$ can never be less than

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$$4$$
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$$5$$
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$$4.5$$
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$$7$$

The number of values of $$\displaystyle k$$ for which $$\displaystyle \left ( x^{2} - \left ( k - 2 \right )x + k^{2} \right ) \left ( x^{2} + kx + \left ( 2k - 1 \right ) \right )$$ is a perfect square

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$$\displaystyle 1$$
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$$\displaystyle 2$$
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$$\displaystyle 3$$
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$$\displaystyle 0$$

Find the root(s) of the equation $$y+\sqrt{y+5}=7$$

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11
0%
4
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4 and 11
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$$\pm 4$$
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none of these

If $$\alpha, \beta$$ are the roots of $$\displaystyle ax^{2}+2bx+c=0$$ and that of $$\displaystyle Ax^{2}+2Bx+C=0$$ be $$\displaystyle \alpha+\delta, \beta +\delta $$ then the value of $$\displaystyle \frac{b^{2}-ac}{B^{2}-AC} $$ is

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$$\displaystyle \left ( \frac{a}{A} \right )^{2}$$
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$$\displaystyle \left ( \frac{A}{a} \right )^{2}$$
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$$0$$
0%
$$1$$

Explanation

$$\Rightarrow$$ $$\alpha$$ and $$\beta$$ are the roots of quadratic equation $$ax^2+2bx+c=0$$.

$$\Rightarrow$$ $$\alpha\beta=\dfrac{c}{a}$$ ----- ( 1 )

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

$$\Rightarrow$$ $$(\alpha+\beta)^2=\alpha^2+\beta^2+2\alpha\beta$$

$$\Rightarrow$$ $$\left(\dfrac{-2b}{a}\right)^2=\alpha^2+\beta^2+2\times\dfrac{c}{a}$$

$$\therefore$$ $$\alpha^2+\beta^2=\dfrac{4b^2}{a^2}-\dfrac{2c}{a}$$ ----- ( 2 )

$$\Rightarrow$$ $$(\alpha-\beta)^2=\alpha^2+\beta^2-2\alpha\beta$$

$$\Rightarrow$$ $$(\alpha-\beta)^2=\dfrac{4b^2}{a^2}-\dfrac{2c}{a}-\dfrac{2c}{a}$$ [ From ( 1 ) and ( 2 ) ]

$$\Rightarrow$$ $$(\alpha-\beta)^2=\dfrac{4b^2-4ac}{a^2}$$

$$\therefore$$ $$\alpha-\beta=\sqrt{\dfrac{4b^2-4ac}{a^2}}$$ ------ ( 3 )

$$\Rightarrow$$ Now, $$\alpha+\delta$$ and $$\beta+\delta$$ are roots of quadratic equation $$Ax^2+2Bx+C=0$$

$$\Rightarrow$$ $$(\alpha+\delta)(\beta+\delta)=\dfrac{C}{A}$$

$$\Rightarrow$$ $$\alpha\beta+\alpha\delta+\beta\delta+\delta^2=\dfrac{C}{A}$$

$$\therefore$$ $$\alpha\beta=\dfrac{C}{A}-\alpha\delta-\beta\delta-\delta^2$$ -------- ( 4 )

$$\Rightarrow$$ $$(\alpha+\delta)+(\beta+\delta)=\dfrac{-2B}{A}$$

$$\Rightarrow$$ $$\alpha+\beta+2\delta=\dfrac{-2B}{A}$$

$$\Rightarrow$$ $$\alpha+\beta=\dfrac{-2B}{A}-2\delta$$ -------- ( 5 )

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

$$\therefore$$ $$\alpha^2+\beta^2=\left(\dfrac{-2B}{A}-2\delta\right)^2-2\alpha\beta$$ -------- ( 6 )

$$\Rightarrow$$ $$[(\alpha+\delta)-(\beta+\delta)]^2=(\alpha-\beta)^2$$

$$\Rightarrow$$ $$(\alpha-\beta)^2=\alpha^2+\beta^2-2\alpha\beta$$

$$\Rightarrow$$ $$(\alpha-\beta)^2=\left(\dfrac{-2B}{A}-2\delta\right)^2-2\alpha\beta-2\alpha\beta$$ [ From ( 6 ) ]

$$\Rightarrow$$ $$(\alpha-\beta)^2=\left(\dfrac{-2B}{A}-2\delta\right)^2-4\alpha\beta$$

$$\Rightarrow$$ $$(\alpha-\beta)^2=\dfrac{4B^2}{A^2}+\dfrac{8B\delta}{A}+4\delta^2-\dfrac{4C}{A}+4\alpha\delta+4\beta\delta+4\delta^2$$ [ From ( 4 ) ]

$$\Rightarrow$$ $$(\alpha-\beta)^2=\dfrac{4B}{A^2}-\dfrac{4C}{A}+\dfrac{8B\delta}{A}+8\delta^2+4\delta(\alpha+\beta)$$

$$\Rightarrow$$ $$(\alpha-\beta)^2=\dfrac{4B^2-4AC}{A^2}+\dfrac{8B\delta}{A}+8\delta^2-\dfrac{8B\delta}{A}-8\delta^2$$ [ From ( 5 ) ]

$$\Rightarrow$$ $$(\alpha-\beta)^2=\dfrac{4B^2-4AC}{A^2}$$

$$\therefore$$ $$(\alpha-\beta)=\sqrt{\dfrac{4B^2-4AC}{A^2}}$$ ------- ( 7 )

$$\Rightarrow$$ $$\dfrac{\alpha-\beta}{\alpha-\beta}=\dfrac{\sqrt{\dfrac{4b^2-4ac}{a^2}}}{\sqrt{\dfrac{4B^2-4AC}{A^2}}}$$ [ Dividing ( 3 ) by ( 7 )]

$$\Rightarrow$$ $$1=\dfrac{\sqrt{\dfrac{4b^2-4ac}{a^2}}}{\sqrt{\dfrac{4B^2-4AC}{A^2}}}$$

$$\Rightarrow$$ $$\dfrac{4B^2-4AC}{A^2}=\dfrac{4b^2-4ac}{a^2}$$

$$\therefore$$ $$\dfrac{b^2-ac}{B^2-AC}=\dfrac{a^2}{A^2}=\left(\dfrac{a}{A}\right)^2$$

The roots of the equation $$\displaystyle \left ( x-a \right )\left ( x-b \right )+\left ( x-b \right )\left ( x-c \right )+\left ( x-c \right )\left ( x-a \right )=0$$ are

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positive
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negative
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real
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imaginary

If the quadratic equation $$\displaystyle x^{2}-mx-4x+1=0$$ has real and distinct roots, then the values of $$m$$ are
A. $$\displaystyle (-\infty , -6)$$
B. $$\displaystyle (-\infty,-3)$$

C. $$(-2,\infty )$$

D. $$\displaystyle \left ( 2,\infty \right )$$

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A or C
0%
A or D
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B or C
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B or D

Let $$\displaystyle f(x)= x^{2}-3x+4 $$, then the value of $$x$$ which satisfies $$\displaystyle f(1)+f(x)= f(1)f(x) $$ is

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$$1$$
0%
$$2$$
0%
$$1 \ or\ 2$$
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$$1\ and\ 0$$

The equation $$\displaystyle 9y^{2}(m+3)+6(m-3)y+(m+3)=0 $$, where $$m$$ is real has real roots then

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$$\displaystyle m< 0$$
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$$\displaystyle m> 0$$
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$$\displaystyle m\leq 0$$
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$$\displaystyle m\geq 0$$

If $$\displaystyle x^{2}+(a-b)x+(1-a-b)= 0, $$ where $$\displaystyle a,b \in R,$$ the value of $$a$$ such that the equation has distinct real roots for all value of $$b$$ are

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$$\displaystyle a> 1 $$
0%
$$\displaystyle a< 1 $$
0%
$$\displaystyle a> 2 $$
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$$\displaystyle a< 2 $$

A ray emanating from (6,2) is incident on ellipse $$\displaystyle \frac{(x-1)^{2}}{45}+\frac{(y-2)^{2}}{20}=1$$ at (4.6) The equation of reflected ray (after 1st reflection ) is

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x - 2y + 8 =0
0%
x + 2y + 8 =0
0%
x + 2y - 8 = 0
0%
x - 2y - 8 =0

The number of real roots of the equation $$(x - 1)^2 + (x - 2)^2 + (x- 3)^2 = 0$$ is :

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

Given that $$r$$ and $$s$$ are constants , the solution of the quadratic equation, $$r{x}^{2}=\dfrac{1}{s}x+3$$ , is:

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$$x=\dfrac { 1 }{ 2sr } \pm \dfrac { \sqrt { \dfrac { 1 }{ { s }^{ 2 } } +12r } }{ 2r } $$
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$$x=\dfrac { 1 }{ 2sr } \pm \dfrac { \sqrt { -\dfrac { 1 }{ { s }^{ 2 } } -12r } }{ 2sr } $$
0%
$$x=\dfrac { s }{ 2r } \pm \dfrac { \sqrt { \dfrac { 1 }{ { s }^{ 2 } } -12r } }{ 2r } $$
0%
$$x=\dfrac { s }{ 2r } \pm \dfrac { \sqrt { { s }^{ 2 }-12sr } }{ 2sr } $$

If the expression $$x^2+2(a+b+c)x+3(bc+ca+ab)$$ is a perfect square, then

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$$a=b=c$$
0%
$$a=\pm b=\pm c$$
0%
$$a=b\neq c$$
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None of the above

Number of possible value(s) of integer '$$a$$' for which the quadratic equation $$x^2+ax+16=0$$ has equal roots, is

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$$4$$
0%
$$6$$
0%
$$2$$
0%
none of these

If, for a positive integer n, the quadratic equation, $$ x ( x + 1 ) + ( x + 1 ) ( x + 2 ) + \ldots + \left( x + \frac { 1 } { n - 1 } \right) ( x + n ) = 10 n $$ has two consecutive integral solutions, then n is equal to

0%
11
0%
12
0%
9
0%
10

If the root of the equation $$x^{2}+px+q=0$$ differ from the roots of the equation $$x^{2}+qx+p=0$$ by the same quantity then

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$$p+q+1=0$$
0%
$$p+q+2=0$$
0%
$$p+q+4=0$$
0%
None of these

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

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Practice Class 10 Maths Quiz Questions and Answers

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

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Practice Class 10 Maths Quiz Questions and Answers

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