MCQ Questions for Class 10 Maths Quadratic Equations Quiz 4

Which of the following is non-quadratic polynomial.

0%
$x ^ { 2 } + 2$
0%
$x ^ { 2 } + 3 x$
0%
$4 x ^ { 4 } + 3 y$
0%
$x ^ { 2 }$

Explanation

$x^2+2, x^2+3x$ and

$x^2$ are all quadratic polynomials. But

$4x^4+3y$ is not a quadratic polynomial because this polynomial has a degree

$4$ .

The discriminent of graph is

0%
$b ^ { 2 } - 4 a c = 0$
0%
$b ^ { 2 } - 4 a c > 0$
0%
$b ^ { 2 } - 4 a c < 0$
0%
$b ^ { 2 } +4 a c < 0$

Explanation

Since the graph is touching the

$x-$ axis at only one point

so the quadratic equation has equal roots

$\implies \mathrm{discriminant}=0$

$\implies b^2-4 {a}{c}=0$

Which of the following must be added to

$x^2-6x+5$ to make it a perfect square ?

0%
$3$
0%
$4$
0%
$5$
0%
$6$

Explanation

We've,

$x^2-6x+5$

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

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

So to make it perfect square i.e.

$(x-3)^2$ , we are to add

$4$ to add.

The discriminant of the quadratic equation

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

0%
$\Delta = b^2 + 4ac$
0%
$\Delta = b^2 - 4ac$
0%
$\Delta =4abc$
0%
$\Delta = b^2 \times 4ac$

Explanation

The discriminant of the quadratic equation

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

$\Delta = b^2 - 4ac$

Values of

$k$ for which the quadratic equation

$2x^2+ kx + k = 0$ has equal roots.

0%
$4, 8$
0%
$0, 4$
0%
$4,-8$
0%
$0,8$

Explanation

For equal roots, discriminant D of equation

$2x^2+kx+k=0$ , should be zero.

$D =k^2-4(2)(k)=0$

$k^2-8k=0$

$k(k-8)=0$

$k=0, 8$

Which of the following is a quadratic equation?

0%
$x^3 + 2x + 1 = x^2 + 3$
0%
$-2x^2=(5-x)\left ( 2x-\cfrac{2}{5}\right )$
0%
$(k+1)x^2+\cfrac{3}{2}x=7,$ where $k=-1$
0%
$x^3 x^2 = (x)^3$

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

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

0%
$\displaystyle \frac{1}{64}$
0%
$\displaystyle \frac{1}{576}$
0%
$\displaystyle \frac{1}{144}$
0%
$\displaystyle \frac{127}{64}$

Explanation

$9x^2 + \dfrac 34x +2 = 0$

Dividing by

$9$ on both sides we get,

$x^2 + \dfrac{x}{12} + \dfrac 29 = 0$

$x^2 + \dfrac{x}{12} = -\dfrac 29$

Third term

$= \left(\dfrac 12 \times\text{coefficient of }x\right)^2$

$= \left(\dfrac 12 \times\dfrac {1}{12}\right)^2$ =

$\left(\dfrac {1}{24}\right)^2 =\dfrac{1}{576}$

Adding

$\dfrac {1}{576}$ on both sides we get,

$\therefore x^2 + \dfrac{x}{12} + \dfrac {1}{576}$ =

$\dfrac{1}{576} - \dfrac 29$

$\left(x + \dfrac{1}{24}\right)^2 =$

$\dfrac{1}{576} - \dfrac 29$

Hence, option

$B$ .

Which of the following is not a quadratic equation :

0%
$(x-2)^{2}+1=2x-3$
0%
$x(x+1)+8=(x+2)(x-2)$
0%
$x(2x+3)=x^{2}+1$
0%
$(x+2)^{3}=x^{3}-4$

Explanation

$x(x+1)+8=(x+2)(x-2)$

$\therefore x^2+x+8=x^2-4$

$\therefore x+12=0$

The highest power of

$x$ in this equation is

$1$ .

So, this is not a quadratic equation.

The answer is option (B)

(Note that, in option

$D$ ,

$(x+2)^3 = (x^3-4)$ also reduces to quadratic, since the

$x^3$ terms cancel out.)

If the expression

$(a-2)x^{2}+2(2a-3)x+(5a-6)$ is positive for all real values of

$x$ , then

0%
$a$ can be any real number
0%
$a>1$
0%
$a>3$
0%
$a=3$

Explanation

The expression is always positive that means the discriminant of the equation is always less than zero .
So,

$4(2a-3)^2-4(a-2)(5a-6)<0$

$(16a^2+36-48a)-(20a^2-64a+48)<0$

$(-4a^2+16a-12)<0$

$(-4)(a^2-4a+3)<0$

$(a-3)(a-1)>0$
So, the expression will be positive for

$a>3$ and

$a<1$
Option C is correct .

lf the roots of

${p}{x}^{2}+2{q}{x}+{r}=0$ and

$qx^{2}-2\sqrt{pr}x+q=0$ are simultaneously real, then

0%
$p=q$ ; $r\neq 0$
0%
$2q=\sqrt{pr}$
0%
$pr=q^{2}$
0%
${p}{r}={q}$

Explanation

Since roots are real,

$\Rightarrow (2q)^{2}-4pr\geq 0$ and

$(2\sqrt{pr})^{2}-4q^{2}\geq 0$

$4q^{2}\geq 4pr$ and

$4pr\geq 4q^{2}$

To hold above two equations simultenously

$\therefore q^{2}=pr$

$a > 0$ , then the expression

$ax^{2}+bx+c$ is positive for all values of

$x$ provided

0%
$b^{2}-4ac> 0$
0%
$b^{2}-4ac< 0$
0%
$b^{2}-4ac= 0$
0%
$b^{2}-ac< 0$

Explanation

Consider the equation,

$ax^{2}+bx+c$

Now if

$b^{2}-4ac<0$ , then it does not have any real roots and hence it will not intersect the

$x$ axis at any point.

$a>0$ then

$y=ax^{2}+bx+c$ will lie completely above

$x$ axis and

$a<0$ then

$y=ax^{2}+bx+c$ will lie completely below

$x$ axis.

It is given that the above equation is always greater than zero,

$ax^{2}+bx+c>0$ for

$\epsilon R$ .

This can only occur is

$b^{2}-4ac<0$ and

$a>0$ .

$a=0$ , then the equation

$\displaystyle \frac{x-a-1}{x-a}=a +1-\displaystyle \frac{1}{x-a}$ has

0%
one root.
0%
two roots.
0%
many roots.
0%
no roots.

Explanation

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

$1-\dfrac{1}{x-a}=a+1-\dfrac{1}{x-a}$
Since, given

$a=0$ , equation holds for all

$x$ except

$x=a$

$\therefore$ The equation has many roots.

Which of the following is not a quadratic equation?

0%
$2(x + 1)^2 = 4x^2 + 2x + 1$
0%
$2x + x^2 = 2x^2 + 5$
0%
$( \sqrt{2} \times \sqrt{3}x)^2+ x^2 = 3x^2 + 5x$
0%
$(x^2 + 2x)^2 = x^5 + 3 + 4x^3$

Explanation

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

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

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

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

$-2x^{2}+2x+1=0$
is a quadratic equation
B.

$2x+x^2=2x^2+5$

$-x^2+2x=5$
is a quadratic equation
C.

$[\sqrt{2}x\sqrt{3}]^2+x^2=3x^2+5x$

$(6x^2)+x^2-3x^2-5x=0$

$4x^2-5x=0$
is a quadratic equation
D.

$(x^2+2x)^2=x^5+4x^3+3$
As degree of

$x$ is

$5$ , it is not quadratic.

The equation

$(a+2){x}^{2}+(a-3)x=2a-1,a\neq -2$ has rational roots for

0%
all rational values of $a$ except $a=-2$
0%
all real values of $a$ except $a=-2$
0%
rational values of $a> \displaystyle \frac{1}{2}$
0%
none of these

$14$ is the maximum of

$-\lambda x^{2} + \lambda x + 8$ , then the value of

$\lambda$ is

0%
$24$
0%
$\sqrt[6]{3}$
0%
$\sqrt[-6]{3}$
0%
$- 12$

Explanation

$-\lambda x^{2} + \lambda x + 8$

$=-\lambda(x^2 - x - \cfrac{8}{\lambda})$

$=- \lambda({(x - \cfrac{1}{2})^2 - \cfrac{1}{4} - \cfrac{8}{\lambda})})$

$=-\lambda(x - \cfrac{1}{2})^2) + \cfrac{\lambda}{4} + 8$

Maximum value of the expression is:

$8 + \cfrac{\lambda}{4}$

$8 + \cfrac{\lambda}{4} = 14$

$\lambda= 24$

If equation

$x^2-(2+m) x+1(m^2-4m+4)=0$ has coincident roots, then :

0%
$m=0$
0%
$m = 6$
0%
$m=2$
0%
$m \displaystyle = \frac{2}{3}$

Explanation

Given:

$x^2-(2+m)x+1(m^2-4m+4)=0$ has coincident roots

Therefore

$b^2 - 4ac=0$

here

$a=1, \ b=-(2+m),\ c=(m^2-4m+4)$

$\Rightarrow [-(2+m)]^2-4\times 1\times (m^2-4m+4)=0$

$\Rightarrow 4+m^2+4m-4m^2-16m-16=0$

$\Rightarrow -3m^2+20m-12=0$

$\Rightarrow 3m^2 - 20m +12=0$

$\Rightarrow 3m^2 - 2m -18m+12=0$

$\Rightarrow m(3m - 2)-6(3m -2)=0$

$\Rightarrow (m-6)(3m-2)=0$

$\therefore\ \ m = \cfrac{2}{3} , 6$

If the roots of the equation

$x^{2} + a^{2} = 8x + 6a$ are real, then:

0%
$a \epsilon \left [ 2,8 \right ]$
0%
$a \epsilon \left [ -2,8 \right ]$
0%
$a \epsilon (2,8)$
0%
$a \epsilon (-2,8)$

Explanation

The given equation can be written as

$x^{2} - 8x + a^{2} -6a = 0$
Since the roots of the above equation are real,

$\therefore B^{2} -4AC \geq 0 \Rightarrow 64-4(a^{2}-6a)\geq 0$

$\Rightarrow a^{2} -6a -16 \leq 0$

$\Rightarrow (a+2)(a-8) \leq 0$

$\Rightarrow a\ \epsilon\ \left [ -2,8 \right ]$

Which of the following is not a quadratic equation?

0%
$2(x-1)^2=4x^2-2x+1$
0%
$(x^2+1)^2=x^2+3x+9$
0%
$(x^2+2x)^2=x^4+3+4x^3$
0%
$x^2+9=3x^2-5x$

Find the roots of the following quadratic equation

$z^2\,+\,6z\,-\,8\,=\,0$

0%
$z\,=\,-3\,\pm\,\sqrt{19}$
0%
$z\,=\,-3\,\pm\,\sqrt{17}$
0%
$z\,=\,-2\,\pm\,\sqrt{17}$
0%
$z\,=\,-3\,\pm\,\sqrt{15}$

Explanation

The roots of the quadratic equation

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

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

So, the roots of

$z^2+6z-8=0$ are

$z=\dfrac{-6\pm\sqrt{6^2-4(1)(-8)}}{2}$

$\therefore z=\dfrac{-6\pm\sqrt{68}}{2}$

$\therefore z=\dfrac{-6\pm2\sqrt{17}}{2}$

$\therefore z=-3\pm \sqrt{17}$

Find the discriminant for the given quadratic equation:

$4x^2\,-\,kx\,+\,2\,=\,0$

0%
$k^2\,-\,21$
0%
$k^2\,-\,24$
0%
$k^2\,-\,28$
0%
$k^2\,-\,32$

Explanation

Given equation is,

$4x^2 - kx + 2 = 0$
We know,

$D = b^2 - 4ac$

$a=4, b=-k, c=2$

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

$\therefore D = k^2 - 32$

If both 'a' and 'b' belong to the set

$\left \{ 1, 2, 3, 4 \right \}$ , then the number of equations of the form

$ax^{2}+bx+1=0$ having real roots is:

0%
$10$
0%
$7$
0%
$6$
0%
$12$

Explanation

For a quadratic equation to have a real roots, discriminant must be greater than or equal to zero

$\Rightarrow b^{2}-4a\geq0$

Since,

$a,b \in \{1,2,3,4\}$

(1) When

$a=1$

$\Rightarrow b^{2}\geq 4$

$\Rightarrow b=2,3,4$

(2) When

$a=2$

$\Rightarrow b^{2}\geq 8$

$\Rightarrow b=3,4$

(3) When

$a=3$

$\Rightarrow b^{2}\geq 12$

$\Rightarrow b=4$

(3) When

$a=4$

$\Rightarrow b^{2}\geq 16$

$\Rightarrow b=4$

Therefore, total

$7$ possibilities are there.

Hence the answer is

$7$ .

If the equation

$4x^{2} + x(p + 1) + 1 = 0$ has exactly two equal roots, then one of the values of

$p$ is

0%
$5$
0%
$-3$
0%
$0$
0%
$3$

Explanation

For the equation

$4x^{2} + x(p + 1) + 1 = 0$ to have two equal roots, the condition is Discriminant

$=0$

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

$(p+1)^{2}-4\times 4\times 1 = 0$

$p^{2}+2p+1-16 = 0$

$p^{2}+2p-15 = 0$

$(p+5)(p-3) = 0$

$\therefore p = -5$ or

$p = 3$
Hence, one of the values of

$p$ is

$3$ .

Option D is correct.

The value of

$x^{2} - 6x + 13$ can never be less than

0%
$4$
0%
$5$
0%
$4.5$
0%
$7$

Explanation

We know that a perfect square can never be less than

$0$ .

$\because x^{2}-6x+13 = x^{2}-6x+9+4$

$= (x-3)^{2}+4$
because

$(x-3)^{2}$ can never be less than zero

$\therefore$ least value of

$x^{2}-6x+13\ is \ 4$

$\displaystyle ax^{2}+bx+1= 0, a \in R, b\in R,$ does not have distinct real roots, then the maximum value of

$b^2$ is

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

Explanation

$The\quad equation\quad a{ x }^{ 2 }-bx+1=0\quad has\quad no\quad distinct\quad real\quad roots,\\ This\quad implies\quad D=b^2-4ac\le 0,\\ b^2-4a\le 0 \\ b^2\le 4a$ .

$\displaystyle x^{2}-bx+c=0$ has equal integral roots, then

0%
$b$ and $c$ are integers
0%
$b$ and $c$ are even integers
0%
$b$ is an even integer and $c$ is $a$ perfect square of $a$ positive integer
0%
none of these

Explanation

The roots of this equation will be

$\alpha =\dfrac{b\pm \sqrt{b^{2}-4c}}{2}$

Both roots are equal so

$\dfrac{b+\sqrt{b^{2}-4c}}{2}=\dfrac{b-\sqrt{b^{2}-4c}}{2}$

$b=\pm 2\sqrt{c}$

Therefore,

$c$ should be perfect square integer and

$b$ is even number.

Hence, options 'A' and 'C' are correct.

$\displaystyle ax^{2}+bx+6=0$ does not have two distinct real roots, where

$a\in R,b\in R$ , then the least value of

$3a+b$ is

0%
$4$
0%
$-1$
0%
$1$
0%
$-2$

Explanation

$ax^{ 2 }+bx+6=0$
Substitute

$b=k-3a$ to obtain

$ax^{ 2 }+\left( k-3a \right) x+6=0$
Since, the equation does not have real distinct roots.
Therefore,

$D={ \left( k-3a \right) }^{ 2 }-24a\le 0$

$\Rightarrow 9a^{ 2 }-6a\left( 4+k \right) +{ k }^{ 2 }\le 0$
Above quadratic equation has real roots.
Therefore,

$D=36\left[ { \left( 4+k \right) }^{ 2 }-{ k }^{ 2 } \right] \ge 0$

$\Rightarrow k\ge -2$

$\Rightarrow 3a+b\ge -2$ .

Ans: D

The number of values of

$k$ for which

$\displaystyle \left \{x^{2}-(k-2)x+k^{2}\right\}+ \left \{x^{2}+kx+(2k-1)\right \}$ is a perfect square is/are

0%
$1$
0%
$2$
0%
$0$
0%
none of these

Explanation

Consider

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

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

$p\left( x \right)$ is perfect square when roots of the equation

$p\left( x \right) =0$ are equal
Therefore,

$D={ b }^{ 2 }-4ac=4-8\left( { k }^{ 2 }+2k-1 \right) =0$

$\Rightarrow 2{ k }^{ 2 }-4k-3=0$
Therefore, number of values of

$k$ are 2.

Ans: B

If at least one of the equations

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

${ x^{ 2 } }+rx+s=0$ has real roots, then

0%
$qs =(p+r)$
0%
$pr =(q+s)$
0%
$pr =2(q+s)$
0%
None of these.

Explanation

Given at least one of the equation

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

$x^{ 2 }+rx+s=0$ has real roots.
Let

$D_1$ and

$D_2$ be the discriminant values.
At least one of the equation has real roots when

$D_1+D_2 \geq 0$ .

$\Rightarrow p^2-4q+r^2-4s \geq 0$

$\Rightarrow (p-r)^2+2pr-4(q+s) \geq 0$
Above inequality holds only if

$2pr-4(q+s)=0$ .

$\therefore pr=2(q+s)$ .
Hence, option C.

$x$ is real, the expression

$\displaystyle \frac { \left( x-a \right) \left( x-c \right) }{ \left( x-b \right) }$ is capable of assuming all values if

0%
$a > b > c$
0%
$a < b < c$
0%
$a < b > c$
0%
Cannot say

Explanation

$y=\dfrac { \left( x-a \right) \left( x-c \right) }{ \left( x-b \right) }$

$(x-b)y={ x }^{ 2 }-(a+c)x+ac$

${ x }^{ 2 }-(a+c+y)x+ac+by=0$

$D\ge 0$

${ (a+c+y) }^{ 2 }-4(ac+by)\ge 0$

${ y }^{ 2 }+{ a }^{ 2 }+{ c }^{ 2 }+2ac+2cy+2ay-4ac-4by\ge 0$

${ y }^{ 2 }+(2a+2c-4b)y+{ (a-c) }^{ 2 }\ge 0$
So

$D\le 0$
So

$4{ (a+c-2b) }^{ 2 }-4{ (a-c) }^{ 2 }\le 0$

$(a+c-2b+a-c)(a+c-2b-a+c)\le 0$

$(a-b)(c-b)\le 0$

$a < b < c$

$a > b > c$

For the quadratic equation

$\displaystyle 2x^{2}+6\sqrt{2}x+1=0$

0%
roots are rational
0%
if one root is $p+\sqrt{q}$ , then the other is $-p+\sqrt{q}$
0%
roots are irrational
0%
if one root is $p+\sqrt{q}$ , then the other is $p-\sqrt{q}$

Explanation

Here,

$2x^{ 2 }+6\sqrt { 2x } +1=0$

$\displaystyle \Rightarrow x=\frac { -b\pm \sqrt { { b }^{ 2 }-4ac } }{ 2a } =\frac { -6\sqrt { 2 } \pm \sqrt {64} }{ 4 } =\frac { -3\sqrt { 2 } \pm 4 }{ 2 }$

Therefore, roots are irrational.

$p=-\dfrac{3\sqrt{2}}{2}$ and

$q={4}$

One root is

$p+\sqrt{q}$ , then the other root is

$p-\sqrt{q}$

Ans: C,D

CBSE Questions for Class 10 Maths Quadratic Equations Quiz 4 - MCQExams.com

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Practice Class 10 Maths Quiz Questions and Answers

CBSE Questions for Class 10 Maths Quadratic Equations Quiz 4 - MCQExams.com

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Practice Class 10 Maths Quiz Questions and Answers

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