MCQ Questions for Class 10 Maths Quadratic Equations Quiz 6

Find the roots of the following quadratic equations by using the quadratic formula

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

0%
$1, 2$
0%
$1, -\cfrac{1}{2}$
0%
$\cfrac{1}{2},-2$
0%
$-2, 1$

Explanation

Given equation is $2\left ( \dfrac{x}{x+1} \right )^{2}-5\left ( \dfrac{x}{x+1} \right )+2=0$

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

Then $2a^{2}-5a+2=0$

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

$\Rightarrow 2a(a-2)-1(a-2)=0$

$\Rightarrow (2a-1)(a-2)=0$

$\therefore a-2=0\Rightarrow a=2$ , put the value $a=\dfrac{x}{x+1}$

If $2a-1=0\Rightarrow a=\dfrac{1}{2}$ , put $a=\dfrac{x}{x+1}$

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

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

$\displaystyle\alpha, \beta$ are roots of

$\displaystyle ax^{2}-2bx+c=0$ then

$\displaystyle a^{3}\beta ^{3}+a^{2}\beta ^{2}+a^{3}\beta^2$ is

0%
$\displaystyle \frac{c^{2}\left ( c+2b \right )}{a^{3}}$
0%
$\displaystyle \frac{bc^{2}}{a^{3}}$
0%
$\displaystyle \frac{c^{2}}{a^{3}}$
0%
None

Explanation

The given expression

$\Rightarrow {\alpha}^3{\beta}^3 + {\alpha}^2{\beta}^3 + {\alpha}^3{\beta}^2$ can be written as

$\Rightarrow (\alpha\beta)^3 + (\alpha\beta)^2 (\alpha + \beta)$ ....

$(1)$

Now as we know that

$\alpha$ and

$\beta$ are the roots of the quadrtic equation

$ax^2 -bx +c=0$

So sum of the roots

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

Also the product of the roots

$\alpha. \beta = \dfrac{c}{a}$

By putting the value of the terms

$\alpha + \beta$ and

$\alpha.\beta$ into equation

$(1)$

We get,

$\Rightarrow (\alpha\beta)^3 + (\alpha\beta)^2 (\alpha + \beta)$

$= (\dfrac{c}{a})^3 + (\dfrac{c}{a})^2 (\dfrac{2b}{a})$

$\Rightarrow (\alpha\beta)^3 + (\alpha\beta)^2 (\alpha + \beta)$

$= \dfrac{c^2(c +2b)} {a^3}$

Hence the Correct option is

$A$

Solve

$9m^2-12m+2=0$ by method of completing square.

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

Explanation

$9m^2-12m+2=0$

$9m^2-12m=-2$

Dividing both sides by

$9$ , we get

$m^2-\dfrac {12}{9} = \dfrac {-2}{9}$

$m^2-\dfrac {4}{3} = \dfrac {-2}{9}$

Third term =

$\left(\dfrac 12 \times \mbox{coefficient of m}\right)^2$

$=\left (\dfrac 12 \times \dfrac {-4}{3}\right)^2 = \dfrac 49$

Adding

$\dfrac 49$ on both sides, we get

$m^2-\dfrac 43 m+ \dfrac 49 = \dfrac {-2}{9} + \dfrac 49$

$\left(m-\dfrac 23\right)^2 = \dfrac 29$

Taking square roots on both sides we get,

$\left(m-\dfrac 23\right) =\pm \dfrac{\sqrt{2}}3$

$m=\dfrac {2+\sqrt 2}{3}$ or

$m=\dfrac {2-\sqrt 2}{3}$

The roots of the equation

$\displaystyle x^{2}-px+q=0$ are consecutive integers. Find the discriminate of the equation.

0%
$-1$
0%
$0$
0%
$1$
0%
None of these

Explanation

For the given equation, let the roots be

$a, a+1$

So sum of roots

$= a + a + 1 = 2a + 1= -\cfrac {-p}{1} = p$

Product of roots

$= a \times (a+1) = {a}^{2} + a= \cfrac {q}{1} = q$

For an equation

$a{x}^{2} + bx + c = 0$ , the discriminant is

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

For the given eqn, the discriminant is

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

Solve:

$\displaystyle x^{2}+3x+1=0$

0%
$\displaystyle x=\frac{3\pm\sqrt{5}}{2}$
0%
$\displaystyle x=\frac{3+\sqrt{5}}{2},\frac{-3+\sqrt{5}}{2}$
0%
$\displaystyle x=\frac{\pm{3}-\sqrt{5}}{2}$
0%
$\displaystyle x=\frac{-3+\sqrt{5}}{2},\frac{-3-\sqrt{5}}{2}$

Explanation

We have

$\displaystyle x^{2}+3x+1=0$
Add and subtract

$\displaystyle \left ( \frac{1}{2}\mbox{coefficient of x} \right )^{2}$ in LHS and get

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

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

$\displaystyle \Rightarrow \left ( x+\frac{3}{2} \right )^{2}-\frac{5}{4}=0$

$\displaystyle \Rightarrow \left ( x+\frac{3}{2} \right )^{2}=\left ( \frac{\sqrt{5}}{2} \right )^{2}$

$\displaystyle \Rightarrow x+\frac{3}{2}=\pm \frac{\sqrt{5}}{2}$

This gives

$\displaystyle x=\frac{-3+\sqrt{5} }{2}$ or

$x=\dfrac{-3-\sqrt{5}}{2}$

Therefore,

$\displaystyle x=\frac{-3+\sqrt{5}}{2},\frac{-3-\sqrt{5}}{2}$ are the solutions of the given equation

Which of the statements given below is correct, if discriminant for equation

$5x^2 - 4x +2 =0$ is D ?

0%
D > 0
0%
D = 0
0%
D < 0
0%
D is not defined

Explanation

The Discriminant(

$D$ ) of the quadratic equation

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

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

Here,

$b=-4,a=5,c=2$

Then,

$D=(-4)^{2}-4(5)(2)$

$D=16-40=-24$

$\Rightarrow$

$D<0$

$\therefore$ Option C is correct.

Solve the equation using the quadratic formula

$3{ x }^{ 2 }=-10x-5$

0%
$\left\{ \displaystyle\frac { -5\pm \sqrt { 10 } }{ 3 } \right\}$
0%
$\left\{ \displaystyle\frac { -10\pm \sqrt { 10 } }{ 3 } \right\}$
0%
$\left\{ \displaystyle\frac { -10\pm \sqrt { 10 } }{ 6 } \right\}$
0%
$\left\{ \displaystyle\frac { -5\pm \sqrt { 10 } }{ 6 } \right\}$

Explanation

Given equation

$3x^2 = -10x-5$

$\implies 3x^2 +10x +5=0$ .

When this equation is compared with general equation

$ax^2 + bx +c =0$

$a= 3$ ,

$b=10$ and

$c=5$

Quadratic formula given by sri dharacharya,

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

on substituting values we get,

$x =\dfrac { -5 \pm \sqrt{ 10 } } {3}.$

The roots of the equation

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

0%
$1,-4$
0%
$-3,\cfrac {1}{\sqrt {3}}$
0%
$\cfrac{\sqrt {33}-1}{4},\cfrac{-\sqrt {33}-1}{4}$
0%
None

Explanation

The given equation

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

$D={b}^{2}-4ac=1+4\times 4\times 2=33>0$

${x}_{1}=\cfrac{-b+\sqrt {D}}{2a}, {x}_{2}=\cfrac{-b-\sqrt {D}}{2a}$

${x}_{1}=\cfrac{-1+\sqrt {33}}{4}, {x}_{2}=\cfrac{-\sqrt {33}-1}{4}$

Solve the equation using the quadratic formula

$2=-\displaystyle\frac { 10 }{ x } -\displaystyle\frac { 5 }{ { x }^{ 2 } }$

0%
$\left\{ \displaystyle\frac { -5\pm \sqrt { 15 } }{ 2 } \right\}$
0%
$\left\{ \displaystyle\frac { -5\pm \sqrt { 15 } }{ 4 } \right\}$
0%
$\left\{ \displaystyle\frac { -5\pm \sqrt { 35 } }{ 2 } \right\}$
0%
$\left\{ \displaystyle\frac { -10\pm \sqrt { 15 } }{ 2 } \right\}$

Explanation

Given equation

$2 = - \dfrac{10}{x}-\dfrac{5}{x^2}$

$\Rightarrow 2x^2 +10x +5=0$ .

When this equation is compared with general equation

$ax^2 + bx +c =0$

$a= 3$ ,

$b=10$ and

$c=5$

Quadratic formula given by sri dharacharya,

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

On substituting values, we get

$x =\dfrac { -5 \pm \sqrt{ 15 } } {2}.$

When solved by the method of completing the square, the solutions to the equation

${-8x = 4x^{2}-1}$ are

0%
${\dfrac {-2+\sqrt{5}}{2}}$
0%
${\dfrac {2+\sqrt{5}}{2}}$
0%
${\dfrac {-2+\sqrt{2}}{2}}$
0%
${\dfrac {-2-\sqrt{5}}{2}}$

Explanation

Given,

${-8x = 4x^{2}-1}$

$\Rightarrow 4x^2+8x=1$

$\Rightarrow 4(x^2+2x)=1$

$\Rightarrow 4(x^2+2x+1)=1+4=5$

$\Rightarrow [2(x+1)]^2=(\sqrt{5})^2$

$\Rightarrow 2(x+1)= \pm \sqrt{5}$

$\Rightarrow x+1=\pm \cfrac{\sqrt{5}}{2}$

$\Rightarrow x=\cfrac{-2\pm\sqrt{5}}{2}$
Hence, options A and D are correct.

The roots of the equation

$(q -r)x^2+ (r -p)x + (p -q) =0$ are

0%
$\dfrac {p-q}{q-r}, 1$
0%
$\dfrac {q-r}{p-q}, 1$
0%
$\dfrac {r-p}{p-q}, 1$
0%
$\dfrac {r-p}{q-r}, 1$

Explanation

$(q-r)x^2+(r-p)x+(p-q)=0$
Sum of the coefficients,

$q -r + r -p + p -q = 0$ . Hence, 1 is a root.
Synthetic division

$1 | (q-r) + (r-p) + (p-q)$

$|\underline {0 + q-r + q-p}$

$q-r + q-p |\underline 0$
Quotient

$(q -r)x + (q -p) = 0$ gives

$x=\frac {p-q}{q-r}$ .

Find the number of real roots in the equation,

$\displaystyle { x }^{ 2 }+3x-9=0$

0%
$1$
0%
$2$
0%
$0$
0%
None

Explanation

$\displaystyle \Delta \ge 0$ for real roots

$\displaystyle { b }^{ 2 }-4ac$

$=\displaystyle 9+36$

$=\displaystyle 45\ge 0$ which is true .

$\displaystyle \therefore$ it has 2 real roots.

Find the discriminant of

$\displaystyle { x }^{ 2 }-5x-10=0$

0%
${65}$
0%
${-65}$
0%
$10$
0%
None of the above

Explanation

Discriminant is given as:

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

$\implies D={ \left( -5 \right) }^{ 2 }-4\left( -10 \right)$

$\displaystyle =25+40$

$\displaystyle =65$

Which of the following is not a quadratic equation

0%
$\displaystyle x-\frac { 3 }{ 2x } =5$
0%
$\displaystyle 4x-\frac { 5 }{ 8 } ={ x }^{ 2 }$
0%
$\displaystyle x+\frac { 1 }{ x } =9$
0%
$\displaystyle 4x-\frac { 2 }{ 3x } =4{ x }^{ 2 }$

Explanation

Lets, simplify all the equations given in options and convert them to standard forms,

(A)

$\displaystyle x-\frac { 3 }{ 2x } =5$

Now,

$\displaystyle x-\frac { 3 }{ 2x } =5$

$\Rightarrow \displaystyle \frac {2x^2 - 3 }{ 2x } =5$

$\Rightarrow \displaystyle 2x^2 - 3 =10x$

$\Rightarrow \displaystyle 2x^2 -10x- 3 =0$

The highest power of variable

$x$ is

$2$ . So, it is an quadratic equation.

(B)

$\displaystyle 4x-\frac { 5 }{ 8 } =x^2$

Now,

$\displaystyle 4x-\frac { 5 }{ 8 } =x^2$

$\Rightarrow \displaystyle \frac {32x - 5 }{ 8 } =x^2$

$\Rightarrow \displaystyle 32x - 5 =8x^2$

$\Rightarrow \displaystyle 8x^2 -32x +5 =0$

The highest power of variable

$x$ is

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

(C)

$\displaystyle x+\frac { 1 }{ x } =9$

Now,

$\displaystyle x+\frac { 1 }{ x } =9$

$\Rightarrow \displaystyle \frac {x^2 + 1 }{ x } =9$

$\Rightarrow \displaystyle x^2 + 1 =9x$

$\Rightarrow \displaystyle x^2 -9x+ 1 =0$

The highest power of variable

$x$ is

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

(D)

$\displaystyle 4x-\frac { 2 }{ 3x } =4x^2$

Now,

$\displaystyle 4x-\frac { 2 }{ 3x } =4x^2$

$\Rightarrow \displaystyle \frac {12x^2 - 2 }{ 3x } =4x^2$

$\Rightarrow \displaystyle 12x^2 - 2 =12x^3$

$\Rightarrow \displaystyle 12x^3 -12x^2+2 =0$

$\Rightarrow \displaystyle 6x^3 -6x^2+1 =0$

Here, the highest power of variable

$x$ is

$3$ . So, it is not a quadratic equation.

Hence,

$Op-D$ is correct.

Given that

${ z }^{ 2 }-10z+25=9$ , what is

$z$ ?

0%
${3,4}$
0%
${1,6}$
0%
${2,6}$
0%
${2,8}$

Explanation

Since you recognize that the left hand side of the equation is a perfect square quadratic, you will factor the left side of the equation first, instead of trying to set everything equal to

$0$

${ z }^{ 2 }-10z+25=9$

${(z-5)}^{2}=9$

$\implies z-5=\pm 3$

$\implies z=5\pm 3$

$\therefore z= 2$ or

$z=8$

$\displaystyle xy\neq 0$ and

$\displaystyle { x }^{ 2 }{ y }^{ 2 }-xy=6$ , which of the following could be y in terms of x ?
I.

$\displaystyle \frac { 1 }{ 2x }$
II.

$\displaystyle -\frac { 2 }{ x }$
III.

$\displaystyle \frac { 3 }{ x }$

0%
I only
0%
II only
0%
I and II only
0%
I and III only
0%
II and III

Explanation

Solving the quadratic in y

$x^2y^2 -xy - 6 = 0$

$\rightarrow$ y =

$\cfrac{x +/- \sqrt {x^2 + 24x^2}}{2x^2}$

$\rightarrow$ y = (

$x$ +/-

$5x$ )/(2

$x^2$ )

y =

$\cfrac{3}{x}$ or y =

$\cfrac{-2}{x}$

Option E.

Find the discriminant of

$\displaystyle pqr{ x }^{ 2 }-8pqx+pr=0$

0%
$\displaystyle { p }^{ 2 }{ q }^{ 2 }-4{ p }^{ 2 }{ qr }^{ 2 }$
0%
$\displaystyle 64{ p }^{ 2 }{ q }^{ 2 }-{ p }^{ 2 }{ r }^{ 2 }$
0%
$\displaystyle 64{ p }^{ 2 }{ q }^{ 2 }-4{ p }^{ 2 }{ qr }^{ 2 }$
0%
$\displaystyle 64{ p }^{ 2 }{ q }^{ 2 }-4{ p }^{ 2 }{ r }^{ 2 }$

Explanation

$\displaystyle D={ b }^{ 2 }-4ac$
Here,

$\displaystyle b=-8pq,a=pqr,c=pr$

$\displaystyle D={ \left( -8pq \right) }^{ 2 }-4\left( pq \right) \left( pr \right)$

$\displaystyle =64{ p }^{ 2 }{ q }^{ 2 }-4{ p }^{ 2 }{ qr }^{ 2 }$

Find the number of real roots in the equation ,

$\displaystyle 4{ x }^{ 2 }+3x+5=0$

0%
$0$
0%
$1$
0%
$2$
0%
None

Explanation

$\displaystyle \Delta \ge 0$ for real roots .

$\displaystyle { b }^{ 2 }-4ac$

$=\displaystyle 9-(80)$

$\displaystyle -71\lt 0$

$\displaystyle \therefore$

it has no real roots.

Which of the following is correct, if the roots of quadratic equation are equal?

0%
$\displaystyle D>0$
0%
$\displaystyle D<0$
0%
$\displaystyle D=0$
0%
None

Explanation

If both roots of a quadratic equation are equal and real, then

$\displaystyle D=0$ ,

$\displaystyle \propto =\beta$

The roots of the equation

$\displaystyle 2{ y }^{ 2 }+y-2=0$ are

0%
$\displaystyle \frac { -1-\sqrt { 17 } }{ 2 } ,\frac { -1-\sqrt { 17 } }{ 2 }$
0%
$\displaystyle \frac {- 1-\sqrt { 17 } }{ 4 } ,\frac { -1+\sqrt { 17 } }{ 4 }$
0%
$\displaystyle -1-\sqrt { 17 } ,-1+\sqrt { 17 }$
0%
None

Explanation

Using quadratic formula,

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

$\Rightarrow \dfrac{-1\pm\sqrt{(-1)^2-4\times 2\times (-2)}}{2\times 2}$

$\Rightarrow \dfrac{-1\pm\sqrt{17}}{4}$

So, the roots are

$\dfrac{-1-\sqrt{17}}{4}, \dfrac{-1+\sqrt{17}}{4}$

Which of the following equations are not quadratic?

0%
$x(2x+3)=x+2$
0%
$(x-2)^2+1=2x-3$
0%
$y(8y+5)=y^2+3$
0%
$y(2y+15)=2(y^2+y+8)$

Explanation

As we know that in quadratic equation the highest degree of the polynomial is

$2$ . so all the option have degree of the polynomail

$2$ except option D. because in option D the degree of the polynimail is

$1$ .so the correct option is D

Find the roots of equation by completing the square method:

$\displaystyle 3{ x }^{ 2 }-12qx+12{ q }^{ 2 }=0$

0%
$2q, 2q$
0%
$-2q, -2q$
0%
$-2q, 2q$
0%
None

Explanation

$\displaystyle 3{ x }^{ 2 }-12qx+12{ q }^{ 2 }=0$

$\displaystyle { x }^{ 2 }-4qx+4{ q }^{ 2 }=0$

$\displaystyle { x }^{ 2 }-2*2qx+4{ q }^{ 2 }=0$

$\displaystyle \left( x-2q \right) ^2 =0$

$\displaystyle \left( x-2q \right) \left( x-2q \right) =0$

$\displaystyle x=2q\quad or\quad x=2q$

Solve

$x^2+8x+15=0$ by method of completing square.

0%
$x=3, -5$
0%
$x=-3, -5$
0%
$x=3, 5$
0%
$x=-3, 5$

Explanation

Given:

$x^{2}+8x+15=0$

$x^{2}+8x=-15$

Add

$16$ on both sides,

$x^{2}+8x+16=-15+16$

$x^{2}+2\times4x+4^{2}=1$

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

$x+4=\pm1$

Hence,

$x=-4+1=-3$ or

$x=-4-1=-5$

$\alpha, \beta$ are the roots of

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

$\alpha +k, \beta +k$ are the roots of

$px^2+qx+r=0$ , then

$\displaystyle\frac{b^2-4ac}{q^2-4pr}$ is equal to:

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

Explanation

Absolute difference of both the roots will be same,

$\Rightarrow \left|\dfrac{\sqrt {D_1}}{a}\right|=\left|\dfrac{\sqrt{D_2}}{p} \right|$

Where

$D_1,D_2$ are the discriminant of the respective quadratic

$\Rightarrow \dfrac{D_1}{D_2}=\dfrac{a^2}{p^2}$ , square above equation

$\Rightarrow \dfrac{b^2-4ac}{q^2-4pr}=\left(\dfrac{a}{p}\right)^2$

Which of the following is correct if one root of quadratic equation is real?

0%
$\displaystyle D\le 0$
0%
$\displaystyle D=0$
0%
$\displaystyle D\ge 0$
0%
None

Explanation

$\displaystyle D\ge 0\quad \because$ both are real

$\displaystyle D=0\quad \because$ both are equal

$\displaystyle D\le 0\quad \because$ both are imaginary

If a, b and c are real, then both the roots of the equation

$\left( x-b \right) \left( x-c \right) +\left( x-c \right) \left( x-a \right) +\left( x-a \right) \left( x-b \right) =0$ are always

0%
Positive
0%
Negative
0%
Real
0%
Imaginary

Explanation

Given equation is

$(x-b)(x-c)+(x-c)(x-a)+(x-a)(x-b)=0$

We can further simplify this equation by multiplying terms and taking common powers of

$x$ .

$(x^2-bx-cx+bc)+(x^2-cx-ax+ac)+(x^2-ax-bx+ab)=0$

$\Rightarrow 3x^2-2ax-2bx-2cx+ab+bc+ca$

$\Rightarrow 3{ x }^{ 2 }-2x\left( a+b+c \right) +ab+bc+ca=0$
We know that, discriminant of a quadratic equation of the form

$ax^2+bx+c=0$ is given by:

$D=b^2-4ac$

$\therefore \ D=[-2(a+b+c)]^2-4\times 3\times (ab+bc+ca)$

$\Rightarrow D=4{ \left( a+b+c \right) }^{ 2 }-4\times 3\left( ab+bc+ca \right)$

$\Rightarrow 4\left( { a }^{ 2 }+{ b }^{ 2 }+{ c }^{ 2 }-ab-bc-ca \right)$

$\Rightarrow 2\left[ { \left( a-b \right) }^{ 2 }+{ \left( b-c \right) }^{ 2 }+{ \left( c-a \right) }^{ 2 } \right]$
Given that,

$a,\ b\text { and }c$ are real.

Hence,

$D\ge 0$
Therefore, roots are real.

Hence, option C is correct.

Solve:

$p^2-12p+32=0$

0%
$p=-8, 4$
0%
$p=8, 4$
0%
$p=8, -4$
0%
$p=-8, -4$

In the given equation, find the negative value of p.

$\displaystyle p-3=\frac { 10 }{ p }$

0%
$-2$
0%
$-5$
0%
$-10$
0%
None

Explanation

$\displaystyle p-3=\frac { 10 }{ p }$

$\displaystyle { p }^{ 2 }-3p=10p$

$\displaystyle { p }^{ 2 }-3p-10=0$

Here,

$a=1, b=-3, c=-10$ by comparing with the general quadratic equation

$ax^2+bx+c$ .

Using quadratic formula we get,

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

$p=\dfrac { -(-3)\pm \sqrt { (-3)^2-4\times 1\times (-10) } }{ 2 }$

$p=\dfrac { +3\pm \sqrt { 9+40 } }{ 2 }$

$\displaystyle p=\dfrac { 3\pm 7 }{ 2 }$

$p=\dfrac { 3+7 }{ 2 } =5$

and

$\displaystyle p=\dfrac { 3-7 }{ 2 } =-2$

Therefore the negative value of

$p=-2$ .

Which of the following is a quadratic equation?

0%
$(a-1)x^{3}+(b -2)x^{2} + c$
0%
$(a-1)x^{2}+(b -2)x + 5c$
0%
$(a-1)x^{4}+(b -2)x^{2} + 2c$
0%
$(x-1)x^{4}+(b -2)x^{2} + 2c$

Explanation

A quadratic equation can be expressed in the form

$ax^{2} + bx + c = 0$ .
Here, the only equation that has a second degree variable is

$(a-1)x^{2}+(b -2)x + 5c$ and all other equations have third and fourth degree variables.
So, the quadratic equation is

$(a-1)x^{2}+(b -2)x + 5c$ .

The solution of which of the following equations is same as that of

$40-6x=x^2$ ?

0%
$y=(x-6)^2-40$
0%
$y=(x-6)^2+40$
0%
$y=(x+3)^2-49$
0%
$y=(x+3)^2+49$

Explanation

Given,

$40-6x=x^2$

$\Rightarrow x^{2}+6x-40=0$

$\Rightarrow (x+3)^{2}-9-40=0$

$\Rightarrow (x+3)^{2}-49=0$

Hence, the correct option is C.

CBSE Questions for Class 10 Maths Quadratic Equations Quiz 6 - MCQExams.com

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Practice Class 10 Maths Quiz Questions and Answers

CBSE Questions for Class 10 Maths Quadratic Equations Quiz 6 - MCQExams.com

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Practice Class 10 Maths Quiz Questions and Answers

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