MCQ Questions for Class 10 Maths Quadratic Equations Quiz 5

Find all values of parameter

$a$ for which the quadratic equation

$\left( a+1 \right) { x }^{ 2 }+2\left( a+1 \right) x+a-2=0$ has two distinct roots.

0%
$a \in (-1,\infty)$
0%
$a \in (-\infty,-1)$
0%
$a \in (-1,1)$
0%
$a \in (-\infty,\infty)$

Explanation

Discriminant of the given quadratic equation is,

$D = 4\left( a+1 \right) ^{ 2 }-4\left( a+1 \right) \left( a-2 \right) =4\left( a+1 \right) \left( a+1-a+2 \right) =12\left( a+1 \right)$
Now for two distinct roots,

$D>0\Rightarrow a+1>0$

$\Rightarrow a \in (-1,\infty)$ .

Find the discriminant for the given quadratic equation:

$x^2\,+\,x\,+\,1\,=\,0$

0%
$-3$
0%
$-5$
0%
$-7$
0%
$-9$

Explanation

Given equation is,

$x^2 + x + 1 = 0$
We know,

$D = b^2 - 4ac$

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

$\therefore D = (1)^2 - 4(1)(1)$

$\therefore D = -3$

$D$ is the discriminant of

$x^2\,+\,4x\,+\,1\,=\,0$ , then the value of

$D^2$ , is

0%
$100$
0%
$12$
0%
$144$
0%
$10$

Explanation

$x^2 + 4x + 1 = 0$

$D = b^2 - 4ac$

$D = (4)^2 - 4(1)(1)$

$D = 16 - 4$

$D = 12$

$D^2 = {12}^2 = 144$

Find

$m$ , if the quadratic equation

$(m\, -\, 1)\, x^{2}\, -\, 2\, (m\, -\, 1)\, x\, +\, 1\, =\, 0$ has real equal roots.

0%
$1$
0%
$2$
0%
$3$
0%
$4$

Explanation

For the quadratic equation,

$(m\, -\, 1)\, x^{2}\, -\, 2\, (m\, -\, 1)\, x\, +\, 1\, =\, 0$
Discriminant =

${(-2(m-1))}^2 - 4\times(m-1)= 0$ ...... Since roots are real equal roots

$4m^2 -8m + 4 - 4m + 4 = 0$

$4m^2 - 12m + 8 =0$

$m^2 - 3m + 2 = 0$

$\therefore m = 1, 2$
Neglect

$m= 1$ as the equation will not be quadratic anymore.
Hence,

$m = 2$

Find c, if the quadratic equation

$x^{2}\, -\, 2\, (c\, +\, 1)\, x\, +\, c^{2}\, =\, 0$ has real and equal roots.

0%
$c=\cfrac{1}{2}$
0%
$c=-\cfrac{1}{2}$
0%
$c=2$
0%
$c=-2$

Explanation

Since, the roots are real and equal,

$D = 0$

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

$(-2(c+1))^2 - 4c^2 = 0$

$4(c^2 + 2c + 1) - 4c^2 = 0$

$8c + 4 = 0$

$\therefore c = \displaystyle \frac{-1}{2}$

Find the value of discriminant for the following equation.

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

0%
$\Delta\, =\, k^{2}\, -16$
0%
$\Delta\, =\, k^{2}\, + 32$
0%
$\Delta\, =\, k^{2}\, -32$
0%
$\Delta\, =\, k^{2}\,+16$

Explanation

Given equation is:

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

Discriminant =

$b^2 - 4ac$

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

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

$= k^2 - 32$

$\therefore \Delta= k^2-32$

Solve the equation using formula.

$2x^{2}\, +\, \displaystyle \frac{x\, -\, 1}{5}\, =\, 0$

0%
$x\, =\, \displaystyle \frac{-1\, \pm\, \sqrt{10}}{4}$
0%
$x\, =\, \displaystyle \frac{-1\, \pm\, \sqrt{41}}{20}$
0%
$x\, =\, \displaystyle \frac{1\, \pm\, \sqrt{41}}{20}$
0%
$x\, =\, \displaystyle \frac{1\, \pm\, \sqrt{10}}{4}$

Explanation

Given equation is

$2x^{2}\, +\, \displaystyle \frac{x\, -\, 1}{5}\, =\, 0$

$\therefore 10x^2 + x - 1 = 0$
Applying qudratic formula,

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

$a=10, b=1, c=-1$

$\therefore x =\displaystyle \frac{-1 \pm \sqrt{1 + 40}}{20}$

$\therefore x =\displaystyle \frac{ -1 \pm \sqrt{41}}{20}$

$\therefore$ roots of the given equation are

$\displaystyle \frac {-1 \pm \sqrt {41}}{20}$

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

$4{ x }^{ 2 }-\sqrt { 3 } x-5=0$ by the method of completing the square?

0%
$\displaystyle\frac{9}{10}$
0%
$\displaystyle\frac{3}{16}$
0%
$\displaystyle\frac{3}{4}$
0%
$\displaystyle\frac{\sqrt{3}}{4}$

Explanation

We have,

$4{ x }^{ 2 } - \sqrt { 3 } x - 5 = 0$
Divide whole equation by

$4$ ,

$\therefore { x }^{ 2 } - \displaystyle\frac { \sqrt { 3 } }{ 4 } x - \displaystyle\frac { 5 }{ 4 } = 0$

$\therefore { \left( x -\displaystyle\frac { \sqrt { 3 } }{ 8 } \right) }^{ 2 } - \displaystyle\frac { 5 }{ 4 } = \displaystyle\frac { 3 }{ 64 }$

$\Rightarrow 4{ \left( x -\displaystyle\frac { \sqrt { 3 } }{ 8 } \right) }^{ 2 } - 5 - \displaystyle\frac { 3 }{ 16 } = 0$

$\therefore \displaystyle \frac { 3 }{ 16 }$ must be added to make

$4{ x }^{ 2 } - \sqrt { 3 } x - 5 = 0$ a complete square

${b}^{2} -4ac\ge 0$ then the roots of quadratic equation

$a{x}^{2} + bx + c =0$ is-

0%
$\displaystyle\frac{b}{2a}\pm \frac{\sqrt{{b}^{2}-4ac}}{2a}$
0%
$\displaystyle -\frac{b}{2a}\pm \frac{\sqrt{{b}^{2}+4ac}}{2a}$
0%
$\displaystyle\frac{b}{2a}\pm \frac{\sqrt{{b}^{2}+4ac}}{2a}$
0%
$\displaystyle -\frac{b}{2a}\pm \frac{\sqrt{{b}^{2}-4ac}}{2a}$

Explanation

$b^{2}-4ac\geq 0$
Then the roots are real.
Hence
By using quadratic formula, we get the roots as

$\dfrac{-b\pm\sqrt{b^{2}-4ac}}{2a}$
Hence, option 'D' is correct.

State true or false:

$5m^{2}\, +\, m\, =\, 3$ , then

$m\, =\, \displaystyle \frac{-1\, \pm\, \sqrt{61}}{10}$ .

0%
True
0%
False

Explanation

$5m^2 + m = 3$

Dividing by

$5$ on both sides we get,

$\therefore m^2 + \dfrac m5 = \dfrac 35$

Third term =

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

$\left(\dfrac 12 \times \dfrac 15\right)^2 = \dfrac {1}{100}$

Adding

$\dfrac {1}{100}$ on both sides we get,

$m^2 + \dfrac m5 + \dfrac {1}{100} = \dfrac 35 + \dfrac {1}{100}$

$\therefore \left(m+\dfrac {1}{10}\right)^2 = \dfrac{61}{100}$

$\therefore \left (m + \dfrac {1}{10} \right)$ = +

$\dfrac{\sqrt {61}}{10}$

$m = \dfrac{-1 + \sqrt{61}}{10}$ OR

$m = \dfrac{-1 - \sqrt{61}}{10}$

The value of

$p$ for which the equation

$x^2+4=(P+2)x$ has equal roots?

0%
$2,-6$
0%
$-2, 6$
0%
$-2,-6$
0%
$2,6$

Explanation

Given Equation is:

$x^{2}+4=(P+2)x\Rightarrow x^{2}-(P+2)x+4=0\Rightarrow a=1, b=-(P+2),C=4$

Since given equation has equal roots

Therefore

$b^{2}-4ac=0\Rightarrow [-P+2]^{2}-4(4)=0$

$P+2=\pm 4\Rightarrow P=2,-6$

Solve the following quadratic equation by completing square method.

$3p^{2}\, +\, 4\, =\, -7p$ .

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

Explanation

The given equation is

$3p^2+4=-7p$

$\Rightarrow$

$3p^2+7p+4=0$

$\Rightarrow$

$p^2+\dfrac{7}{3}p+\dfrac{4}{3}=0$ [ Dividing both sides by

$3$ ]

$\Rightarrow$

$p^2+\dfrac{7}{3}p=-\dfrac{4}{3}$

$\Rightarrow$

$p^2+2\times p \times\dfrac{7}{6}=-\dfrac{4}{3}$

$\Rightarrow$

$p^2+2\times p \times\dfrac{7}{6}+\left(\dfrac{7}{6}\right)^2=-\dfrac{4}{3}+\left(\dfrac{7}{6}\right)^2$

$\Rightarrow$

$\left(p+\dfrac{7}{6}\right)^2=-\dfrac{4}{3}+\dfrac{49}{36}$

$\Rightarrow$

$\left(p+\dfrac{7}{6}\right)^2=\dfrac{-48+49}{36}$

$\Rightarrow$

$\left(p+\dfrac{7}{6}\right)^2=\dfrac{1}{36}$

$\Rightarrow$

$\left(p+\dfrac{7}{6}\right)^2=\left(\dfrac{1}{6}\right)^2$

$\Rightarrow$

$p+\dfrac{7}{6}=\pm\dfrac{1}{6}$

$\Rightarrow$

$p=\pm\dfrac{ 1}{6}-\dfrac{7}{6}$

$\Rightarrow$

$p=\dfrac{1}{6}-\dfrac{7}{6}$ or

$p=\dfrac{-1}{6}-\dfrac{7}{6}$

$\Rightarrow$

$p=\dfrac{-6}{6}$ or

$p=\dfrac{-8}{6}$

$\Rightarrow$

$p=-1$ or

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

Hence the solution is

$-1$ and

$\dfrac{-4}{3}$

Solve the following quadratic equation by completing square.

$2y^{2}\, +\, 5y\, +\, 1\, =\, 0$ , then

$y\, =\, \displaystyle \frac{-5\, \pm\, \sqrt{23}}{2}$ . State true or false.

0%
True
0%
False

Explanation

Given,

$2y^2+5y+1=0$
Compare the given equation with

$ax^2+bx+c=0$
Since,

$x=\frac {-b\pm\sqrt {b^2-4ac}}{2a}$
In given quadratic equation

$a=2 , b=5 , c=1$

$\therefore y=\frac {-5\pm\sqrt {(5)^2-4\times 2 \times 1}}{2\times 2}$

$\therefore y=\frac {-5\pm\sqrt {17}}{4}$
equation is false.
Answer is 0.

Consider the quadratic equations

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

$\displaystyle \left ( a+c \right )\left ( ax^{2}+2bx+c \right )-2\left ( ac-b^{2} \right )\left ( x^{2}+1 \right )=0$ If the roots of one are real (complex) then the roots of the other are complex (real.)

0%
True
0%
False

Explanation

$\displaystyle B^{2}-4AC> 0$ for 1st equation.

$\displaystyle \therefore 4\left ( c^{2}-ab \right )> 0 \Rightarrow c^{2}-ab$ is + ive.

$\displaystyle B^{2}-4AC$
for the 2nd equation is

$\displaystyle 4\left ( a+b \right )^{2}-4\left ( a^{2}+b^{2}+2c^{2} \right )$

$\displaystyle =4\left [ 2ab-2c^{2} \right ]=-8\left ( c^{2}-ab \right )=-ive.$ as

$\displaystyle =c^{2}-ab$ is+ive.
Hence the roots of the second equation are imaginary.

The ratio of the roots of the equation

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

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

${ D }_{ 1 }$ and

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

$a{ x }^{ 2 }+bx+c=0$ and

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

${ D }_{ 1 }:{ D }_{ 2 }$ is equal to

0%
$\displaystyle \frac { { a }^{ 2 } }{ { p }^{ 2 } }$
0%
$\displaystyle \frac { { b }^{ 2 } }{ { q }^{ 2 } }$
0%
$\displaystyle \frac { { c }^{ 2 } }{ { r }^{ 2 } }$
0%
None of these

Explanation

Let

$\displaystyle { \alpha }_{ 1 },{ \beta }_{ 1 }$ be the roots of

$a{ x }^{ 2 }+bx+c=0$ and

${ \alpha }_{ 2 },{ \beta }_{ 2 }$ be the roots of

$p{ x }^{ 2 }+qx+r=0$ . Then

$\displaystyle \frac { { \alpha }_{ 1 } }{ { \beta }_{ 1 } } =\frac { { \alpha }_{ 2 } }{ { \beta }_{ 2 } }$ (given)

$\displaystyle \frac { { \alpha }_{ 1 }+{ \beta }_{ 1 } }{ { \alpha }_{ 1 }-{ \beta }_{ 1 } } =\frac { { \alpha }_{ 2 }+{ \beta }_{ 2 } }{ { \alpha }_{ 2 }-{ \beta }_{ 2 } }$ (Applying componendo and dividendo)

$\displaystyle \frac { { \left( { \alpha }_{ 1 }+{ \beta }_{ 1 } \right) }^{ 2 } }{ { \left( { \alpha }_{ 1 }-{ \beta }_{ 1 } \right) }^{ 2 } } =\frac { { \left( { \alpha }_{ 2 }+{ \beta }_{ 2 } \right) }^{ 2 } }{ { \left( { \alpha }_{ 2 }-{ \beta }_{ 2 } \right) }^{ 2 } }$

$\displaystyle\frac { { \left( { \alpha }_{ 1 }+{ \beta }_{ 1 } \right) }^{ 2 } }{ { \left( { \alpha }_{ 1 }+{ \beta }_{ 1 } \right) }^{ 2 }-4{ \alpha }_{ 1 }{ \beta }_{ 1 } } =\frac { { \left( { \alpha }_{ 2 }+{ \beta }_{ 2 } \right) }^{ 2 } }{ { \left( { \alpha }_{ 2 }+{ \beta }_{ 2 } \right) }^{ 2 }-4{ \alpha }_{ 2 }{ \beta }_{ 2 } }$

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

$\Rightarrow \dfrac { { b }^{ 2 } }{ { D }_{ 1 } } =\dfrac { { q }^{ 2 } }{ { D }_{ 2 } }$

$\Rightarrow \dfrac { { D }_{ 1 } }{ { D }_{ 2 } } =\dfrac { { b }^{ 2 } }{ { q }^{ 2 } }$

The equations to a pair of opposite sides of a parallelogram are

$\displaystyle x^{2}-5x+6=0 \ and \ y^{2}-6y+5=0$ the equation of a diagonal can be

0%
$\displaystyle x+4y=13$
0%
$\displaystyle 4x+y=13$
0%
$\displaystyle y=4x-7$
0%
$\displaystyle y=4x+7$

Explanation

$\displaystyle x^{2}-5x+6=0$

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

$\Rightarrow x=2,x=3$

and

$\displaystyle y^{2}-6y+5=0$

$\Rightarrow y=1, y=5$

Vertices of the parallelogram are

$A(2, 1), \ B(3, 1),$

$C(3, 5),\ D(2, 5)$

equation of

$AC$ is

$\displaystyle y-1=\frac{5-1}{3-2}\left ( x-2 \right )$

$\Rightarrow y=4x-7$

equation of

$BD$ is

$\displaystyle y-1=\frac{5-1}{3-2}\left ( x-3 \right )$

$\Rightarrow 4x+y=13$

Find the roots of each of the following quadratic equations by the method of completing the squares

$2x^2 - 5x + 3 = 0$

0%
$x=2,\,\,x=-7$
0%
$x=-1,\,\,x=3$
0%
$x=\displaystyle 1,x= \frac{3}{2}$
0%
$x=\displaystyle 1, x=\frac{1}{2}$

Explanation

Dividing the entire equation by 2, we get

$x^{2}-\dfrac{5x}{2}+\dfrac{3}{2}=0$

$(x-\dfrac{5}{4})^{2}-\dfrac{25}{16}+\dfrac{24}{16}=0$

$(x-\dfrac{5}{4})^{2}=\dfrac{1}{16}$

$x-\dfrac{5}{4}=\pm\dfrac{1}{4}$

$x=\dfrac{5\pm1}{4}$
Hence

$x=\dfrac{3}{2}$ and

$x=1$

Find the solutions of

$3x^2 - 2\sqrt 6x + 2 = 0$ by the method of completing the squares when

$x$ is an irrational number.

0%
$x=\pm\sqrt{\dfrac{7}{2}}$
0%
$x=\pm\sqrt{\dfrac{2}{3}}$
0%
$x=\sqrt{\dfrac{2}{3}}$
0%
$x=\sqrt{\dfrac{7}{2}}$

Explanation

$3x^{2}-2\sqrt{6}x+2=0$
Dividing the entire equation, by

$3$ , we get

$x^{2}-\dfrac{2\sqrt{6}}{3}+\dfrac{2}{3}=0$

$\left (x-\dfrac{\sqrt{6}}{3}\right)^{2}-\dfrac{6}{9}\dfrac{2}{3}=0$

$\left (x-\dfrac{\sqrt{6}}{3}\right)^{2}=0$

$x=\dfrac{\sqrt{6}}{3}$
Hence, no rational roots.

Find

$x$ by solving the given equation:

$\displaystyle \frac{x + 3}{x + 2} = \frac{3x - 7}{2x - 3}$

0%
$5,-1$
0%
$-5,-1$
0%
$5,1$
0%
$-5,1$

Explanation

Given,

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

$\Rightarrow 3x^2-7x+6x-14=2x^2-3x+6x-9$

$\Rightarrow 3x^2-2x^2-7x+6x+3x-6x-14+9=0$

$\Rightarrow x^2-4x-5=0$

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

$\Rightarrow x(x-5)+1(x-5)=0$

$\Rightarrow (x-5)(x+1)=0$

$\Rightarrow x-5=0$ and

$x+1=0$

$\Rightarrow x=5$ and

$x=-1$

Solve for

$y$ :

$\sqrt 7 y^2 - 6y - 13 \sqrt 7 = 0$

0%
$\sqrt 7, 2 \sqrt 7$
0%
$\displaystyle 3, \frac{2}{\sqrt 7}$
0%
$\displaystyle \frac{13}{\sqrt 7}, -\sqrt 7$
0%
None of these

Explanation

Given quadratic equation is $\sqrt 7 y^2 - 6y - 13 \sqrt 7 = 0$

Comparing it with the standard form of equation $ax^2+bx+c$ we get $a=7,b=6,c=13\sqrt 7$
Therefore, $x= \dfrac { -b\pm \sqrt { b^ 2-4ac } }{ 2a }$
$= \dfrac { -(-6)\pm \sqrt { (-6)^{ 2 }-4\times \sqrt { 7 } \times \left( -13\sqrt { 7 } \right) } }{ 2\sqrt { 7 } }$
$=\dfrac { 6\pm \sqrt { 36+364 } }{ 2\sqrt { 7 } }$
$= \dfrac { 6\pm \sqrt { 400 } }{ 2\sqrt { 7 } }$
$=\dfrac { 6\pm 20 }{ 2\sqrt { 7 } }$
$=\dfrac { 6- 20 }{ 2\sqrt { 7 } }$
$=\dfrac { \\- 14 }{ 2\sqrt { 7 } }$
$=\dfrac { \\ -7 }{ \sqrt { 7 } }$
$=\dfrac { \\ -7\times \sqrt { 7 } }{ \sqrt { 7 } \times \sqrt { 7 } }$
$=-\sqrt 7$
$=\dfrac { 6+ 20 }{ 2\sqrt { 7 } }$
$=\dfrac { \\ 26 }{ 2\sqrt { 7 } }$
$=\dfrac { \\ 13 }{ \sqrt { 7 } }$

Find the roots of each of the following quadratic equations by the method of completing the sqaures:

$x^2 - 6x + 4 = 0$

0%
$9\pm \sqrt 8$
0%
$6 \pm \sqrt 7$
0%
$3 \pm \sqrt 5$
0%
$1\pm \sqrt 11$

Explanation

Given,

$x^2 - 6x + 4 = 0$

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

$\Rightarrow (x-3)^2=5$

$\Rightarrow x-3=\pm \sqrt 5$

$\Rightarrow x=3\pm \sqrt 5$

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

$2x^2 - 2\sqrt 2x + 1 = 0$

0%
$\displaystyle -\frac{1}{\sqrt 2}, \frac{1}{\sqrt{2}}$
0%
$\displaystyle \frac{1}{\sqrt 2}, \frac{1}{\sqrt{2}}$ .
0%
$\displaystyle -\frac{1}{ 2}, \frac{1}{{2}}$
0%
None of these

Explanation

The quadratic equation is $2x^2 - 2\sqrt 2x + 1 = 0$ .
Comparing it with standard form of quadratic equation $ax^2+bx+c=0$ ,we get $a=2,b=- 2\sqrt 2,c=1$ Therefore the roots of the quadratic equation are,
$x= \dfrac { -b\pm \sqrt { b^{ 2 }-4ac } }{ 2a }$
$= \dfrac { -\left( -2\sqrt { 2 } \right) \pm \sqrt { \left( -2\sqrt { 2 } \right) ^{ 2 }-4\times 2\times 1 } }{ 2\times 2 }$
$=\dfrac { \left( 2\sqrt { 2 } \right) \pm \sqrt { 8-8 } }{ 4 }$
$=\dfrac { 2\sqrt { 2 } \pm 0 }{ 4 }$
$=\dfrac { 2\sqrt { 2 } }{ 4 }$
$=\dfrac { \sqrt { 2 } }{ 2 }$
$=\dfrac{ 1 }{ \sqrt { 2 } }$

Determine the value of

$k$ for which the quadratic equation

$4x^2 - 3kx + 1 = 0$ has equal roots.

0%
$\displaystyle \pm \frac{2}{3}$
0%
$\displaystyle \pm \frac{4}{3}$
0%
$\displaystyle \pm 4$
0%
$\displaystyle \pm 6$

Explanation

Nature of the roots for a quadratic equation

$ax^2+bx+c=0$ , can be determined by its discriminant

$D=b^2-4ac$
Here, given quadratic equation is

$4x^2 - 3kx + 1 = 0$
Putting the value of

$a,b,c$ , we get

$D=(-3k)^2-4\times 4\times 1$

$\Rightarrow 9k^2-16=0$

$\Rightarrow 9k^2=16$

$\Rightarrow k^2=\cfrac{16}{9}$

$\Rightarrow k=\pm \cfrac{4}{3}$

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

$x^2 - 16x + 64 = 0$

0%
$-8,-8$
0%
$8,8$
0%
$16,16$
0%
$-16,-16$

Explanation

Given equation is

$x^2-16x+6=0$

$\Rightarrow x^{2}-2(8)x+8^{2}=0$
This is the case of

$(x-a)^{2}=x^{2}-2ax+a^{2}$
Therefore,

$(x-8)^{2}=0$

$\Rightarrow x=8,8$

Find the solutions of

$3x^2 - 2\sqrt 6x + 2 = 0$ by the method of completing the squares when

$x$ is a real number.

0%
$x=\displaystyle- \sqrt{\frac{2}{3}}$
0%
$x=\displaystyle \pm \sqrt{\frac{2}{3}}$
0%
Cannot be determined
0%
None of these

Explanation

Given,

$3x^2 - 2\sqrt 6x + 2 = 0$

$\Rightarrow x^2-\dfrac{2\sqrt 6}{3}x=-\dfrac{2}{3}$

$\Rightarrow x^2-\dfrac{2\sqrt 6}{3} x+\left(\dfrac{\sqrt 6}{3}\right)^2=\left(\dfrac{\sqrt 6}{3}\right)^2-\dfrac{2}{3}$

$\Rightarrow x^{ 2 }-2 x\times \dfrac { \sqrt { 6 } }{ 3 } +\left(\dfrac { \sqrt { 6 } }{ 3 } \right)^{ 2 }=\dfrac { 6 }{ 9 } -\dfrac { 2 }{ 3 }$

$\Rightarrow \left(x-\dfrac { \sqrt { 6 } }{ 3 } \right)^{ 2 }=\dfrac { 6-6 }{ 9 }$

$\Rightarrow x-\dfrac { \sqrt { 6 } }{ 3 } =0$

$\Rightarrow x=\dfrac { \sqrt { 6 } }{ 3 }$

$\Rightarrow x=\dfrac { \sqrt { 2 } \times \sqrt { 3 } }{ \sqrt { 3 } \times \sqrt { 3 } }$

$\Rightarrow x=\sqrt { \dfrac { 2 }{ 3 } }$

Determine k such that the quadratic equation

$x^2 + 7(3 + 2k)+(1 + 3k)x = 0$ has equal roots

0%
$2, 7$
0%
$7, 5$
0%
$2,$ $\displaystyle - \frac{10}{9}$
0%
None of these

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

$\displaystyle \frac{x}{x - 1} + \frac{x - 1}{x} = 4, x \neq 0, 1$

0%
$\displaystyle \frac{2 \pm \sqrt 3}{4}$
0%
$\displaystyle \frac{-1 \pm \sqrt 3}{2}$
0%
$\displaystyle \frac{1 \pm \sqrt 3}{1}$
0%
$\displaystyle \frac{-2 \pm \sqrt 3}{4}$

Explanation

Given, $\dfrac{x}{x - 1} + \dfrac{x - 1}{x} = 4$
$\Rightarrow \dfrac { x^{ 2 }+(x-1)^{ 2 } }{ x(x-1) } =4$
$\Rightarrow x^2+x^2-2x+1=4x^2-4x$
$\Rightarrow 2x^2-4x^2-2x+4x+1=0$
$\Rightarrow -2x^2+2x+1=0$
$\Rightarrow 2x^2-2x-1=0$
Comparing $2x^2-2x-1=0$ with $ax^2+bx+c=0$ , we get $a=2, b=-2. c=-1$
Now the roots are $=$ $\dfrac { -b \pm \sqrt { b^{ 2 }-4ac } }{ 2a }$
$=$ $\dfrac { -(-2)\pm \sqrt { (-2)^ 2-4\times 2\times (-1) } }{ 2\times 1 }$
$=$ $\dfrac { 2\pm \sqrt { 4+8 } }{ 2 }$
$=$ $\dfrac { 2\pm \sqrt { 12 } }{ 2 }$
$=$ $\dfrac { 2\pm 2\sqrt { 3 } }{ 2 }$
$=$ $1\pm \sqrt 3$

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

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

0%
$-1, 3, 1\pm \sqrt{2}$
0%
$1, 3, -1\pm \sqrt{2}$
0%
$1, 3, 1\pm \sqrt{2}$
0%
None of these

Explanation

Given,

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

Let

$(x^{2}-2x)$ =y

Then

$y^{2}-4y+3=0$

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

$\Rightarrow y(y-3)-1(y-3)=0$

$\Rightarrow y=3$ &

$y = 1$

Place the

$(x^{2}-2x)$

$=y$ , If

$y = 3$

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

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

$\Rightarrow x(x-3)+(x-3)=0$

$\Rightarrow x=3 ,x=1$

When

$y = 1$

Then

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

Then

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

Then roots are

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

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

Then roots are

$1,3, 1\pm\sqrt2$ .

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

$\displaystyle x + \frac{1}{x} = 3, x \neq 0$

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$\displaystyle \frac{3 \pm \sqrt {13}}{2}$
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$\displaystyle \frac{3 \pm \sqrt 5}{2}$
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$\displaystyle \frac{-3 + \sqrt 5}{2}$
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$\displaystyle \frac{-3 \pm \sqrt {13}}{2}$

Explanation

Given, $x+\dfrac{1}{x}=3$
$\Rightarrow \dfrac{x+1}{x}=3$
$\Rightarrow x^2-3x+1=0$
Comparing $x^2-3x+1=0$ with $ax^2+bx+c=0$ , we get $a=1, b=3. c=14$
Now the roots are $=$ $\dfrac { -b \pm \sqrt { b^{ 2 }-4ac } }{ 2a }$
$=$ $\dfrac { -(-3)\pm \sqrt { (3)^ 2-4\times 1\times 1 } }{ 2\times 1 }$
$=$ $\dfrac { 3\pm \sqrt { 9-4 } }{ 2 }$
$=$ $\dfrac { 3\pm \sqrt { 5 } }{ 2 }$
$=$ $\dfrac { 3+ \sqrt { 5 } }{ 2 }$ or $\dfrac { 3- \sqrt { 5 } }{ 2 }$

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

$\displaystyle \frac{x - 3}{x + 3} - \frac{x + 3}{x - 3} = 6\frac{6}{7}, x \neq - 3, 3$

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$4,\displaystyle \frac{9}{4}$
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$-4,\displaystyle \frac{4}{9}$
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$-4,-\displaystyle \frac{4}{9}$
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$-4,\displaystyle \frac{9}{4}$

Explanation

Given,

$\dfrac{x-3}{x+3}-\dfrac{x+3}{x-3}=6\dfrac{6}{7}$

$\Rightarrow (x-3)^{2}-(x+3)^{2}=\dfrac{48}{7}(x+3)(x-3)$

$\Rightarrow 7\left [ \left ( x^{2}-6x+9-x^{2}-6x-9 \right ) \right ]=48\left ( x^{2}-9 \right )$

$\Rightarrow 7(-12x)=48x^{2}-432$

$\Rightarrow 48x^{2}+84x-432=0$

$\Rightarrow 4x^{2}+7x-36=0$

$\Rightarrow 4x^{2}+16x-9x-36=0$

$\Rightarrow 4x(x+4)-9x(x+4)=0$

$\Rightarrow (x+4)(4x-9)=0$

$\Rightarrow x+4=0$

$\Rightarrow x=-4$

$\Rightarrow 4x-9=0$

$\Rightarrow x=\dfrac{9}{4}=2\dfrac{1}{4}$

CBSE Questions for Class 10 Maths Quadratic Equations Quiz 5 - MCQExams.com

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Practice Class 10 Maths Quiz Questions and Answers

CBSE Questions for Class 10 Maths Quadratic Equations Quiz 5 - MCQExams.com

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Practice Class 10 Maths Quiz Questions and Answers

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