Quadratic Equations Test

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Quadratic Equations Test - 30

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Weekly Quiz Competition

Question 1
1 / -0

The roots of the equation are $$lx^2 + nx+n=0$$ are in the ratio $$p:q$$. Then

A
$$\displaystyle\sqrt{ \frac {p} {q}} +\sqrt{\frac {q} {p}} +\sqrt{\frac {n} {l}} =0$$

B
$$\displaystyle\sqrt{ \frac {p} {q}} -\sqrt{\frac {q} {p}} +\sqrt{\frac {n} {l}} =0$$

C
$$\displaystyle \sqrt{\frac {p} {q}} -\sqrt{\frac {q} {p}} -\sqrt{\frac {n} {l}} =0$$

D
$$\displaystyle \sqrt{\frac {p} {q}} +\sqrt{\frac {q} {p}} +\sqrt{\frac {n} {l}} =1$$

Solution

$$l{ x }^{ 2 }+nx+n=0$$
Roots are in ratio $$p:q$$
Let roots be $$ap,aq$$
$$ap+pq=\cfrac { -n }{ l } ,{ a }^{ 2 }pq=\cfrac { n }{ l } $$
$$a=\sqrt { \cfrac { n }{ pql } } $$
$$\therefore \left( p+q \right) \sqrt { \cfrac { n }{ pql } } =\cfrac { -n }{ l } $$
$$\sqrt { \cfrac { n }{ l } } \left( \cfrac { p+q }{ \sqrt { pq } } \right) =-{ \left( \sqrt { \cfrac { n }{ l } } \right) }^{ 2 }$$
$$\sqrt { \cfrac { p }{ q } } +\sqrt { \cfrac { q }{ p } } =-\sqrt { \cfrac { n }{ l } } $$
$$\therefore \sqrt { \cfrac { p }{ q } } +\sqrt { \cfrac { q }{ p } } +\sqrt { \cfrac { n }{ l } } =0$$
Question 2
1 / -0

$$ (x - a)(x - b) + (x - c)(x - a) + (x - c)(x - b) = 0$$ then its roots are

A
Real

B
Complex

C
Imaginary

D
Rational

Solution

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

$$\Rightarrow 3x^2 - 2(a + b + c)x + ab + bc + ca = 0$$

$$D = 4(a + b + c)^2 - 12(ab + bc + ca)$$

$$D = 4a^2 + 4b^2 + 4c^2+8ab+8bc+8ac - 12(ab + bc + ca)$$

$$= 4(a^2 + b^2+ c^2 - ab - bc - ca)$$

$$=2[{ (a-b)^{ 2 } } +{ (b-c)^{ 2 } } +{ (c-a)^{ 2 } } ]\ge 0$$

So roots are real
Question 3
1 / -0

Find the roots of the following quadratic equation by using the quadratic formula
$$2x^2 - 2\sqrt 2x + 1 = 0$$

A
$$\displaystyle -\frac{1}{\sqrt 2}, \frac{1}{\sqrt{2}}$$

B
$$\displaystyle \frac{1}{\sqrt 2}, \frac{1}{\sqrt{2}}$$.

C
$$\displaystyle -\frac{1}{ 2}, \frac{1}{{2}}$$

D
None of these

Solution

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,
$$x= \dfrac { -b\pm \sqrt { b^{ 2 }-4ac } }{ 2a } $$
$$= \dfrac { -\left( -2\sqrt { 2 } \right) \pm \sqrt { \left( -2\sqrt { 2 } \right) ^{ 2 }-4\times 2\times 1 } }{ 2\times 2 } $$
$$=\dfrac { \left( 2\sqrt { 2 } \right) \pm \sqrt { 8-8 } }{ 4 } $$
$$=\dfrac { 2\sqrt { 2 } \pm 0 }{ 4 } $$
$$=\dfrac { 2\sqrt { 2 } }{ 4 } $$
$$=\dfrac { \sqrt { 2 } }{ 2 } $$
$$=\dfrac{ 1 }{ \sqrt { 2 } } $$
Question 4
1 / -0

Find the roots of the following quadratic equation by using the quadratic formula
$$\displaystyle x + \frac{1}{x} = 3, x \neq 0$$

A
$$\displaystyle \frac{3 \pm \sqrt {13}}{2}$$

B
$$\displaystyle \frac{3 \pm \sqrt 5}{2}$$

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

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

Solution

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 $$\dfrac { 3- \sqrt { 5 } }{ 2 }$$
Question 5
1 / -0

Find the roots of the following quadratic equation by using the quadratic formula
$$x^2 - 16x + 64 = 0$$

A
$$-8,-8$$

B
$$8,8$$

C
$$16,16$$

D
$$-16,-16$$

Solution

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, $$(x-8)^{2}=0$$
$$\Rightarrow x=8,8$$
Question 6
1 / -0

Find the roots of the following quadratic equations by using the quadratic formula
$$\displaystyle \frac{x - 3}{x + 3} - \frac{x + 3}{x - 3} = 6\frac{6}{7}, x \neq - 3, 3$$

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

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

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

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

Solution

Given, $$\dfrac{x-3}{x+3}-\dfrac{x+3}{x-3}=6\dfrac{6}{7}$$
$$\Rightarrow (x-3)^{2}-(x+3)^{2}=\dfrac{48}{7}(x+3)(x-3)$$
$$\Rightarrow 7\left [ \left ( x^{2}-6x+9-x^{2}-6x-9 \right ) \right ]=48\left ( x^{2}-9 \right )$$
$$\Rightarrow 7(-12x)=48x^{2}-432$$
$$\Rightarrow 48x^{2}+84x-432=0$$
$$\Rightarrow 4x^{2}+7x-36=0$$
$$\Rightarrow 4x^{2}+16x-9x-36=0$$
$$\Rightarrow 4x(x+4)-9x(x+4)=0$$
$$\Rightarrow (x+4)(4x-9)=0$$
$$\Rightarrow x+4=0$$
$$\Rightarrow x=-4$$
$$\Rightarrow 4x-9=0$$
$$\Rightarrow x=\dfrac{9}{4}=2\dfrac{1}{4}$$
Question 7
1 / -0

The discrimiant of the quadratic equation $$\displaystyle 3\sqrt{3}x^{2}+10x+\sqrt{3}=0$$ is

A
$$8$$

B
$$64$$

C
$$\displaystyle -\frac{1}{3\sqrt{3}}$$

D
$$\displaystyle -\sqrt{3}$$

Solution

Given equation is:
$$\displaystyle 3\sqrt{3}x^{2}+10x+\sqrt{3}=0$$
$$D=b^2-4ac$$
$$\therefore \displaystyle D=(10)^{2}-4(3\sqrt{3})(\sqrt{3})$$
$$= 100 - 36$$
$$= 64$$
Question 8
1 / -0

Find the solutions of $$3x^2 - 2\sqrt 6x + 2 = 0$$ by the method of completing the squares when $$x$$ is a real number.

A
$$x=\displaystyle- \sqrt{\frac{2}{3}}$$

B
$$x=\displaystyle \pm \sqrt{\frac{2}{3}}$$

C
Cannot be determined

D
None of these

Solution

Given, $$ 3x^2 - 2\sqrt 6x + 2 = 0$$
$$\Rightarrow x^2-\dfrac{2\sqrt 6}{3}x=-\dfrac{2}{3}$$
$$\Rightarrow x^2-\dfrac{2\sqrt 6}{3} x+\left(\dfrac{\sqrt 6}{3}\right)^2=\left(\dfrac{\sqrt 6}{3}\right)^2-\dfrac{2}{3}$$
$$\Rightarrow x^{ 2 }-2 x\times \dfrac { \sqrt { 6 } }{ 3 } +\left(\dfrac { \sqrt { 6 } }{ 3 } \right)^{ 2 }=\dfrac { 6 }{ 9 } -\dfrac { 2 }{ 3 } $$
$$\Rightarrow \left(x-\dfrac { \sqrt { 6 } }{ 3 } \right)^{ 2 }=\dfrac { 6-6 }{ 9 }$$
$$\Rightarrow x-\dfrac { \sqrt { 6 } }{ 3 } =0$$
$$\Rightarrow x=\dfrac { \sqrt { 6 } }{ 3 }$$
$$\Rightarrow x=\dfrac { \sqrt { 2 } \times \sqrt { 3 } }{ \sqrt { 3 } \times \sqrt { 3 } }$$
$$\Rightarrow x=\sqrt { \dfrac { 2 }{ 3 } } $$
Question 9
1 / -0

Let $$\displaystyle \alpha ,\beta $$ and $$\displaystyle \gamma $$ be the roots of the equation $$( x-1) \displaystyle (x^{2}+x-3)=0.$$ Then, the value of $$\displaystyle (\alpha +\beta )$$ is

A
$$-1$$

B
$$0$$

C
$$3$$

D
$$2$$

Solution

Roots of the equation are $$1,\dfrac{-1+\sqrt{13}}{2},\dfrac{-1-\sqrt{13}}{2}$$

As given options are real, we will take 2nd and 3rd roots as $$\alpha, \beta $$

$$\Rightarrow \alpha +\beta =\dfrac{-1+\sqrt{13}}{2}+\dfrac{-1-\sqrt{13}}{2}$$

$$\therefore \alpha +\beta =-1$$
Question 10
1 / -0

If $$\displaystyle\alpha, \beta $$ are roots of $$\displaystyle ax^{2}-2bx+c=0$$ then $$\displaystyle a^{3}\beta ^{3}+a^{2}\beta ^{2}+a^{3}\beta^2$$ is

A
$$\displaystyle \frac{c^{2}\left ( c+2b \right )}{a^{3}}$$

B
$$\displaystyle \frac{bc^{2}}{a^{3}}$$

C
$$\displaystyle \frac{c^{2}}{a^{3}}$$

D
None

Solution

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 $$A$$

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Quadratic Equations Test - 30

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Quadratic Equations Test - 30

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Question 1

$$\displaystyle\sqrt{ \frac {p} {q}} +\sqrt{\frac {q} {p}} +\sqrt{\frac {n} {l}} =0$$

$$\displaystyle\sqrt{ \frac {p} {q}} -\sqrt{\frac {q} {p}} +\sqrt{\frac {n} {l}} =0$$

$$\displaystyle \sqrt{\frac {p} {q}} -\sqrt{\frac {q} {p}} -\sqrt{\frac {n} {l}} =0$$

$$\displaystyle \sqrt{\frac {p} {q}} +\sqrt{\frac {q} {p}} +\sqrt{\frac {n} {l}} =1$$

Solution

Question 2

Real

Complex

Imaginary

Rational

Solution

Question 3

$$\displaystyle -\frac{1}{\sqrt 2}, \frac{1}{\sqrt{2}}$$

$$\displaystyle \frac{1}{\sqrt 2}, \frac{1}{\sqrt{2}}$$.

$$\displaystyle -\frac{1}{ 2}, \frac{1}{{2}}$$

None of these

Solution

Question 4

$$\displaystyle \frac{3 \pm \sqrt {13}}{2}$$

$$\displaystyle \frac{3 \pm \sqrt 5}{2}$$

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

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

Solution

Question 5

$$-8,-8$$

$$8,8$$

$$16,16$$

$$-16,-16$$

Solution

Question 6

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

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

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

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

Solution

Question 7

$$8$$

$$64$$

$$\displaystyle -\frac{1}{3\sqrt{3}}$$

$$\displaystyle -\sqrt{3}$$

Solution

Question 8

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

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

Cannot be determined

None of these

Solution

Question 9

$$-1$$

$$0$$

$$3$$

$$2$$

Solution

Question 10

$$\displaystyle \frac{c^{2}\left ( c+2b \right )}{a^{3}}$$

$$\displaystyle \frac{bc^{2}}{a^{3}}$$

$$\displaystyle \frac{c^{2}}{a^{3}}$$

None

Solution

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