Quadratic Equations Test

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Quadratic Equations Test - 40

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Weekly Quiz Competition

Question 1
1 / -0

If $$x^{2}-5x+1=0$$, then $$x^{5}+\dfrac {1}{x^{5}}$$ is equal to

A
$$2424$$

B
$$3232$$

C
$$2525$$

D
None of these

Solution

We have,
$$\begin{aligned}{}{x^2} - 5x + 1 &= 0\\{x^2} + 1 &= 5x\\\frac{{{x^2} + 1}}{x} &= 5\\x + \frac{1}{x}& = 5\quad\quad\quad\dots(i)\end{aligned}$$

On squaring both side of equation $$(i)$$,
$${x}^{2}+\cfrac{1}{x^2}+2=25$$
$$\Rightarrow {x}^{2}+\cfrac{1}{{x}^{2}}=25-2$$
$$\Rightarrow {x}^{2}+\cfrac{1}{{x}^{2}}=23\quad\quad\quad\dots(ii)$$

Multiplying equation $$(i)$$ and $$(ii)$$ we get,
$$\left(x+\cfrac{1}{x}\right)\left({x}^{2}+\cfrac{1}{{x}^{2}}\right)=5\times 23$$
$$\Rightarrow {x}^{3}+\cfrac{1}{{x}^{3}}+x+\cfrac{1}{x}=115$$
$$\Rightarrow {x}^{3}+\cfrac{1}{{x}^{3}}=115-5$$
$$\Rightarrow {x}^{3}+\cfrac{1}{{x}^{3}}=110\quad\quad\quad\dots(iii)$$

Multiplying $$(ii)$$ and $$(iii)$$ we get,
$$\left({x}^{2}+\cfrac{1}{{x}^{2}}\right)\left({x}^{3}+\cfrac{1}{{x}^{3}}\right)=23\times 110$$
$$\Rightarrow {x}^{5}+\cfrac{1}{{x}^{5}}+x+\cfrac{1}{x}=2530$$
$$\Rightarrow {x}^{5}+\cfrac{1}{{x}^{5}}+5=2530$$
$$\Rightarrow {x}^{5}+\cfrac{1}{{x}^{5}}=2525$$
Question 2
1 / -0

If the equation $$(a^2+b^2)x^2-2(ac+bd)x+c^2+d^2=0$$ has equal roots, then

A
$$ab=cd$$

B
$$ad=bc$$

C
$$ad=\sqrt{bc}$$

D
$$ab=\sqrt{cd}$$

Solution

For equation $$\left( { a }^{ 2 }+{ b }^{ 2 } \right) { x }^{ 2 }-2\left( ac+bd \right) x+\left( { c }^{ 2 }+{ d }^{ 2 } \right) =0$$ to has equal real roots, then the condition to be satisfied is,
$${ B }^{ 2 }-4AC=0$$

$$ \Rightarrow 4{ \left( ac+bd \right) }^{ 2 }-4\left( { a }^{ 2 }+{ b }^{ 2 } \right) \left( { c }^{ 2 }+{ d }^{ 2 } \right) =0$$

$$ \Rightarrow 8abcd-4{ a }^{ 2 }{ d }^{ 2 }-4{ b }^{ 2 }{ c }^{ 2 }=0$$

$$ \Rightarrow 4{ \left( ad-bc \right) }^{ 2 }=0$$

$$ \Rightarrow ad=bc$$
Question 3
1 / -0

If the discriminant of quadratic equation $${ b }^{ 2 }-4ac=0$$, then the roots are:

A
Real and equal

B
Roots are equal

C
No real roots

D
Roots are unequal and irrational

Solution

Given discriminant:
$$D= b^{2}-4ac=0$$

Let, $$ax^{2}+bx+c=0$$ be a standard quadratic
equation then roots are -

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

Hence, roots are real and equal.
Question 4
1 / -0

If $$\dfrac{\sqrt{1296}}{x}=\dfrac{x}{2.25}$$ then value $$x$$ is:

A
$$0.9$$

B
$$0.09$$

C
$$9$$

D
$$1.9$$

Solution

we know that, $$\sqrt {1296} = 36$$

$$\dfrac {36}{x} = \dfrac{x}{2.25}$$

$$x ^ 2 = 36 \times 2.25$$

$$x ^ 2 = 81$$

$$x = \pm \sqrt {81}$$

$$x =\pm 9$$
Question 5
1 / -0

Find the value of $$x:$$
$$x^2-4x+4=0$$

A
$$2$$

B
$$1$$

C
$$0$$

D
None of the above

Solution

Given equation is:
$$x^2-4x+4=0$$
It can also be written as:
$$x^2-2(2)(x)+(2)^2=0\dots (1)$$
Above equation is of the form of following identity:
$$(a-b)^2=a^2-2ab+b^2$$
Hence, the equation $$(1)$$ can be written as follows:
$$(x-2)^2=0$$ or
$$(x-2)(x-2)=0$$
$$x=2,2$$
Question 6
1 / -0

If an angle A of a $$\triangle ABC$$ satisfies 5cosA 7.3=0. than the roots of the quadratic equation $${ 9x }^{ 2 }+27x+20=0$$ are :

A
sinA , secA

B
secA , tanA

C
tanA , cosA

D
secA , cotA

Solution
Question 7
1 / -0

If $$\alpha ,\quad \beta $$ are the roots of the equation $${ x }^{ 2 }-2x+4=0,$$ then the value of $${ \alpha }^{ 6 }+{ \beta }^{ 6 }$$ is :

A
64

B
128

C
256

D
None of these
Question 8
1 / -0

If $$x ^ { 2 } - 5 x + 1 = 0$$ then $$\frac { x ^ { 10 } + 1 } { x ^ { 5 } }$$ is equal to:-

A
2424

B
3232

C
2525

D
None of these
Question 9
1 / -0

$$(x+1)^2+(y-1)^2 +(x-5)^2+(y-1)^2$$ has the value equal to

A
$$16$$

B
$$25$$

C
$$36$$

D
$$49$$
Question 10
1 / -0

Which of the following is a quadratic equation ?

A
$$6x^2 = 20 -x^3$$

B
$$x^2-\left ( \dfrac{1}{x^2} \right ) $$ = $$\dfrac{7}{2}$$

C
$$\dfrac{3}{x}$$ $$ = 4$$ $$x^2$$

D
$$5x^2 + 7 = 3x$$

Solution

For $$(A)$$
$$6x^2=20-x^3$$

$$\Rightarrow x^3+6x^2-20=0$$

In this, the maximum power of $$x$$ is $$3$$.
So, it is not a quadratic equation.

For $$(B)$$
$$x^2-\left(\dfrac{1}{x^2}\right)=\dfrac{7}{2}$$

$$\Rightarrow 2x^4-2=7x^2$$

$$\Rightarrow 2x^4-7x^2-2=0$$

In this, the maximum power of $$x$$ is $$4$$.
So, it is not a quadratic equation.

For $$(C)$$
$$\dfrac{3}{x}=4x^2$$

$$\Rightarrow 3=4x^3$$

In this, the maximum power of $$x$$ is $$3$$.
So, it is not a quadratic equation.

For $$(D)$$
$$5x^2+7=3x$$
$$\Rightarrow 5x^2-3x+7=0$$

In this, the maximum power of $$x$$ is $$2$$.
So, it is a quadratic equation.

Hence, $$Op-D$$ is correct.

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Quadratic Equations Test - 40

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Quadratic Equations Test - 40

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SHARING IS CARING

Question 1

$$2424$$

$$3232$$

$$2525$$

None of these

Solution

Question 2

$$ab=cd$$

$$ad=bc$$

$$ad=\sqrt{bc}$$

$$ab=\sqrt{cd}$$

Solution

Question 3

Real and equal

Roots are equal

No real roots

Roots are unequal and irrational

Solution

Question 4

$$0.9$$

$$0.09$$

$$9$$

$$1.9$$

Solution

Question 5

$$2$$

$$1$$

$$0$$

None of the above

Solution

Question 6

sinA , secA

secA , tanA

tanA , cosA

secA , cotA

Solution

Question 7

64

128

256

None of these

Question 8

2424

3232

2525

None of these

Question 9

$$16$$

$$25$$

$$36$$

$$49$$

Question 10

$$6x^2 = 20 -x^3$$

$$x^2-\left ( \dfrac{1}{x^2} \right ) $$ = $$\dfrac{7}{2}$$

$$\dfrac{3}{x}$$ $$ = 4$$ $$x^2$$

$$5x^2 + 7 = 3x$$

Solution

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