Quadratic Equations Test

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Quadratic Equations Test - 35

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Weekly Quiz Competition

Question 1
1 / -0

The number of values of $$a$$ for which $$(a^2-3a+2)x^2+(a^2-5a+6)x+a^2-4=0$$ is an identity in $$x$$ is-

A
$$0$$

B
$$2$$

C
$$1$$

D
$$3$$

Solution

Equation is a identity if all its coefficients are equal to zero simultaneously

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

$$\Rightarrow a=2,1$$

$$\Rightarrow a^{2}-5a+6=0$$

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

$$\Rightarrow a=\pm 2$$

$$\Rightarrow$$coefficients are equal to zero simultaneously when $$a=2$$

Therefore only one value of 'a' the equation is identity in $$x$$
Question 2
1 / -0

Solve the following quadratic equation by completing the square: $$ x^2+(\sqrt{3}+1)x+\sqrt{3}=0$$

A
$$\left \{ \sqrt{3}, 1\right \}$$

B
$$\left \{ \sqrt{3}, 2\right \}$$

C
$$\left \{ \sqrt{3}, 3\right \}$$

D
None of these

Solution

Given equation is $$x^2-(\sqrt{3}+1)x+\sqrt{3}=0$$

$$\Rightarrow$$ $$x^2-(\sqrt{3}+1)x=-\sqrt{3}$$

Now, adding $$\left(\dfrac{\sqrt{3}+1}{2}\right)^2$$ on both sides,

$$\Rightarrow$$ $$x^2-2\times x\times \left(\dfrac{\sqrt{3}+1}{2}\right)+\left(\dfrac{\sqrt{3}+1}{2}\right)^2=-\sqrt{3}+\left(\dfrac{\sqrt{3}+1}{2}\right)^2$$

$$\Rightarrow$$ $$\left[x-\dfrac{(\sqrt{3}+1)}{2}\right]^2=\left[\dfrac{(\sqrt{3}+1)^2}{4}-\sqrt{3}\right]$$

$$\Rightarrow$$ $$\left[x-\dfrac{(\sqrt{3}+1)}{2}\right]^2=\left[\dfrac{3+2\sqrt{3}+1-4\sqrt{3}}{4}\right]$$

$$\Rightarrow$$ $$\left[x-\dfrac{(\sqrt{3}+1)}{2}\right]^2=\left[\dfrac{3-2\sqrt{3}+1}{4}\right]$$

$$\Rightarrow$$ $$\left[x-\dfrac{(\sqrt{3}+1)}{2}\right]^2=\left[\dfrac{\sqrt{3}-1}{2}\right]^2$$

Taking square root on both sides,

$$\Rightarrow$$ $$\left[x-\dfrac{(\sqrt{3}+1)}{2}\right]=\pm\dfrac{\sqrt{3}-1}{2}$$

$$\Rightarrow$$ $$\left[x-\dfrac{(\sqrt{3}+1)}{2}\right]=\dfrac{\sqrt{3}-1}{2}$$
and $$\left[x-\dfrac{(\sqrt{3}+1)}{2}\right]=-\dfrac{\sqrt{3}-1}{2}$$

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

$$\therefore$$ $$x=\sqrt{3}$$ and $$x=1$$
Question 3
1 / -0

Solve the following quadratic equation by completing the square: $$ 4x^2+4bx-(a^2-b^2)=0$$

A
$$\left \{ \dfrac{a-2}{2}, -\left ( \dfrac{a+2}{2} \right ) \right \}$$

B
$$\left \{ \dfrac{a+2}{2}, -\left ( \dfrac{a+2}{2} \right ) \right \}$$

C
$$\left \{ \dfrac{a-2}{2}, -\left ( \dfrac{a-2}{2} \right ) \right \}$$

D
None of these

Solution

Given equation is $$4x^2+4bx-(a^2-b^2)=0$$

$$\Rightarrow$$ $$x^2+bx-\dfrac{(a^2-b^2)}{4}=0$$

$$\Rightarrow$$ $$x^2+bx=\dfrac{(a^2-b^2)}{4}$$

Now, $$\left(\dfrac{b}{2}\right)^2=\dfrac{b^2}{4}$$

Adding $$\dfrac{b^2}{4}$$ on bothe sides,

$$\Rightarrow$$ $$x^2+bx+\dfrac{b^2}{4}=\dfrac{a^2}{4}-\dfrac{b^2}{4}+\dfrac{b^2}{4}$$

$$\Rightarrow$$ $$\left(x+\dfrac{b}{2}\right)^2=\dfrac{a^2}{4}$$

Taking square root on both sides

$$\Rightarrow$$ $$x+\dfrac{b}{2}=\pm\dfrac{a}{2}$$

$$\Rightarrow$$ $$x+\dfrac{b}{2}=\dfrac{a}{2}$$ and $$x+\dfrac{b}{2}=-\dfrac{a}{2}$$

$$\Rightarrow$$ $$x=\dfrac{a-b}{2}$$ and $$x=-\left(\dfrac{a+b}{2}\right)$$
Question 4
1 / -0

............... is true for discriminate of quadratic equation $$x^2 + x + 1 = 0$$.

A
D = 0

B
D < 0

C
D > 0

D
D is a perfect square

Solution

The polynomial is $$x^2 + x + 1 = 0$$.

Comparing with $$ax^2 + bx + c = 0$$, we get

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

$$D = b^2 - 4ac = 1 - 4(1) (1) = -3 < 0$$
Question 5
1 / -0

The discriminant value of equation $$5{x}^{2}-6x+1=0$$ is ...............

A
$$16$$

B
$$\sqrt { 56 } $$

C
$$4$$

D
$$56$$

Solution

The given equation is $$5{x}^{2}-6x+1=0$$ ........... $$(i)$$

General form of quadratic equation is given by $$ax^2+bx+c=0$$ ....... $$(ii)$$

Comparing equation $$(i)$$ with $$(ii)$$ we have,

$$a=5,b=-6,c=1$$

The discriminant $$D$$ is given by

$$D={b}^{2}-4ac={(-6)}^{2}-4(5)(1)=36-20=16$$

$$\therefore$$ Discriminant value is $$D=16$$
Question 6
1 / -0

Solve the following quadratic equation by completing the square: $$\dfrac{5x+7}{x-1}=3x+2$$

A
$$\left \{ -1, 3 \right \}$$

B
$$\left \{ 1, 3 \right \}$$

C
$$\left \{ -1, -3 \right \}$$

D
None of these

Solution

Given $$\dfrac{5x+7}{x-1}=3x+2$$

$$\Rightarrow$$ $$5x+7=(3x+2)(x-1)$$

$$\Rightarrow$$ $$5x+7=3x^2-3x+2x-2$$

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

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

$$\Rightarrow$$ Here, $$a=1,\,b=-2,\,c=-3$$

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

Now, $$\left(\dfrac{b}{2}\right)^2=\left (\dfrac{-2}{2} \right )^2=1$$

Adding $$1$$ on both sides,

$$\Rightarrow$$ $$x^2-2x+1=3+1$$

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

Taking square root on both sides

$$\Rightarrow$$ $$x-1=\pm2$$

$$\Rightarrow$$ $$x-1=2$$ and $$x-1=-2$$

$$\therefore$$ $$x=3$$ and $$x=-1$$
Question 7
1 / -0

If the roots of the equation $$x^{2} + px + c = 0$$ are $$(2, -2)$$ and the roots of the equation $$x^{2} + bx + q = 0$$ are $$(-1, -2)$$, then the roots of the equation $$x^{2} + bx + c = 0$$ are

A
$$-3, -2$$

B
$$-3, 2$$

C
$$1, -4$$

D
$$-5, 1$$

Solution

Given, roots of the equation $$x^{2} + px + c = 0$$ are $$2$$ and $$-2$$.

$$\therefore -p = 2 - 2 \Rightarrow p = 0$$

and $$(2)\cdot (-2) = c\Rightarrow c = -4$$

Again, roots of the equation $$x^{2} + bx + q = 0$$ are $$-1$$ and $$-2$$.

$$\therefore -b = -1 -1 \Rightarrow b = 3$$

and $$(1-) \cdot (-2) = q\Rightarrow q = 2$$

$$\therefore x^{2} + bc + c \equiv x^{3} + 3x - 4 =0$$

$$\Rightarrow (x - 1)(x + 4) = 0$$

So, the required roots are $$1$$ and $$-4$$.
Question 8
1 / -0

Let $$\alpha$$ be the root of the equation $$25\cos^{2}\theta + 5\cos \theta - 12 = C$$, where $$\dfrac {\pi}{2} < \alpha < \pi$$.
What is $$\tan \alpha$$ equal to?

A
$$-\dfrac {3}{4}$$

B
$$\dfrac {3}{4}$$

C
$$-\dfrac {4}{3}$$

D
$$-\dfrac {4}{5}$$

Solution

Given equation is $$25\cos^2\theta+5\cos\theta-12=0$$

$$\Rightarrow 25\cos^2\theta+20\cos\theta-15\cos\theta-12=0$$

$$\Rightarrow 5\cos\theta(5\cos\theta+4)-3(5\cos\theta+4)=0$$

$$\Rightarrow (5\cos\theta-3)(5\cos\theta+4)=0$$

So, $$\cos\theta=\dfrac{3}{5}$$ or $$\cos\theta=\dfrac{-4}{5}$$

Since $$\alpha$$ is the root of the equation and it lies between $$\dfrac{\pi}{2}<\alpha<\pi\Rightarrow \cos \alpha\ $$ is '-ve'.

Thus $$\alpha=\cos^{-1}\left (\dfrac{-4}{5}\right)$$

Since it lies in the range $$\dfrac{\pi}{2}<\alpha<\pi$$, so $$\tan \alpha$$ is '-ve'

$$\tan\alpha=\tan\left [\cos^{-1}\left (\dfrac{-4}{5}\right)\right]$$

$$=\tan \left [\tan^{-1}\left (-\dfrac{3}{4}\right)\right]$$

$$=-\dfrac{3}{4}$$
Question 9
1 / -0

Sum of the roots of the equation $${ \left| x-3 \right| }^{ 2 }+\left| x-3 \right| -2=0$$ is

A
$$2$$

B
$$4$$

C
$$6$$

D
$$16$$

E
$$-2$$

Solution

Given equation is $${ \left| x-3 \right| }^{ 2 }+\left| x-3 \right| -2=0$$

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

$$\Rightarrow \left| x-3 \right| \left( \left| x-3 \right| +2 \right) -1\left( \left| x-3 \right| +2 \right) =0$$

$$\Rightarrow \left( \left| x-3 \right| +2 \right) \left( \left| x-3 \right| -1 \right) =0$$

Therefore, $$ \left| x-3 \right| =-2,1$$

But $$\left| x-3 \right| \neq -2$$

Hence, $$\left| x-3 \right| =1$$

$$\Rightarrow x-3=\pm 1$$

$$\Rightarrow x=3\pm 1$$

$$\Rightarrow x=4, 2$$

Now, sum of the roots $$=4+2=6$$.
Question 10
1 / -0

The equation $${ e }^{ \sin { x } }-{ e }^{ -\sin { x } }-4=0$$ has

A
No solution

B
Two solutions

C
Three solutions

D
None of these

Solution

Given, $${ e }^{ \sin { x } }-{ e }^{ -\sin { x } }-4=0$$

multiply throughout by $$e^{sinx}$$

$$\Rightarrow { e }^{ 2\sin { x } }-4{ e }^{ \sin { x } }-1=0$$

this is quadratc equation in $$e^{sinx}$$

converting from exponential to logarithmic form.

$${ e }^{ \sin { x } }=\dfrac { 4\pm \sqrt { 16+4 } }{ 2 } =2\pm \sqrt { 5 }$$

$$\Rightarrow \sin { x } =\log { \left( 2+\sqrt { 5 } \right) } $$ [$$ \because \log { \left( 2-\sqrt { 5 } \right) }$$ is not defined ]

Since, $$2+\sqrt { 5 } > e \Rightarrow \log { \left( 2+\sqrt { 5 } \right) } > 1$$

$$\Rightarrow \sin { x } > 1$$, which is not possible.
Hence, no solution exist.

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Re-Attempt Weekly Quiz Competition

Quadratic Equations Test - 35

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Quadratic Equations Test - 35

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SHARING IS CARING

Question 1

$$0$$

$$2$$

$$1$$

$$3$$

Solution

Question 2

$$\left \{ \sqrt{3}, 1\right \}$$

$$\left \{ \sqrt{3}, 2\right \}$$

$$\left \{ \sqrt{3}, 3\right \}$$

None of these

Solution

Question 3

$$\left \{ \dfrac{a-2}{2}, -\left ( \dfrac{a+2}{2} \right ) \right \}$$

$$\left \{ \dfrac{a+2}{2}, -\left ( \dfrac{a+2}{2} \right ) \right \}$$

$$\left \{ \dfrac{a-2}{2}, -\left ( \dfrac{a-2}{2} \right ) \right \}$$

None of these

Solution

Question 4

D = 0

D < 0

D > 0

D is a perfect square

Solution

Question 5

$$16$$

$$\sqrt { 56 } $$

$$4$$

$$56$$

Solution

Question 6

$$\left \{ -1, 3 \right \}$$

$$\left \{ 1, 3 \right \}$$

$$\left \{ -1, -3 \right \}$$

None of these

Solution

Question 7

$$-3, -2$$

$$-3, 2$$

$$1, -4$$

$$-5, 1$$

Solution

Question 8

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

$$\dfrac {3}{4}$$

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

$$-\dfrac {4}{5}$$

Solution

Question 9

$$2$$

$$4$$

$$6$$

$$16$$

$$-2$$

Solution

Question 10

No solution

Two solutions

Three solutions

None of these

Solution

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