Quadratic Equations Test

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Quadratic Equations Test - 41

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Weekly Quiz Competition

Question 1
1 / -0

Solve the quadratic equation by completing the square method: $${x}^{2}+8x-9=0$$

A
$$-9,-1$$

B
$$2,1$$

C
$$-9,1$$

D
$$9,1$$

Solution

$$x^2+8x-9=0$$
$$\Rightarrow x^2+2\times \left( 4\right) \times x +16-25=0$$
$$\left( x+4\right)^2=25$$
$$\left( x+4\right)^2=5^2$$
$$x+4=\pm 5$$
$$\Rightarrow x=1$$ and $$-9$$
Hence, the answer is $$1,-9.$$
Question 2
1 / -0

If the discriminant of $$3 x ^ { 2 } - 2 x + K = 0$$ is zero then $$\mathrm { K } =$$ _____

A
$$3$$

B
$$\dfrac { 1 } { 3 }$$

C
$$-3$$

D
$$\dfrac { -1 } { 3 }$$

Solution
Question 3
1 / -0

Which constant should be added and subtracted to solve the quadratic equation $${ 4x }^{ 2 }-\sqrt { 3x } -5=0$$ by the method of completing the square ?

A
$$\dfrac { 9 }{ 16 } $$

B
$$\dfrac { 3 }{ 16 } $$

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

D
$$\dfrac { \sqrt { 3 } }{ 16 } $$

Solution

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

$$\Rightarrow (2x)^2-\dfrac{\sqrt{3}}{2}2x-5=0$$

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

$$\Rightarrow \left(2x-\dfrac{\sqrt{3}}{4}\right)^2-\dfrac{3}{16}-5=0$$ ...{$$\because$$ We know, $$a^2-2ab+b^2=(a-b)^2$$}

Hence, we had to add $$\dfrac{3}{16}$$ to complete the square.
Question 4
1 / -0

Factors of $${ x }^{ 2 }+x\left( a+b \right) +ab$$ are

A
$$x+a$$

B
$$x+b$$

C
both A & B

D
None

Solution

$$x^{2}+x(a+b)+ab$$

comparing with $$ax^{2}+bx+c=0$$

$$x^{2}+(a+b)x+ab=0$$

comparing with $$Ax^{2}+Bx+C=0$$

$$A=1$$ $$B=a+b$$ $$C=ab$$

$$B^{2}-4ac= (a+b)^{2}-4\times 1\times ab$$

$$=a^{2}+2ab+b^{2}-4ab$$

$$=a^{2}-2ab+b^{2}$$

$$B^{2}-4AC=(a-b)^{2}$$

$$x=\dfrac{-B\pm \sqrt{B^{2}-4ac}}{2A}=\dfrac{-(a+b)\pm \sqrt{(a-b)^{2}}}{2\times 1}$$

$$x=\dfrac{-a-b+a-b}{2}$$ OR $$x=\dfrac{-a-b-(a-b)}{2}$$

$$x=-b$$ OR $$x=-a$$

$$(x+b)=0$$ OR $$(x+a)=0$$

are the factors
Question 5
1 / -0

Quadratic equation among the following is:

A
$${ x }^{ 2 }+1={ x }^{ 2 }+2x$$

B
$$x\left( x+1 \right) =x\left( x-3 \right) $$

C
$${ x }^{ 2 }-2x=3\left( 5x+2 \right) $$

D
$${ x }^{ 2 }+2x+3={ x }^{ 2 }+5x$$

Solution
Question 6
1 / -0

Which constant must be added and subtracted to solve the Quadratic equation $$9 x ^ { 2 } + \dfrac { 3 } { 4 } x - \sqrt { 2 }$$ by the method of completion the square.

A
$$\dfrac { 1 } { 8 }$$

B
$$\dfrac { 1 } { 4 }$$

C
$$\dfrac { 1 } { 64 }$$

D
$$\dfrac { 9 } { 64 }$$

Solution
Question 7
1 / -0

For what value of $$c$$, the root of $$ (c-2) x^{2}+2(c-2) x+2=0$$ are not real

A
$$( 1,2)$$

B
$$( 2,3)$$

C
$$( 3,4)$$

D
$$(2,4)$$

Solution

Given the quadratic equation is $$ (c-2) x^{2}+2(c-2) x+2=0$$.
Now the roots of the equation will be not real if the discriminant of the equation is less than $$0$$.
Then,
$$\{2(c-2)\}^2-4.2.(c-2)<0$$
or, $$4\{(c-2)(c-4)\}<0$$
or, $$(c-2)(c-4)<0$$
or, $$2<c<4$$.
So $$c\in (2,4)$$.
Question 8
1 / -0

Which constant must be added and subtracted to solve the quadratic equation $$9x^2 + \dfrac{3}{4}x - \sqrt{2} = 0$$ by the method of completing the square ?

A
$$\dfrac{1}{8}$$

B
$$\dfrac{1}{64}$$

C
$$\dfrac{1}{4}$$

D
$$\dfrac{9}{64}$$

Solution

Given equation is $$9x^2 + \dfrac{3}{4}x - \sqrt{2} = 0$$
$$\Rightarrow (3x)^2 + 2\left(\dfrac{1}{8}\right)(3x) - \sqrt{2} = 0$$

Hence, to make it a perfect square we must add and subtract $$\left(\dfrac{1}{8}\right)^2=\dfrac1{64}$$
Question 9
1 / -0

If $$(1+m^2)x^2-2(1+3m)x+(1+8m)=0$$. Then numbers of value(s) of m for which this equation has no solution.

A
$$3$$

B
$$\infty$$

C
$$2$$

D
$$1$$

Solution

We know that $$D=b^2-4ac$$

$$D=4(1+3m)^2-4(1+m^2)(1+8m)$$

$$=4(1+9m^2+6m-(1+8m+m^2+8m^3))$$

$$=4(8m^2-2m-8m^3)$$

$$=-8(4m^3-4m^2+m)$$

$$=-8m(4m^2-4m+1)$$

$$=-8m(2m-1)^2 < 0$$.........$$\because (2m-1)^2 \, is \, always +ve$$

Hence infinitely many values.
Question 10
1 / -0

Discriminant is ................. for the equation $$5x-6=-\frac{1}{x}$$

A
-56

B
16

C
-16

D
0

Solution

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Quadratic Equations Test - 41

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Quadratic Equations Test - 41

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Question 1

$$-9,-1$$

$$2,1$$

$$-9,1$$

$$9,1$$

Solution

Question 2

$$3$$

$$\dfrac { 1 } { 3 }$$

$$-3$$

$$\dfrac { -1 } { 3 }$$

Solution

Question 3

$$\dfrac { 9 }{ 16 } $$

$$\dfrac { 3 }{ 16 } $$

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

$$\dfrac { \sqrt { 3 } }{ 16 } $$

Solution

Question 4

$$x+a$$

$$x+b$$

both A & B

None

Solution

Question 5

$${ x }^{ 2 }+1={ x }^{ 2 }+2x$$

$$x\left( x+1 \right) =x\left( x-3 \right) $$

$${ x }^{ 2 }-2x=3\left( 5x+2 \right) $$

$${ x }^{ 2 }+2x+3={ x }^{ 2 }+5x$$

Solution

Question 6

$$\dfrac { 1 } { 8 }$$

$$\dfrac { 1 } { 4 }$$

$$\dfrac { 1 } { 64 }$$

$$\dfrac { 9 } { 64 }$$

Solution

Question 7

$$( 1,2)$$

$$( 2,3)$$

$$( 3,4)$$

$$(2,4)$$

Solution

Question 8

$$\dfrac{1}{8}$$

$$\dfrac{1}{64}$$

$$\dfrac{1}{4}$$

$$\dfrac{9}{64}$$

Solution

Question 9

$$3$$

$$\infty$$

$$2$$

$$1$$

Solution

Question 10

-56

16

-16

0

Solution

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