Quadratic Equations Test

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Quadratic Equations Test - 21

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Weekly Quiz Competition

Question 1
1 / -0

Choose the best possible option.
$$\displaystyle { x }^{ 2 }+\frac { 1 }{ 4{ x }^{ 2 } } -8=0$$ is a quadratic equation.

A
Yes

B
No

C
Can't predict

D
None

Solution

$$\displaystyle { x }^{ 2 }+\frac { 1 }{ 4{ x }^{ 2 } } -8=0$$
$$\displaystyle \frac { 4{ x }^{ 4 }+1-8\times 4{ x }^{ 2 } }{ 4{ x }^{ 2 } } =0$$
$$\displaystyle 4{ x }^{ 4 }+1-32{ x }^{ 2 }=0$$
The degree of the equation is 4
$$\displaystyle \therefore \quad { x }^{ 2 }+\frac { 1 }{ 4{ x }^{ 2 } } -8=0$$ is not a quadratic equation
Question 2
1 / -0

Determine whether the equation $$\displaystyle 5{ x }^{ 2 }=5x$$ is quadratic or not.

A
Yes

B
No

C
Complex equation

D
None

Solution

The degree of the equation is 2
$$\displaystyle \therefore \quad 5{ x }^{ 2 }=5x$$
$$\displaystyle =\quad 5{ x }^{ 2 }-5x=0$$ is quadratic equation.
Question 3
1 / -0

Choose best possible option.
$$\displaystyle \left( x+\frac { 1 }{ 2 } \right) \left( \frac { 3x }{ 2 } +1 \right) =\frac { 6 }{ 2 } \left( x-1 \right) \left( x-2 \right) $$ is quadratic.

A
Yes

B
No

C
Complex equation

D
None

Solution

$$\displaystyle \left( x+\frac { 1 }{ 2 } \right) \left( \frac { 3x }{ 2 } +1 \right) =\frac { 6 }{ 2 } \left( x-1 \right) \left( x-2 \right) $$
$$\displaystyle \left( 2x+2 \right) \left( 3x+2 \right) =12\left( x-1 \right) \left( x-2 \right) $$
$$\displaystyle 12{ x }^{ 2 }-36x+24-6{ x }^{ 2 }-7x-2=0$$
$$6x^2$$$$-43x$$ +$$22$$=0
The degree of equation is 2
Therefore, it is a quadratic equation.
Question 4
1 / -0

Choose the best possible answer
$$\displaystyle 32{ x }^{ 2 }-6=\left( 4x+10 \right) \left( 10x-6 \right) $$ is quadratic equation

A
Yes

B
No

C
Complex

D
None

Solution

$$\displaystyle 32{ x }^{ 2 }-6=\left( 4x+10 \right) \left( 10x-6 \right) $$
$$\displaystyle 32{ x }^{ 2 }-6=40{ x }^{ 2 }-24x+100x-60$$
$$\displaystyle 8{ x }^{ 2 }+76x-54=0$$
The degree of the equation is 2
$$\displaystyle \therefore $$ It is a quadratic equation
Question 5
1 / -0

Calculate the zeroes of the quadratic equation $$(x+3)^2=49$$.

A
$$x=-10, x=4$$

B
$$x=-10, x=10$$

C
$$x=-4, x=10$$

D
$$x=3\pm 2\sqrt{3}$$

Solution

$$(x+3)^2=49$$, given equation
$$\Rightarrow x+3=\pm \sqrt{49}=\pm 7$$
$$\Rightarrow x=\pm 7-3$$
$$\Rightarrow x =-7-3, 7-3$$
Therefore, the values are $$x =-10,4$$
Question 6
1 / -0

Which of the following is not a quadratic equation?

A
$$x^{2}+ 6y + 2$$

B
$$(x - 2) + (x + 2)^{2} + 3$$

C
$$(3x - 2)^{3} + \dfrac{1}{2}x - 4$$

D
$$3x^{2} - 6x + \dfrac{1}{2}$$

Solution

A quadratic equation can be expressed in the form $$ax^{2} + bx + c = 0$$.
Here, the only equation that has a third degree variable is $$(3x - 2)^{3} + \dfrac{1}{2}x - 4$$ and all other equations have two degree variables.
So, $$(3x - 2)^{3} + \dfrac{1}{2}x - 4$$ is not a quadratic equation.
Question 7
1 / -0

Discriminant of a quadratic equation $$\displaystyle p{ x }^{ 2 }+qx+r=0$$ is given by.

A
$$\displaystyle { \left( { { q }^{ 2 }-4pr } \right) }$$

B
$$\displaystyle \sqrt { { q }^{ 2 }+4qr } $$

C
$$\displaystyle \sqrt { { q }^{ 2 }-4pr } $$

D
$$\displaystyle { q }^{ 2 }+4{ q }^{ 2 }$$

Solution

Given equation is $$px^2+qx+r=0$$
Here $$a = p$$, $$b=q$$ and $$c=r$$
Discriminant $$D\displaystyle = {{ b }^{ 2 }-4ac}$$ $$=\displaystyle {{ q }^{ 2 }-4pr}$$
Hence, option A is correct.
Question 8
1 / -0

Identify the standard format of quadratic equation.

A
$$ax^{2}+bx + c = 0$$

B
$$ax^{3}+ bx^{2} + 2$$

C
$$ax^{3}+ bx^{3}+13$$

D
$$a(x - 1)^{2}+ bx^{3}+ c$$

Solution

A quadratic equation can be expressed in the standard form is $$ax^{2} + bx + c = 0$$ and all other equations have third degree variables.
Hence, option A is correct.
Question 9
1 / -0

Choose the best possible option.
$$\displaystyle (x+5)(x-8)=0$$ is quadratic equation.

A
Yes

B
No

C
Complex equation

D
None

Solution

$$\displaystyle \left( x+5 \right) \left( x-8 \right) =0$$
$$\displaystyle { x }^{ 2 }-8x+5x-40=0$$
$$\displaystyle { x }^{ 2 }-3x-40=0$$ is quadratic , because the degree of equation is 2
Question 10
1 / -0

Solution of the equation $$2x^{2} = 4x + 3$$ is

A
$$\dfrac {2\pm \sqrt {10}}{2}$$

B
$$\dfrac {2\pm 5\sqrt {2}}{2}$$

C
$$1\pm \sqrt {5}$$

D
$$1\pm \sqrt {10}$$

Solution

We know the formula, $$\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}$$
Here the equation, $$2x^2-4x-3 =0$$
$$a = 2, b = -4, c = -3$$
Substituting the values, we get
$$=$$ $$\dfrac{4\pm\sqrt{(-4)^2-4\times 2\times -3}}{2\times 2}$$
$$=$$ $$\dfrac{4\pm\sqrt{16+24}}{4}$$
$$=$$ $$\dfrac{4\pm\sqrt{40}}{4}$$
Take $$2$$ as common, we get
$$=$$ $$\dfrac{2\pm\sqrt{10}}{2}$$
$$\therefore$$ solution of the equation $$ 2x^2-4x-3 $$ is $$ \dfrac{2\pm\sqrt{10}}{2}$$.

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Quadratic Equations Test - 21

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Quadratic Equations Test - 21

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SHARING IS CARING

Question 1

Yes

No

Can't predict

None

Solution

Question 2

Yes

No

Complex equation

None

Solution

Question 3

Yes

No

Complex equation

None

Solution

Question 4

Yes

No

Complex

None

Solution

Question 5

$$x=-10, x=4$$

$$x=-10, x=10$$

$$x=-4, x=10$$

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

Solution

Question 6

$$x^{2}+ 6y + 2$$

$$(x - 2) + (x + 2)^{2} + 3$$

$$(3x - 2)^{3} + \dfrac{1}{2}x - 4$$

$$3x^{2} - 6x + \dfrac{1}{2}$$

Solution

Question 7

$$\displaystyle { \left( { { q }^{ 2 }-4pr } \right) }$$

$$\displaystyle \sqrt { { q }^{ 2 }+4qr } $$

$$\displaystyle \sqrt { { q }^{ 2 }-4pr } $$

$$\displaystyle { q }^{ 2 }+4{ q }^{ 2 }$$

Solution

Question 8

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

$$ax^{3}+ bx^{2} + 2$$

$$ax^{3}+ bx^{3}+13$$

$$a(x - 1)^{2}+ bx^{3}+ c$$

Solution

Question 9

Yes

No

Complex equation

None

Solution

Question 10

$$\dfrac {2\pm \sqrt {10}}{2}$$

$$\dfrac {2\pm 5\sqrt {2}}{2}$$

$$1\pm \sqrt {5}$$

$$1\pm \sqrt {10}$$

Solution

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