Quadratic Equations Test

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Quadratic Equations Test - 19

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Weekly Quiz Competition

Question 1
1 / -0

Is the following equation quadratic?
$$(x\, +\, 3) (x\, -\, 4)\, =\, 0$$

A
Yes

B
No

C
Ambiguous

D
Data insufficient

Solution

$$Answer=1$$
The equation $$(x\, +\, 3) (x\, -\, 4)\, =\, 0$$ can be formed as
$$x^2 - x -12 =0 $$ has the highest power as 2. Thus, it is quadratic equation.
Question 2
1 / -0

The mentioned equation is in which form?
$$\dfrac{q^{2}- 4}{q^{2}}\, =\, -3$$

A
Linear

B
Quadratic

C
Cubic

D
none

Solution

Given equation is $$\dfrac{q^2 - 4}{q^2} = -3$$
$$\Rightarrow q^2 - 4 = -3q^2$$
$$\Rightarrow 4q^2 = 4$$
$$\Rightarrow q^2 = 1$$
The highest power of $$q$$ is $$2$$. Hence, it is a quadratic equation.
Question 3
1 / -0

The mentioned equation is in which form?
$$3y^{2}\,-\, 7\, =\, \sqrt{3}\,y$$

A
linear

B
Quadratic

C
Cubic

D
None

Solution

The highest power of y is 2. Hence, the equation is quadratic.
Question 4
1 / -0

Is the following equation quadratic?
$$\displaystyle -\frac{5}{3}\, x^{2}\, =\, 2x\, +\, 9$$

A
Yes

B
No

C
Ambiguous

D
Data insufficient

Solution

$$Answer=1$$
The equation, $$\displaystyle -\frac{5}{3}\, x^{2}\, =\, 2x\, +\, 9$$ has the highest power as 2.
Thus, the equation is quadratic.
Question 5
1 / -0

Is the following equation quadratic?
$$13\, =\, -5y^{2}\, -\, y^{3}$$

A
Yes

B
No

C
Ambiguous

D
Data insufficient

Solution

Answer=0
The equation $$13 = -5y^2 - y^3$$ has the highest power as 3. Thus, it is not quadratic equation.
Question 6
1 / -0

Find discriminant of $$5x^2+2x+1=0$$

A
$$-13$$

B
$$12$$

C
$$-16$$

D
$$14$$

Solution

The discriminant $$D={ b }^{ 2 }- 4ac = 4 - 4(5)(1) = -16$$
Question 7
1 / -0

Find a quadratic equation, from the equations given below, having same discriminant as $$5x^2+2x+1=0$$

A
$$-x^2+4x-8$$

B
$$x^2+3x-7$$

C
$$4x^2-5$$

D
$$-2x^2-3x+6$$

Solution

Given: set of quadratic equations
To find: the equation having same discriminant as $$5x^2+2x+1=0$$
Sol: The equation $$5x^2+2x+1=0$$ is of form $$ax^2+bx+c=0$$
$$\therefore a=5, b=2, c=1$$
The discriminant of this equation becomes
$$b^2-4ac=2^2-4(5)(1)=4-20=-16$$
(i) $$-x^2+4x-8$$ this is also of the form $$ax^2+bx+c$$
$$\therefore a=-1, b=4, c=-8$$
The discriminant of this equation becomes
$$b^2-4ac=4^2-4(-1)(-8)=16-32=-16$$
(ii) $$x^2+3x-7$$ this is also of the form $$ax^2+bx+c$$
$$\therefore a=1, b=3, c=-7$$
The discriminant of this equation becomes
$$b^2-4ac=3^2-4(1)(-7)=9+28=37$$
(iii) $$4x^2-5\implies 4x^2-0x-5$$ this is also of the form $$ax^2+bx+c$$
$$\therefore a=4, b=0, c=-5$$
The discriminant of this equation becomes
$$b^2-4ac=0^2-4(4)(-5)=0+80=80$$
(iv) $$-2x^2-3x+6$$ this is also of the form $$ax^2+bx+c$$
$$\therefore a=-2, b=-3, c=6$$
The discriminant of this equation becomes
$$b^2-4ac=(-3)^2-4(-2)(6)=9+48=57$$
Question 8
1 / -0

Find the roots of the quadratic equation by applying the quadratic formula$$\displaystyle 2x^2 - 7x + 3 = 0$$

A
$$3,\displaystyle -\frac{1}{2}$$

B
$$-3,\displaystyle \frac{1}{2}$$

C
$$3,\displaystyle \frac{1}{2}$$

D
None of these

Solution

Given equation is $$2x^{2}-7x+3=0$$
Hence, $$a=2,b=-7,c=3$$
Therefore, $$x=\dfrac{-b\pm\sqrt{b^{2}-4ac}}{2a}$$
$$=\dfrac{7\pm\sqrt{25}}{4}$$
$$=\dfrac{7\pm5}{4}$$
Thus, $$x=3$$ and $$x=\dfrac{1}{2}$$
Question 9
1 / -0

Is the following equation a quadratic equation?
$$\displaystyle 3x + \frac{1}{x} - 8 = 0$$

A
Yes

B
No

C
Ambiguous

D
Data insufficient

Solution

Given, $$3x + \dfrac{1}{x} - 8 = 0$$
$$\Rightarrow \dfrac{3x^2+1-8x}{x}=0$$
$$\Rightarrow 3x^2+1-8x=0$$
Comparing it with the standard form of quadratic equation $$x^2+bx+c=0, a\neq 0$$
Here, $$a=3$$
Thus, it is a quadratic equation.
Question 10
1 / -0

Is the following equation a quadratic equation?
$$(x + 2)^3 = x^3 - 4$$

A
Yes

B
No

C
Ambiguous

D
Data insufficient

Solution

Given equation is $$(x + 2)^3 = x^3 - 4$$
$$\rightarrow x^3+2^3+3\times x\times 2(x+2)=x^3-4$$
$$\Rightarrow x^3+8+6x(x+2)=x^3-4$$
$$\Rightarrow x^3+8+6x^2+12x-x^3+4=0$$
$$\Rightarrow 6x^2+12x+12=0$$
Comparing it with the standard form of quadratic equation $$x^2+bx+c=0, a\neq 0$$.
Here, $$a=6$$
Thus, it is a quadratic equation.

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Quadratic Equations Test - 19

Result

Quadratic Equations Test - 19

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Question 1

Yes

No

Ambiguous

Data insufficient

Solution

Question 2

Linear

Quadratic

Cubic

none

Solution

Question 3

linear

Quadratic

Cubic

None

Solution

Question 4

Yes

No

Ambiguous

Data insufficient

Solution

Question 5

Yes

No

Ambiguous

Data insufficient

Solution

Question 6

$$-13$$

$$12$$

$$-16$$

$$14$$

Solution

Question 7

$$-x^2+4x-8$$

$$x^2+3x-7$$

$$4x^2-5$$

$$-2x^2-3x+6$$

Solution

Question 8

$$3,\displaystyle -\frac{1}{2}$$

$$-3,\displaystyle \frac{1}{2}$$

$$3,\displaystyle \frac{1}{2}$$

None of these

Solution

Question 9

Yes

No

Ambiguous

Data insufficient

Solution

Question 10

Yes

No

Ambiguous

Data insufficient

Solution

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