Quadratic Equations Test

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Quadratic Equations Test - 3

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Weekly Quiz Competition

Question 1
1 / -0

Which of the following is a quadratic equation?

1. (x - 3)²= (x - 1)² + 2
2. (2x - 4)³= 8x³
3. x⁴- 3x²+ 5x - 4 = 0
4. x²+ x^1/2- 1 = 0
5. (x - 1/x)² + 2 = 0

A
1

B
2

C
3

D
4

E
5

Solution

8x³- 48x²+ 96x - 64 = 8x³
This implies -48x²+ 96x - 64 = 0.

The degree of the equation is 2. So, it is a quadratic equation.
Question 2
1 / -0

The discriminant of the quadratic equation 2x²- 6x - 3 = 0 is

A
60

B
square root of 60

C
12

D
square root of 12

E
-12

Solution

Discriminant = b² - 4ac = (-6)² - 4 x 2 x -3 = 36 + 24 = 60
Question 3
1 / -0

The roots of the quadratic equation x²- (1/16) = 0 are

A
+1/4 and +1/4

B
-1/8 and +1/2

C
+1/4 and -1/4

D
+4 and -4

E
+1/8 and -1/8

Solution

x²- (1/16) = 0
(x + 1/4)(x - 1/4) = 0
(x + 1/4) = 0
Or (x - 1/4) = 0
x = -1/4 or x = +1/4
So, the roots of the given equation are -1/4 and +1/4. [1]
Question 4
1 / -0

If the equation 6x²+ 4x - a = 0 has equal roots, then choose the correct option.

A
2a - 3 = 0

B
3a - 3 = 0

C
3a + 1 = 0

D
3a - 2 = 0

E
3a + 2 = 0

Solution

For equal roots, discriminant = b² - 4ac = 4² - 4 x 6 x -a = 16 + 24a = 0. This implies 8(2 + 3a) = 0.
This equals 2 + 3a = 0. [1]
Question 5
1 / -0

Choose the quadratic equation with two real roots.

A
x²+ 5x + 12 = 0

B
x²+ 2x - 9 = 0

C
x²+ 9 = 0

D
4x²+ 2x + 1 = 0

E
3x²+ 3 = 0

Solution

For real roots, b²- 4ac > 0.
2²- (4 x 1 x -9) = 4 + 36 = 40 > 0
The equation has two real roots.
Question 6
1 / -0

Choose the quadratic equation with roots 7 and -1.

A
x²- 7x + 6 = 0

B
x²+ 8x + 7 = 0

C
(x + 7)(x - 1) = 0

D
x²+ 6x - 7 = 0

E
x²- 6x - 7 = 0

Solution

Since 7 and -1 are the roots of the equation, the equation is
(x - 7)(x + 1) = 0
x²- 6x - 7 = 0
Question 7
1 / -0

Choose the equation with imaginary roots.

A
x²= 19

B
4x²- 4x + 1 = 0

C
6x²- 1 = 0

D
2x²+ x + 1 = 0

E
10x²- 2x - 3 = 0

Solution

b²- 4ac = 1²- (4 x 2 x 1) = 1 - 8 = -7 < 0.
Since the discriminant b²- 4ac < 0, the equation has imaginary roots.
Question 8
1 / -0

Choose the equation which has two distinct real roots.

A
(x - 3)²= 0

B
2x²- 32 = 0

C
9x²- 30x + 25 = 0

D
x²+ 2x + 1 = 0

E
(x - 1)(x - 1) = 0

Solution

2x²= 32 implies x²= 16.
We know, 16 = 4 x 4 and 16 = -4 x -4
Hence, x = +4, -4
So, the equation has two different roots.
Question 9
1 / -0

The roots of the equation (x - 3)²= 4 are

A
+3, +2

B
+5, +1

C
-5, -1

D
+5, +5

E
+3, +3

Solution

x² + 9 - 6x = 4
Gives x² - 6x + 5 = 0
This implies (x - 5)(x - 1) = 0
Therefore, x - 5 = 0 or (x - 1) = 0
Hence, x = 5 or x = 1.
The root of the equation are 1 and 5.
Question 10
1 / -0

If 5x²- kx - 4 = 0 has imaginary roots, then choose the correct option.

A
k² > -80

B
k²> 100

C
k²> 80

D
k² = 80

E
k² < -80

Solution

5x²- kx - 4 = 0 has imaginary roots.
So, b²- 4ac < 0(-k)²- (4 x 5 x -4) < 0
k²+ 80 < 0
k²< -80
Question 11
1 / -0

The roots of the quadratic equation (2x- 3)(x + 2) = 3x + 9 are

A

B
-12 and -7/2

C
3/2 and -2

D
1 and -3/7

E

Solution

(2x - 3)(x + 2) = 3x + 9
2x²+ 4x - 3x - 6 = 3x + 9
2x²- 2x - 15 = 0

x =

=

=

=
Question 12
1 / -0

The root of which of the following quadratic equations is the reciprocal of the other root?

A
6x²- 54 = 0

B
3x²+ 8x - 3 = 0

C
3x²+ 10x + 3 = 0

D
3x²- 8x - 3 = 0

E
3x²+ 8x + 3 = 0

Solution

x =

=

The roots are reciprocal to each other.
Question 13
1 / -0

Choose the quadratic equation having roots 2/3 and -1/2.

A
6x²+ x + 2 = 0

B
6x²- x + 2 = 0

C
6x²+ 7x + 2 = 0

D
6x²- x - 2 = 0

E
6x²+ x - 2 = 0

Solution

Since the roots are 2/3 and -1/2, the quadratic equation corresponding to them is as follows.
(x - 2/3)(x + 1/2) = 0
(3x - 2)(2x + 1) = 0
6x2 - x - 2 = 0
Question 14
1 / -0

If the quadratic equation 3x²- 12x + 4k = 0 has equal roots, then the value of k is

A
1/3

B
12

C
3

D
2

E
-3

Solution

Since the equation has equal roots,
b²- 4ac = 0
(-12)²- 4 × 3 × 4k = 0
144 - 48k = 0
48k = 144
k = 3
Question 15
1 / -0

Choose the quadratic equation with real roots.

A

B
+ 16 = 0

C
(x + 4)² + 20 = 0

D
7x²+ 1 = 0

E
4x²- 3x + 2 = 0

Solution

The discriminant = b²- 4ac = 3²- 4 × 4 × (-36) = 9 + 576 = 585 > 0

Therefore, the equation has real roots.