Quadratic Equation Test

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Quadratic Equation Test - 8

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Weekly Quiz Competition

Question 1
1 / -0

Which of the following equations has the solution x =?

A
=

B
=

C
=

D

Solution

Concept: To check x = a is a solution of a quadratic equation, put x = a in L.H.S of quadratic equation. If it is equal to R.H.S, then x = a is the solution otherwise not.
Solution: Consider = ………………. (i)
Putting x = . In L.H.S of.. (i)
Then, L.H.S = = = = R.H.S.
Hence, x = is solution of = .
Workout option (A): = Putting x =
L.H.S= = = = R.H.S.
Hence, x = is solution of = .
Missing concept: Take care of adding two fractional numbers.
Workout option (B): = Putting x =
L.H.S. = = = = R.H.S
Hence, x = is solution of = .
Missing concept: Need to take care of squaring a fractional number.
Workout option (D): = , Putting x =
L.H.S = = = = R.H.S
Hence, x = is solution of = .
Missing concept: Need to take care while putting the values.
Question 2
1 / -0

Which of the following equations has the solution y = - 2?

A
(2y + 2)² = - 4

B
(- 3y + 7)² = 1

C
(y + 4)² = 2

D
(2y + 2)² = 4

Solution

Concept: To check x = a is a solution of a quadratic equation, then put x = a in L.H.S of quadratic equation. If it is equal to R.H.S, then x = a is solution otherwise not.

Putting y = - 2 in L.H.S. of (i)
L.H.S. = (2- 2 + 2)² = (- 4 + 2)² = (- 2)² = 4
So, (2y + 2)² = 4 is the required equation.
Workout option (A): (2y + 2)² = - 4. Putting y = - 2
Note that square of any real no. can never be negative.
Workout option (B): (- 3y + 7)² = 1 Putting y = - 2
L.H.S. = {(- 3) (- 2) + 7}² = (- 6 + 7)² = 1 = R.H.S.
Missing concept: Take care of sign while multiplying two negative numbers.
Workout option (C): (y + 4)² = 2. Putting y = - 2
L.H.S. = (- 2 + 4) = (2) = R.H.S.
Missing concept: Take care of square.
Question 3
1 / -0

Find the sum of the roots of 2x² - 5x + 7 = 0.

A
- 5/2

B
5/2

C
7/2

D
2/7

Solution

2x² - 5x + 7 = 0
Sum of roots = -b/a = -
Question 4
1 / -0

If the equation x² – m (2x – 4) + 5 = 0 has equal roots, which of the following is the value of m?

A
5

B
4

C
2

D
1

Solution

x² - m (2x - 4) + 5 = 0
x² - 2mx + 4m + 5 = 0
For equal roots, b² - 4ac = 0
(- 2m)² - 4 (4m + 5) = 0
4 (m² - 4m - 5) = 0
(m - 5) (m + 1) = 0
Thus, m = 5 is the answer.
Question 5
1 / -0

If the roots of the equation 4x² – 4px + 7p – 12 = 0 are equal, find the value of p.

A
3 or 5

B
2 or 5

C
3 or 4

D
2 or 3

Solution

4x² - 4px + 7p - 12 = 0
Roots are equal. Therefore, b² - 4ac = 0
(- 4p)² - 4(4) (7p - 12) = 0
16 (p² - 7p + 12) = 0
(p - 3) (p - 4) = 0
Hence, p = 3, p = 4.
Question 6
1 / -0

Find the product of the roots of the equation 2x² – 3x + 5 = 0.

A
3/2

B
5/2

C
5/3

D
– 3/2

Solution

2x² – 3x + 5 = 0
Product of roots = =
Question 7
1 / -0

Find the sum of the roots of equation 6x² + x - 2 = 0.

A
1/6

B
1/3

C
- 1/3

D
- 1/6

Solution

6x² + x - 6= 0
+ = sum of roots = =
Question 8
1 / -0

If α and β are the roots of the equation 3x² + 4x + 7 = 0, then the value of is

A

B

C

D

Solution

Sum of roots,

Product of roots,

Hence, the desired answer is .
Question 9
1 / -0

Only one of the roots of ax² + bx + c = 0, a ≠ 0, is zero, if

A
c = 0

B
c = 0, b ≠ 0

C
b = 0, c = 0

D
b ≠ 0, c ≠ 0

Solution

When one root is zero and the other is non-zero, then
a(0)² + b(0) + c = 0
i.e. c = 0
Now, b ≠ 0 because if b = 0, then both the roots will be equal to 0.
As the equation will become
ax²= 0
Or x = 0, 0 (not possible, as both the roots are zero)
i.e. c = 0 and b ≠ 0
Question 10
1 / -0

The solution set of the equation is

A

B
{1}

C

D
None of these

Solution

The given equation can be written as:
2x - 3 + x - 1 = 6x² - x - 6, x ≠ 1
Or 6x² - 4x - 2 = 0
Or 3x² - 2x - 1 = 0, x ≠ 1
⇒ x =
But x ≠ 1.
So, the only solution is x = -.
Question 11
1 / -0

The discriminant of the quadratic equation 3x² + 5x + 2 = 0 is _______.

A
0

B
1

C
2

D
none of these

Solution

D = b² - 4ac
= 25 - 4 X 3 X 2
= 25 - 24
= 1
Question 12
1 / -0

The discriminant of the quadratic equation x² + 4x + 4 = 0 is _______.

A
0

B
1

C
2

D
none of these

Solution

x² + 4x + 4 = 0
D = 16 - 4 x 1 x 4
= 16 - 16
= 0
Question 13
1 / -0

What is the product of roots of the equation 2x²- 3x - 4 = 0?

A
- 6

B
- 4

C
- 2

D
None of these

Solution

2x² - 3x - 4 = 0
x = and
= and
= and
Product = -2
Question 14
1 / -0

Find the roots of the equation (2x - 5) (2x + 5) = 0.

A
,

B
,

C
,

D
,

Solution

Roots of equation (2x - 5) (2x + 5) = 0 can be given as:
2x - 5 = 0 and 2x + 5 = 0
x = and x =
Question 15
1 / -0

Which of the following expressions is a factor of the polynomial 6x² + 13x + 5?

A
3x + 5

B
3x – 5

C
2x – 1

D
x + 2

Solution

If x = a is root of an equation, then (x – a) is factor of that equation.
To find the root of 6x² + 13x + 5 = 0, compare it with ax² + bx + c = 0.
We get a = 6, b = 13 and c = 5.
So, x =
=
=
x =
Or x = are the roots of the given equation.
So, and are the factors of given polynomial
Or (2x + 1) and (3x + 5) are the factors of polynomial.
Question 16
1 / -0

Which of the following is a root of the quadratic equation x² + x - 12 = 0?

A
3

B
4

C
-6

D
-3

Solution

We have to find the root of the quadratic equation x² + x - 12 = 0.
On comparing it with ax² + bx + c = 0, we get
a = 1, b = 1 and c = - 12
x =
=
= ^{So, the root of the equation is 3.}Note: By hit and trial method, the value which satisfies the given equation is the root of that equation.
Question 17
1 / -0

Which of the following is a solution of the quadratic equation x² + x – 2?

A
– 1

B
– 2

C
2

D
– 4

Solution

The given quadratic equation is x² + x – 2 = 0.
Comparing it with ax² + bx + c = 0, we get
a = 1, b = 1 and c = – 2
x =
=
So, x = 1, – 2 are the solutions of the given equation.
Note: The value of x, which satisfies the given equation, is the solution of that equation.
Question 18
1 / -0

Which of the following pairs of numbers represents two roots of the equation x² - 6x - 7 = 0?

A
1, -7

B
1, 7

C
-1, 7

D
-1, -7

Solution

The given equation is x² - 6x - 7 = 0.
Comparing it with ax² + bx + c = 0, we get
a = 1, b = -6 and c = -7
_{x =}
₌
So, x = 7, -1
Hence, option 3 represents the roots of the given equation.
Question 19
1 / -0

Which of the following is a root of the quadratic equation x² + 2x - 15 = 0?

A
-3

B
-6

C
5

D
3

Solution

We have to find the roots of the quadratic equation x² + 2x - 15 = 0.
Comparing it with ax² + bx + c = 0, we get a = 1, b = 2 and c = -15.
So, x =
=
= 3, -5
So, x = 3 and x = -5 are the roots of quadratic equation.
Question 20
1 / -0

The roots of equation 2x² - 6 + 9 = 0 are _________.

A
not real

B
real unequal

C
real and equal

D
none of these

Solution

Finding b² - 4ac = 72 - 72 = 0. Hence, roots are real and equal.