Test: Quadratic Equations - 1 - Grade 12 MCQ

Roots are real. ∴ b² – 4ac ≥ 0
5² – 4(1)(1)
25 – 4 = 21 which is greater than 0.
Hence, the discriminant of the equation is greater than zero, so roots are real.

Roots are real and equal. ∴ b² – 4ac = 0
b² = 4ac
b = ±√4ac

Roots are imaginary. ∴ b² – 4ac < 0
(2k)² – 4(9)(1) < 0
4k² – 36 < 0
k² – 9 < 0
k² < 9
k < ±3
-3 < k < 3

To check the nature of the roots, the discriminant must be either equal to zero, less than zero or greater than zero.
Discriminant = b² – 4ac = – 11² – 4 × 5 × 13 = 121 – 260 = – 139
Since discriminant is less than zero, the roots of the equation are imaginary.

To check the nature of the roots, the discriminant must be either equal to zero, less than zero or greater than zero.
Discriminant = b² – 4ac = 10² – 4 × 25 × 1 = 100 – 100 = 0
Since discriminant is equal to zero, the roots of the equation are equal.

To check the nature of the roots, the discriminant must be either equal to zero, less than zero or greater than zero.
Discriminant = b² – 4ac = 10² – 4 × 2 × 9 = 100 – 72 = 28
Since discriminant is greater than zero, the roots of the equation are real and distinct.

The roots of both the equations are real.
Discriminant of 5x² + ax + 5 : b² – 4ac = a² – 4 × 5 × 5 = a² – 100
Since, roots are real; discriminant will be greater than 0.
a² ≥ 100
a ≥ ±10
Now, discriminant of x² – 12x + a : b² – 4ac = -12² – 4 × 1 × a = 144 – 4a
Since, roots are real; discriminant will be greater than 0.
144 ≥ 4a
a ≤ 144/4 = 36
For both the equations to have real roots the value of a must lie between 36 and 10.

Roots are equal. ∴ b² – 4ac = 0
-(2k)² – 4(k – 4)(k + 5) = 0
4k² – 4(k² – 4k + 5k – 20) = 0
4k² – 4(k² + k – 20) = 0
4k² – 4k² – 4k + 80 = 0
-4k = – 80
k = −80/−4 = 20

Given:
x² – (k + 2)x + 121 = 0
Concept Used:
If a quadratic equation (ax² + bx + c = 0) has equal roots, then discriminant should be zero i.e. b² – 4ac = 0
Calculation:
x² – (k + 2)x + 121 = 0
Therefore, b² – 4ac = 0
⇒ (k + 2)² – 4(1)(121) = 0
⇒ k² + 4k + 4 – 484 = 0
⇒ k² + 4k – 480 = 0
⇒ k² + 24k – 20k – 480 = 0
⇒ k(k + 24) – 20(k + 24) = 0
⇒ (k + 24)(k – 20) = 0
⇒ k = -24, 20
Therefore, k = 20 (k > 0)
∴ The value of k is 20  

Given:
x² + ax + b = 0 and x² + bx + a = 0
Concept used:
If two polynomials c(x) = 0 and d(x) = 0 have a common root α, then c(α) = d(α) = 0.
Calculation:
Let common root between c(x) = x2 + ax + b = 0 and d(x) = x2 + bx + a = 0 is α.
Then, we must have c(α) = d(α) = 0.
⇒ α² + aα + b = α² + bα + a = 0
⇒ aα - bα + b - a = 0
⇒ α(a - b) - (a - b) = 0
⇒ (a - b)(α - 1) = 0
⇒ a - b = 0 OR α - 1 = 0
⇒ a = b OR α = 1.
So, the value of |ab| = 1 × 1 = 1
∴ The answer is 1