Question

Find the value of k for which given equation has real and equal roots.
(k – 12)x² + 2(k – 12)x + 2 = 0

Answer 1

Given: (k – 12)x² + 2(k – 12)x + 2 = 0
First, we need to find the discriminant of the given quadratic equation.
We know that the discriminant of a quadratic equation, Δ = b² – 4ac.
On comparing the given equation with ax² + bx + c = 0, we get:
a = k – 12, b = 2(k – 12) and c = 2
Thus,
Δ = (2(k – 12))² – 4(k – 12)(2)
$\Rightarrow$Δ = 4(k² + 144 – 24k) – 8k + 96
$\Rightarrow$Δ = 4k² + 576 – 96k – 8k + 96
$\Rightarrow$Δ = 4k² – 104k + 672
Since the roots of the given quadratic equation are real and equal, Δ = 0.
Thus,
4k² – 104k + 672 = 0
$\Rightarrow$4(k² – 26k + 168) = 0
$\Rightarrow$k² – 26k + 168 = 0
On splitting the middle term –26k as –14k – 12k, we get:
k² –14k – 12k + 168 = 0
$\Rightarrow$k(k – 14) – 12(k – 14) = 0
$\Rightarrow$(k – 14) (k – 12) = 0
We know that if the product of two numbers is zero, then at least one of them must be zero.
Thus,
k – 14= 0 or k – 12= 0
$\Rightarrow$ k = 14 or k = 12
Now, on substituting k = 12 in the given equation, we observe that the coefficient of x² becomes zero, which is not possible for a quadratic equation. Thus, k = 12 is not possible.
Therefore, k = 14.