Question

Find the values of k for which the roots are real and equal in each of the following equation:

x² - 2(k + 1)x + (k + 4) = 0

Answer 1

The given equation is x² - 2(k + 1)x + (k + 4) = 0

This equation is in the form of ax² + bx + c = 0

Here a = 1, b = -2(k + 1) and c = k + 4

⇒ The nature of the roots of this equation is given that it is real and equal

i.e D = b² - 4ac

⇒ [-2(k + 1)]² - 4 × 1 × (k + 4) = 0

⇒ 4(k + 1)² - 4(k + 4) = 0

⇒ 4(k² + 1 + 2k) - 4k - 16 = 0

⇒ 4k² + 4 + 8k - 4k - 16 = 0

⇒ 4k² + 4k - 12 = 0

⇒ 4(k² + k - 3) = 0

⇒ k² + k - 3 = 0

The value of 'k' can be obtained by formula

`k=(-b+-sqrt(b^2-4ac))/(2a)`

where a = 1, b = 1 and c = -3

`k=(-1+sqrt((1)^2-4(1)(-3)))/2=1/2`

`k=(-1-sqrt((1)^2-4(1)(-3)))/2=1/2`

The value of 'k' for the given equation are 1/2