Question

if the roots of x^2+a^2-8x+6a=0 are real then find the value of a?​

Answer 1

Answer: -2 ≤ a ≤ 8

Step-by-step explanation:

x² - 8x + (a² - 6a) = 0

For the quadratic have real roots, then

⇒ the discriminant must be greater than or equal to zero

Solve a:

b² - 4ac ≥ 0

Insert the known values:

(8)² - 4(1)(a² - 6a)  ≥ 0

expand the terms:

64 - 4a² + 24a  ≥ 0

Switch sides:

4a² - 24a - 64 ≤ 0

Divide by 4 through:

a² - 6a - 16 ≤ 0

Factorise:

(a + 2) (a - 8) ≤ 0

Find the range:

⇒ so -2 ≤ a ≤ 8