Question

if the roots of the equation x^2-8x+(a^2-6a)=0 are real then

Answer 1

For real roots b²-4ac>=0

(8)²-4(1)(a²-6a)>=0

64-4a²+24a>=0

Divide by 4

a²-6a-16<=0. Changing side

a²-8a+2a-16<=0

(a+2)(a-8)<=0

By wavy curve plot -2 and 8 number line for value greater then 8 its positive so

A€[-2,8]

Answer 2

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

Answer: -2 ≤ a ≤ 8