Find a is y = 3(a+2)x^2 + 6ax + (4-3a) has no zeros.

Question

Working our please!

Step by step..

Thank you!

Answer 1

A quadratic equation Ax² + Bx + C = 0 has no real zeros when B² - 4AC < 0.

So, we are looking for values of a such that (6a)² - 4(3)(a+2)(4-3a) < 0

36a² - 12(-3a² -2a + 8) < 0

72a² + 24a - 96 < 0

3a² + a - 4 < 0

(3a + 4)(a - 1) < 0

(3a+4)(a-1) = 0 when a = -4/3 or 1

If a < -4/3, then (3a + 4)(a - 1) > 0

If -4/3 < a < 1, then (3a + 4)(a - 1) < 0

If a > 1, then (3a + 4)(a - 1) > 0

So the given equation has no real zeros when -4/3 < a < 1.