Question

The equation y = 3x^2 + 2px + 4q has no real roots. Show that p^2 < 12q

Answer 1

Answer:

See below

Step-by-step explanation:

Given Equation is:

$y = 3x^2 + 2px + 4q$

Comparing it with the quadratic equation, $y = ax^2+bx+c$, we get:

a = 3, b = 2p and c = 4q

Since, the given equation has no real roots, Discriminant < 0

We know that:

Discriminant = $b^2-4ac$

Put this in the above inequality:

$b^2-4ac < 0\\\\Put \ values \ of \ a, \ b \ and \ c\\\\(2p)^2 - 4(3)(4q)<0\\\\4p^2 -48q <0\\\\Add \ 48q \ to \ both \ sides\\\\4p^2 < 48q\\\\Divide \ both \ sides \ by 4\\\\\boxed{p^2 < 12 q}\\\\Hence\ proved.\\\\\rule[225]{225}{2}$

Hope this helped!

~AH1807