Question

If one zero of the polynomial ax 2 + bx +c is double the other then show that 2b 2 = 9ac ? Thanks

Answer 1

Let α,β are the two zeros of the polynomial ax²+bx+c
Then, α+β=-b/a and αβ=c/a
By the given condition, β=2α
∴, α+2α=-b/a
or, 3α=-b/a
or, α=-b/3a -------(1)  and
α.2α=c/a
or, 2α²=c/a
or, α²=c/2a
or, (-b/3a)²=c/2a [using (1)]
or, b²/9a²=c/2a
or, b²/9a=c/2
or, 2b²=9ac (Proved)

Answer 2

We have the following quadratic equation-
ax²+bx+c such that α, 2α are the roots
so, sum of the roots = -b/a = 3α
⇒ α = -b/3a  .......(1)
Products of the roots = c/a = 2α²
Put the value of α from equation (1)
⇒ c/a = 2(-b/3a)²
⇒ c/a = 2b²/9a²
⇒ 2b² = 9ac   Answer