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One root of ax^2 +bx +c=0 is twice other. prove2b^2=9ac

I am struggling here.

Question: One root of ax^2 +bx +c=0 is twice the other root. prove 2b^2 = 9ac

This is what I have so far:

So let α=2β\alpha = 2\beta

α+β=3β=32α\alpha + \beta = 3\beta = \frac{3}{2}\alpha

αβ=2β2=12α2\alpha \beta = 2\beta^2 = \frac{1}{2}\alpha^2

Now the 2β2 2\beta^2 looks like the LHS of equation above I have to prove. but how do I relate that to b, c and a?

I tried thinking about how 2β2 2\beta^2 relates to 3β3\beta , but that is just concerning beta so not sure if I can directly relate to b,c and a. I am confused. Can someone please give me a hint.
Original post by acomber

Can someone please give me a hint.


Consider:

What's the sum of the roots in terms of the coefficients of the quadratic?

Similarly the product.
Reply 2
The sum of the roots is -b/a and the product is c/a.

OK, rethink.

on left hand side I have

ba=3β \frac{-b}{a} = 3\beta

on rhs

ca=2β2 \frac{c}{a} = 2\beta^2

So to equate lhs to rhs I must square lhs and multiply by 2/9.

ie

29(ba)2=ca \frac{2}{9} (\frac{-b}{a})^2 = \frac{c}{a}


29b2a2=ca \frac{2}{9} \frac{b^2}{a^2} = \frac{c}{a}

Multiply both sides by a

29b2a=c \frac{2}{9} \frac{b^2}{a} = c


2b2=9ac 2b^2 = 9ac

Wohoo!


Original post by ghostwalker
Consider:

What's the sum of the roots in terms of the coefficients of the quadratic?

Similarly the product.
(edited 9 years ago)
Original post by acomber
The sum of the roots is -b/a and the product is c/a. Now where to go?


Well you know what they are in terms of alpha, so eliminate alpha between the two equations.

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