An algebra problem by A Former Brilliant Member

Algebra Level 2

Find the constant k k if one of the root of the following equation exceeds the other root by 5 5 .

x 2 − x − 2 k = 0 \large x^2-x-2k=0


The answer is 3.

A Former Brilliant Member by A Former Brilliant Member

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1 solution

let x 1 x_1 and x 2 x_2 be the roots

from the problem, x 1 = x 2 + 5 x_1=x_2+5

by Vieta's Formula , the sum of the roots is x 1 + x 2 = − b a = − ( − 1 1 ) = 1 x_1+x_2=-\dfrac{b}{a}=-\left(\dfrac{-1}{1}\right)=1

substitute: x 2 + 5 + x 2 = 1 x_2+5+x_2=1 ⟹ \implies 2 x 2 = − 4 2x_2=-4 ⟹ \implies x 2 = − 2 x_2=-2

it follows that x 1 = − 2 + 5 = 3 x_1=-2+5=3

by Vieta's Formula , the product of the roots is x 1 ( x 2 ) = c a = − 2 k 1 = − 2 k x_1(x_2)=\dfrac{c}{a}=\dfrac{-2k}{1}=-2k

substitue: 3 ( − 2 ) = − 2 k 3(-2)=-2k ⟹ \implies k = 3 \boxed{k=3}

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