My quadratics

Algebra Level 4

If a + b + c = 0 a+b+c=0 and a a , b b and c c are rational, the roots of the equation ( b + c − a ) x 2 + ( c + a − b ) x + ( a + b − c ) = 0 (b+c-a)x^{2}+(c+a-b)x+(a+b-c)=0 are

irrational imaginary equal rational

Bala Vidyadharan by Bala vidyadharan , 20, India

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1 solution

Tom Engelsman
Oct 3, 2020

Let us substitute the equation a + b + c = 0 a+b+c=0 into the quadratic equation as follows:

− 2 a x 2 − 2 b x − 2 c = 0 ⇒ a x 2 + b x + c = 0 -2ax^2 - 2bx - 2c =0 \Rightarrow ax^2 + bx + c= 0 ;

with roots:

x = − b ± b 2 − 4 a c 2 a = − b ± ( − a − c ) 2 − 4 a c 2 a = − b ± a 2 + 2 a c + c 2 − 4 a c 2 a = − b ± ( a − c ) 2 2 a = − b ± ( a − c ) 2 a x = \frac{-b \pm \sqrt{b^2 -4ac}}{2a} = \frac{-b \pm \sqrt{(-a-c)^2 - 4ac}}{2a} = \frac{-b \pm \sqrt{a^2 + 2ac + c^2 - 4ac}}{2a} = \frac{-b \pm \sqrt{ (a-c)^2}}{2a} = \boxed{\frac{-b \pm (a-c)}{2a}}

which are both rational for all a , b , c ∈ Q . a,b,c \in \mathbb{Q}.

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