An algebra problem by Akeel Howell

Algebra Level 3

If x = 1 x = 1 and x = 2 x = 2 are solutions to the equation x 3 + a x 2 + b x + c = 0 x^3+ax^2+bx+c = 0 and a + b = 1 a+b = 1 , then what is the value of b b ?


The answer is 5.

Akeel Howell by Akeel Howell , 22, USA

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2 solutions

Let r r be the third root. Then by Vieta's we know that

a = − ( 1 + 2 + r ) = − 3 − r a = -(1 + 2 + r) = -3 - r and that b = 1 × 2 + 1 × r + 2 × r = 2 + 3 r b = 1 \times 2 + 1 \times r + 2 \times r = 2 + 3r .

Given that a + b = 1 a + b = 1 we find that a + b = − 3 − r + 2 + 3 r = 1 ⟹ 2 r = 2 ⟹ r = 1 a + b = -3 - r + 2 + 3r = 1 \Longrightarrow 2r = 2 \Longrightarrow r = 1 .

So finally b = 2 + 3 r = 5 b = 2 + 3r = \boxed{5} .

(With a = − 3 − r = − 4 a = -3 - r = -4 and c = − a b r = − 2 c = -abr = -2 the given equation pans out to be

x 3 − 4 x 2 + 5 x − 2 = ( x − 1 ) 2 ( x − 2 ) = 0 x^{3} - 4x^{2} + 5x - 2 = (x - 1)^{2}(x - 2) = 0 .)

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Nice solution!

Akeel Howell - 4 years, 2 months ago

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