An algebra problem by Ilham Saiful Fauzi

Algebra Level pending

x 4 − 8 x 3 + 20 x 2 − 32 x + 16 = 0 x^{4}-8x^{3}+20x^{2}-32x+16=0 Find the sum of its real roots.


The answer is 6.

Ilham Saiful Fauzi by Ilham Saiful Fauzi , 25, Indonesia

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1 solution

Michael Huang
Dec 8, 2016

Method 1 - Factoring out x 2 x^2

Rearranging terms, x 4 − 8 x 3 + 20 x 2 − 32 x + 16 = 1 x 2 ( x 2 − 8 x + 20 − 32 x + 16 x 2 ) Factor each term by x 2 = 1 x 2 ( ( x 2 + 16 x 2 ) − ( 8 x + 32 x ) + 20 ) = 1 x 2 ( ( x 2 + ( 4 x ) 2 + 8 ) − 8 ( x + 4 x ) + 20 − 8 ) For ( x 2 + 16 x 2 ) , complete the squares = 1 x 2 ( ( x + 4 x ) 2 − 8 ( x + 4 x ) + 12 ) = 1 x 2 ( ( x + 4 x ) − 6 ) ( ( x + 4 x ) − 2 ) Treating ( x + 4 x ) a s a v a r i a b l e , f a c t o r = ( x 2 + 4 − 6 x ) ( x 2 + 4 − 2 x ) Distribute 1 x for each factor \begin{array}{rlccl} x^4 - 8x^3 + 20x^2 - 32x + 16 &= \dfrac{1}{x^2}\left(x^2 - 8x + 20 - \dfrac{32}{x} + \dfrac{16}{x^2}\right) & & & {\color{#3D99F6}\text{Factor each term by }x^2}\\ &= \dfrac{1}{x^2}\left(\left(x^2 + \dfrac{16}{x^2}\right) - \left(8x + \dfrac{32}{x}\right) + 20\right)\\ &= \dfrac{1}{x^2}\left(\left(x^2 + \left(\dfrac{4}{x}\right)^2 + 8\right) - 8\left(x + \dfrac{4}{x}\right) + 20 - 8\right) & & & {\color{#3D99F6}\text{For }\left(x^2 + \dfrac{16}{x^2}\right),\text{ complete the squares}}\\ &= \dfrac{1}{x^2}\left(\left(x + \dfrac{4}{x}\right)^2 - 8\left(x + \dfrac{4}{x}\right) + 12\right)\\ &= \dfrac{1}{x^2}\left(\left(x + \dfrac{4}{x}\right) - 6\right)\left(\left(x + \dfrac{4}{x}\right) - 2\right) & & & {\color{#3D99F6}\text{Treating }\left(x + \dfrac{4}{x}\right) as a variable, factor}\\ &= \left(x^2 + 4 - 6x\right)\left(x^2 + 4 - 2x\right) & & & {\color{#3D99F6}\text{Distribute }\dfrac{1}{x}\text{ for each factor}} \end{array} So we evaluate ( x 2 − 6 x + 4 ) ( x 2 − 2 x + 4 ) = 0 \left(x^2 - 6x + 4\right)\left(x^2 - 2x + 4\right) = 0

Method 2 - Combining Like Terms

Factorization can also be done by combining like terms: x 4 − 8 x 3 + 20 x 2 − 32 x + 16 = x 4 + ( − 6 x 3 − 2 x 3 ) + ( 4 x 2 + 12 x 2 + 4 x 2 ) + ( − 8 x − 24 x ) + 16 = ( x 4 − 6 x 3 + 4 x 2 ) + ( − 2 x 3 + 12 x 2 − 8 x ) + ( 4 x 2 − 24 x + 16 ) = x 2 ( x 2 − 6 x + 4 ) − 2 x ( x 2 − 6 x + 4 ) + 4 ( x 2 − 6 x + 4 ) = ( x 2 − 6 x + 4 ) ( x 2 − 2 x + 4 ) \begin{array}{rlccl} x^4 - 8x^3 + 20x^2 - 32x + 16 &= x^4 + \left(-6x^3 - 2x^3\right) + \left(4x^2 + 12x^2 + 4x^2\right) + \left(-8x - 24x\right) + 16\\ &= \left(x^4 - 6x^3 + 4x^2\right) + \left(-2x^3 + 12x^2 - 8x\right) + \left(4x^2 - 24x + 16\right)\\ &= x^2\left(x^2 - 6x + 4\right) - 2x\left(x^2 - 6x + 4\right) + 4\left(x^2 - 6x + 4\right)\\ &= \left(x^2 - 6x + 4\right)\left(x^2 - 2x + 4\right) \end{array}

Firstly, if x 2 − 6 x + 4 = 0 x^2 - 6x + 4 = 0 then by quadratic formula , x = 6 ± 6 2 − 4 ⋅ 1 ⋅ 4 2 = 6 ± 20 2 = 6 ± 2 5 2 = 2 ( 3 ± 5 ) 2 = 3 ± 5 \begin{array}{rl} x &= \dfrac{6 \pm \sqrt{6^2 - 4\cdot 1 \cdot 4}}{2}\\ &= \dfrac{6 \pm \sqrt{20}}{2}\\ &= \dfrac{6 \pm 2\sqrt{5}}{2}\\ &= \dfrac{2\left(3 \pm \sqrt{5}\right)}{2}\\ &= 3 \pm \sqrt{5} \end{array} Also note that it's simple to solve by completing the squares , x 2 − 6 x + 9 + 4 − 9 = 0 ( x − 3 ) 2 = 5 x = 3 ± 5 \begin{array}{rl} x^2 - 6x + 9 + 4 - 9 &= 0\\ (x - 3)^2 &= 5\\ x &= 3 \pm \sqrt{5} \end{array}

Secondly, if x 2 − 2 x + 4 = 0 x^2 - 2x + 4 = 0 since the discriminant is Δ = b 2 − 4 a c = ( − 2 ) 2 − 4 ( 1 ) ( 4 ) = − 12 < 0 \Delta = b^2 - 4ac = \left(-2\right)^2 - 4(1)(4) = -12 < 0 this shows that the roots are imaginary. Because the problem asks for real solutions, we can neglect these roots.

Therefore, the sum of the real roots are 3 + 5 + ( 3 − 5 ) = 6 3 + \sqrt{5} + \left(3 - \sqrt{5}\right) = \boxed{6}

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