Some times the simplest way is a division

Algebra Level 3

x 4 − 6 x 3 + 7 x 2 − 6 x + 1 = 0 \large { x }^{ 4 }-{ 6x }^{ 3 }+{ 7x }^{ 2 }-{ 6x }+1=0

Note that ( x + 1 x ) 2 = x 2 + 1 x 2 + 2 \left(x + \frac1x\right)^2 = x^2 + \frac1{x^2} + 2 and ( x + 1 x ) 3 = x 3 + 1 x 3 + 3 ( x + 1 x ) \left(x + \frac1x\right)^3 = x^3 + \frac1{x^3} + 3\left(x+ \frac1x\right) . With that in mind, find the sum of real values of x x that satisfy the equation above.


The answer is 5.

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David Amato Mantegari by David Amato Mantegari , 24, Brazil

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1 solution

Chew-Seong Cheong
Sep 21, 2015

x 4 − 6 x 3 + 7 x 2 − 6 x + 1 = 0 Divide throughout by x 2 . x 2 − 6 x + 7 − 6 x + 1 x 2 = 0 x 2 + 1 x 2 − 6 x − 6 x + 7 = 0 ( x 2 + 2 + 1 x 2 ) − 6 ( x + 1 x ) + 5 = 0 ( x + 1 x ) 2 − 6 ( x + 1 x ) + 5 = 0 ( x + 1 x − 5 ) ( x + 1 x − 1 ) = 0 \begin{aligned} x^4-6x^3+7x^2-6x+1 & = 0 \quad \quad \color{#3D99F6}{\text{Divide throughout by } x^2. } \\ x^2 - 6 x + 7 - \frac{6}{x} + \frac{1}{x^2} & = 0 \\ x^2 + \frac{1}{x^2} - 6x - \frac{6}{x} + \color{#3D99F6}{7} & = 0 \\ \left(x^2 +\color{#3D99F6}{2} + \frac{1}{x^2} \right) - 6 \left(x + \frac{1}{x} \right) + \color{#3D99F6}{5} & = 0 \\ \left(x + \frac{1}{x} \right)^2 - 6 \left(x + \frac{1}{x} \right) + 5 & = 0 \\ \left(x + \frac{1}{x} - 5 \right) \left(x + \frac{1}{x} - 1 \right) & = 0 \end{aligned}

{ x + 1 x − 5 = 0 ⇒ x 2 − 5 x + 1 = 0 ⇒ x = 5 ± 21 2 x + 1 x − 1 = 0 ⇒ x 2 − x + 1 = 0 No real roots \begin{cases} x + \dfrac{1}{x} - 5 = 0 & \Rightarrow x^2 -5x +1 = 0 & \Rightarrow x = \dfrac{5 \pm \sqrt{21}}{2} \\ x + \dfrac{1}{x} - 1 = 0 & \Rightarrow x^2 -x +1 = 0 & \color{#D61F06}{\text{No real roots}} \end{cases}

Therefore the sum of real roots = 5 = \boxed{5}

In response to Dev Sharma , the plot of the f ( x ) = x 4 − 6 x 3 + 7 x 2 − 6 x + 1 f(x) = x^4-6x^3+7x^2-6x+1 is as follows. It is clearly shown that there are only two real roots 5 − 21 2 = 0.208712153 \dfrac{5 - \sqrt{21}}{2} = 0.208712153 and 5 + 21 2 = 4.791287847 \dfrac{5 + \sqrt{21}}{2} = 4.791287847

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using descarte rule of sign, we find all roots are real and their sum should be 6.

Dev Sharma - 5 years, 8 months ago

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Unlikely, because:

x 4 − 6 x 3 + 7 x 2 − 6 x + 1 = 0 ( x 2 − 5 x + 1 ) ( x 2 − x + 1 ) = 0 x = { 5 ± 21 2 1 ± i 3 2 \begin{aligned} x^4-6x^3+7x^2-6x+1 & = 0 \\ (x^2-5x+1)(x^2 -x +1) & = 0 \\ x & = \begin{cases} \dfrac{5 \pm \sqrt{21}}{2} \\ \dfrac{1 \pm i \sqrt{3}}{2} \end{cases} \end{aligned}

− a 3 = 5 − 21 2 + 5 + 21 2 + 1 − i 3 2 + 1 + i 3 2 = 6 -a_3 = \dfrac{5 - \sqrt{21}}{2} + \dfrac{5 + \sqrt{21}}{2} + \dfrac{1 - i \sqrt{3}}{2} + \dfrac{1 + i \sqrt{3}}{2} = 6

a 0 = 5 − 21 2 × 5 + 21 2 × 1 − i 3 2 × 1 + i 3 2 = 1 a_0 = \dfrac{5 - \sqrt{21}}{2} \times \dfrac{5 + \sqrt{21}}{2} \times \dfrac{1 - i \sqrt{3}}{2} \times \dfrac{1 + i \sqrt{3}}{2} = 1 .

− a 1 = 2 5 − 21 + 2 5 + 21 + 2 1 − i 3 + 2 1 + i 3 = 5 + 21 2 + 5 − 21 2 + 1 + i 3 2 + 1 − i 3 2 = 6 -a_1 = \dfrac{2}{5 - \sqrt{21}} + \dfrac{2}{5 + \sqrt{21}} + \dfrac{2}{1 - i \sqrt{3}} + \dfrac{2}{1 + i \sqrt{3}} \\ \quad \quad = \dfrac{5 + \sqrt{21}}{2} + \dfrac{5 - \sqrt{21}}{2} + \dfrac{1 + i \sqrt{3}}{2} + \dfrac{1 - i \sqrt{3}}{2} = 6

If you do the math, you will find that a 2 = 7 a_2 = 7 .

Chew-Seong Cheong - 5 years, 8 months ago

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