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Algebra Level 2

z = x 4 + 1 x 4 \large z = x^{4}+\frac{1}{x^{4}}

Solve for z z , where z z is a positive integer, if x + 1 x = 3 x + \dfrac 1x = 3 .


The answer is 47.

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3 solutions

x + 1 x = 3 Squaring both sides x 2 + 2 + 1 x 2 = 9 x 2 + 1 x 2 = 7 Squaring both sides x 4 + 2 + 1 x 4 = 49 x 2 + 1 x 2 = 47 z = 47 \begin{aligned} x + \frac 1x & = 3 & \small \color{#3D99F6} \text{Squaring both sides} \\ x^2 + 2 + \frac 1{x^2} & = 9 \\ x^2 + \frac 1{x^2} & = 7 & \small \color{#3D99F6} \text{Squaring both sides} \\ x^4 + 2 + \frac 1{x^4} & = 49 \\ x^2 + \frac 1{x^2} & = 47 \\ \implies z & = \boxed{47} \end{aligned}

Matin Naseri
Apr 5, 2018

x + \text{}x+ 1 x = a \frac{1}{x}=a \implies x 4 + \text{}x^{4}+ 1 x 4 = a 4 4 a 2 + 2 \frac{1}{x^{4}}={a^{4}}-{4{a^{2}}}+2

we have :

x + \text{}x+ 1 x = 3 \frac{1}{x}=3 \implies x 4 + \text{}x^{4}+ 1 x 4 = z \frac{1}{x^{4}}=z

Solve for z z .

For x + \text{}x+ 1 x = 3 \frac{1}{x}=3 \implies x 4 + \text{}x^{4}+ 1 x 4 = 3 4 4 × 3 2 + 2 = 47 \frac{1}{x^{4}}= {3^{4}}-{4×{3^{2}}}+2={\boxed{47}}

x + 1 x = 3 x+\dfrac{1}{x}=3

squaring both sides, we have

( x + 1 x ) 2 = 3 2 \left(x+\dfrac{1}{x}\right)^2=3^2 \color{#D61F06}\implies x 2 + 2 ( x ) ( 1 x ) + ( 1 x ) 2 = 9 x^2+2(x)\left(\dfrac{1}{x}\right)+\left(\dfrac{1}{x}\right)^2=9 \color{#D61F06}\implies x 2 + 2 + 1 x 2 = 9 x^2+2+\dfrac{1}{x^2}=9 \color{#D61F06}\implies x 2 + 1 x 2 = 7 x^2+\dfrac{1}{x^2}=7

squaring both sides, we have

( x 2 + 1 x 2 ) 2 = 7 2 \left(x^2+\dfrac{1}{x^2}\right)^2=7^2 \color{#D61F06}\implies x 4 + 2 ( x 2 ) ( 1 x 2 ) 2 + ( 1 x 2 ) 2 = 49 x^4+2(x^2)\left(\dfrac{1}{x^2}\right)^2+\left(\dfrac{1}{x^2}\right)^2=49 \color{#D61F06}\implies x 4 + 1 x 4 = 49 2 = 47 x^4+\dfrac{1}{x^4}=49-2=\color{#69047E}\large{\boxed{47}}

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