Let's just increase the powers!

Algebra Level 1

If $x+\dfrac{1}{x}=3$ , what is $x^4+\dfrac{1}{x^4}$ ?

1 solution

Jonathan Dou
Dec 26, 2016

Recall that $(a+b)^2 = a^2 + b^2 + 2ab$ .

So, we have $x+\frac{1}{x}=3$

Square both sides, $x^2+\frac{1}{x^2}+2=9$ , then $x^2+\frac{1}{x^2}=9-2=7$

Square again, $x^4+\frac{1}{x^4}+2=49$ , and finally, $x^4+\frac{1}{x^4}=49-2=\boxed{47}$ .

Same method:)

Dan Ley - 4 years, 5 months ago

Nice! Did the same..

Rishu Jaar - 4 years, 5 months ago

By rearranging first equation we have x^2 -3x + 1 = 0 Solving for that quadratic equation we have x = 2.62 or x = 0.32

For x = 2.62 the second expression gives a value of 47.33 For x= 0.32 the second expression gives a value of 95.37

Pierre Arpin - 4 years, 4 months ago

If you don't round $x=2.62$ and just leave it as $x=\frac{3+\sqrt{5}}{2}$ , then the answer to the second expression is exactly 47.

The other root of the quadratic that you formed is $x=0.382$ and not $x=0.32$ .

Dan Ley - 4 years, 4 months ago

(x + 1/x)² = x² + 2x + 1/x².

The answer is wrong!!!

Maggs Naidu - 4 years, 4 months ago

$(x+1/x)^2$ = $x^2+2+1/x^2$

Jonathan Dou - 4 years, 3 months ago

A potência é 4.Por que este resultado?

Francisco Maurílio Fernandes - 4 years, 4 months ago

Please can you deal this question with lucas method. Because someone say it's easy, means we can find out most tough questions through it

Amit Kumar - 3 years, 1 month ago