The Medium Equality

Algebra Level 1

If $x$ is a real number satisfying $x^3 - \dfrac1{x^3} = 14$ , compute $x - \dfrac1x$ .

The answer is 2.

2 solutions

Rishabh Jain
Jan 29, 2016

$14=x^3-\dfrac{1}{x^3}=(x-\dfrac{1}{x})^3 +3(x-\dfrac{1}{x})$ $Put~x-\dfrac{1}{x}=t$ $\large \Rightarrow t^3+3t-14=0$ $\large \Rightarrow (t-2)(t^2+2t+7)=0$ $\large\Rightarrow x-\dfrac{1}{x}=t=\Large \boxed{\color{#D61F06}{\boxed{2}}}$

Very nice! Did the same.

A Former Brilliant Member - 5 years, 4 months ago

cool, same way.

Jason Chrysoprase - 5 years, 4 months ago

I try to explain this to my friend, yet he doesn't understand.

Can you guys ( expecially Risabh Cool and Abhay Kumar ) help me to tell my friend with your clear own word

This is his question :

How you find $t^3 + 3t - 14 = ( t-2)( t^2 +2t+7)$ ?

Please give me a clear and understand word, so I can teach my friend

Jason Chrysoprase - 5 years, 4 months ago

(X^3 + 3X - 14) = 0 Let this expression be f(X) Find values of X when f(X) = 0 by trial and error. Generally you would get a value within few tries. For example try f(–1), f(1), f(2), f(–2) You should get a value for X when f(X) = 0 by now. In this case, when X = 2, f(X) = 0. Therefore (X–2) is a factor of the equation. Now follow me: (X–2)*(AX^2 + BX + C) + D = (X^3 + 3X – 14) This is how we find the factors of a higher order polynomial. For finding a cubic polynomial, multiply the linear factor with a general quadratic factor so that you'll get your highest power as '3'. Now you need to expand the polynomial so that you will get the coefficients of X^3, X^2, X, and the constant in the terms of A, B, C, D. After simplifying it, just compare the coefficients on the LHS with the RHS and you'll get the desired solution! ;) Voilà!

aalekh patel - 5 years, 4 months ago

thx, ur explanation is more explainable

Jason Chrysoprase - 5 years, 4 months ago

@Jason Chrysoprase – My pleasure!

aalekh patel - 5 years, 4 months ago

Yasir Soltani
Feb 4, 2016

Let $x-\frac{1}{x}=a$ then cubing both sides we get $\left(x-\frac{1}{x}\right)^3=a^3$ $x^3-3.x^2.\frac{1}{x}+3.x\frac{1}{x^2}-\frac{1}{x^3}=a^3$ $x^3-\frac{1}{x^3}-3\left(x-\frac{1}{x}\right)=a^3$ $14-3a=a^3$ By inspection it is clear that $a=2$ is solutions so $a-2$ is a root $\Rightarrow (a-2)(a^2+2a+7)=0$ $\therefore x-\frac{1}{x}=2 \quad (\because x-\frac{1}{x} \in \mathbb{R})$

The Medium Equality

The answer is 2.

2 solutions

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