My friend asked me a question!

Algebra Level 1

$x - \frac 1x = \frac{15}4 \quad, \quad \left| x + \frac 1 x \right| = \ ?$

17/4 4 13/4 1/4

3 solutions

Anish Harsha
Jun 20, 2015

A simple problem :

If the x - 1/x = 15/4, then the other side of the equation's denominator will be the same.
So, x = 4 then 4 - 1/4 = 15/4, then 4 + 1/4 = 17/4.

Your reasoning isn't the best. Note it forms a quadratic in $x$

$x-\frac{1}{x}=\frac{15}{4} \implies x^2-\frac{15}{4}x-1=0$

Solving the quadratic equation we get $x=4$ and $x=-\frac{1}{4}$

You can plug both values into $x+\frac{1}{x}$

$4+\frac{1}{4}=\frac{17}{4}$ and $-\frac{1}{4}-4=-\frac{17}{4}$

Isaac Buckley - 5 years, 11 months ago

It will also come -17/4 , but i have only written only one option in it, so it is the answer.

Anish Harsha - 5 years, 11 months ago

Then I recomend you change the question to "Find $| x+\frac{1}{x}|$ " since it removes all ambiguity.

Also your solution does not mention anything about there being a two different values.

Isaac Buckley - 5 years, 11 months ago

Isaac Buckley
Jun 20, 2015

Given that $x-\frac{1}{x}=\frac{15}{4}$

We then square it to get $x^2-2+\frac{1}{x^2}=\frac{225}{16}$

Add $4$ to each side $x^2+2+\frac{1}{x^2}=\frac{289}{16}$

We square root both sides to finally get $x+\frac{1}{x}=\frac{17}{4}$ and $-\frac{17}{4}$

Sriram Venkatesan
Jun 20, 2015

x - 1/x = 15/4 x = 15x +4/4x

Finally , we get the quadratic equation, 4x^2 - 15x +4 = 0

This can be factorised to get the root 1/4 . This can be substituted in the place of x to get

1/4 + 4 = 17/4

$x=\frac{1}{4}$ is not a solution of $4x^2-15x+4=0$

Isaac Buckley - 5 years, 11 months ago