Question

If 81x^4 +1/x^4=3007 then (6x-7)^2 = ?

Answer 1

$(6x-7)^2$= 37

Step-by-step explanation:

Given,

$81x^4+ \frac{1}{x^4} =3007$

$\Leftrightarrow (9x^2)^2 +(\frac{1}{x^2})^2 +2.9x^2.\frac{1}{x^2}=3007 +18$

$\Leftrightarrow (9x^2+\frac{1}{x^2} )^2= 3025$

$\Leftrightarrow (9x^2+\frac{1}{x^2} )=55$

$\Leftrightarrow (9x^2+\frac{1}{x^2} ) - 2.3x.\frac{1}{x} =55 -6$

$\Leftrightarrow (3x-\frac{1}{x} ) ^2=49$

$\Leftrightarrow 3x- \frac{1}{x}=7$

$\Leftrightarrow 3x^2 -7x +1=0$

$\Leftrightarrow 3x^2 -7x =-1$

Then, $(6x-7)^2$

$=36 x^2 - 84 x +49$

$=12(3x^2- 7x) +49$

$= 12(-1) +49$

=37

Answer 2

This may be helpful to you.

It will make question easier to solve.