Question

27x^6-152x^3+125=0
HELP ME AND PLEASE EXPLAIN!

Answer 1

`x=5/3\text{ or } x=1`

Explanation:

We have the equation:

`27x^6-152x^3+125=0`

We can solve this using u-substitution. Let's let u=x³. Therefore:

`27(x^3)^2-152(x^3)+125=0`

Substitute all x³s for u:

`27u^2-152u+125=0`

This is now in quadratic form. So, we can use the quadratic formula:

`u=(-b\pm √(b^2-4ac))/(2a)`

Our a is 27, b is -152, and c is 125. So:

`u=(-(-152)\pm √((-152)^2-4(27)(125)))/(2(27))`

Simplify:

`u=(152\pm√(9604))/(54)`

Simplify:

`u=(152\pm 98)/(54)`

Reduce. Divide everything by 2:

`u=(76\pm49)/(27)`

Split into two cases:

`u=(76+49)/(27)\text{ or } u=(76-49)/(27)`

Solve for each case:

`u=(125)/(27)\text{ or } u=(27)/(27)=1`

Substitute back x³ for our u:

`x^3=(125)/(27)\text{ or } x^3=1`

Take the cube root of each equation:

`x=\sqrt[3]{(125)/(27)}\text{ or } x=\sqrt[3]1`

Evaluate:

`x=5/3\text{ or } x=1`

And that's our two solutions.

And we're done!