How do you solve x^(3/2) -2x^(3/4) +1 = 0?

Jan 25, 2017

$x = 1$

Explanation:

Let $t = {x}^{\frac{3}{4}}$

Then:

${x}^{\frac{3}{2}} = {x}^{\frac{3}{4} \cdot 2} = {\left({x}^{\frac{3}{4}}\right)}^{2} = {t}^{2}$

and our equation becomes:

$0 = {t}^{2} - 2 t + 1 = {\left(t - 1\right)}^{2}$

This has one (repeated) root, namely $t = 1$

So:

${x}^{\frac{3}{4}} = 1$

If $x \ge 0$ then:

$x = {x}^{1} = {x}^{\frac{3}{4} \cdot \frac{4}{3}} = {\left({x}^{\frac{3}{4}}\right)}^{\frac{4}{3}} = {1}^{\frac{4}{3}} = 1$

This is the only Real root.