# Chapter 1 - Equations and Inequalities - Exercise Set 1.6 - Page 179: 51

The solution is $x=1$

#### Work Step by Step

$x^{3/2}-2x^{3/4}+1=0$ Rewrite $x^{3/2}$ as $(x^{1/2})^{3}$ and $x^{3/4}$ as $(x^{1/4})^{3}$: $(x^{1/2})^{3}-2(x^{1/4})^{3}+1=0$ Let $u$ be equal to $x^{1/4}$ If $u=x^{1/4}$, then $u^{2}=x^{1/2}$ Rewrite the equation using the new variable $u$: $(u^{2})^{3}-2u^{3}+1=0$ Simplify: $u^{6}-2u^{3}+1=0$ Let $z$ be equal to $u^{3}$ If $z=u^{3}$, then $z^{2}=u^{6}$ Rewrite the equation again, using the variable $z$: $z^{2}-2z+1=0$ Solve by factoring: $(z-1)^{2}=0$ $\sqrt{(z-1)^{2}}=\sqrt{0}$ $z-1=0$ Solve for $z$: $z=1$ Substitute $z$ back to $u^{3}$ and solve for $u$: $u^{3}=1$ $\sqrt[3]{u^{3}}=\sqrt[3]{1}$ $u=1$ Substitute $u$ back to $x^{1/4}$ and solve for $x$: $x^{1/4}=1$ $(x^{1/4})^{4}=1^{4}$ $x=1$ The solution is $x=1$

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