Roots and Rational Exponents

Recall that $\sqrt[n]{x}$ can be defined as follows:

1. If $n$ is odd, then $\sqrt[n]{x}$ is the number $t$ such that $t^n=x$.

2. If $n$ is even and $x\ge0$, then $\sqrt[n]{x}$ is the number $t\ge0$ such that $t^n=x$. (Notice that $\sqrt[n]{x}$ is undefined if $n$ is even and $x<0$.

We can define rational exponents in the following manner:

If $m/n$ is a fraction reduced to lowest terms, then

$x^{m/n}=(\sqrt[n]{x})^m=\sqrt[n]{x^m}$,

assuming that $x\ge0$ if $n$ is even.

Ex 1 Solve the equation $2\sqrt{x}=x-15$.

Sol Squaring both sides gives $4x=(x-15)^2=x^2-30x+225$, and then subtracting $4x$ from both sides gives $0=x^2-34x+225$. Then factoring gives $(x-9)(x-25)=0$, so $x=9$ or $x=25$. However, $x=9$ does not check in the original equation, so $x=25$ is the only solution.

Ex 2 Solve the equation $x-5\sqrt{x}=-6$.

Sol a Adding 6 to both sides gives $x-5\sqrt{x}+6=0$, and then factoring gives $(\sqrt{x}-3)(\sqrt{x}-2)=0$, so either $\sqrt{x}=3$ or $\sqrt{x}=2$ and therefore $x=9$ or $x=4$.

Sol b Adding $5\sqrt{x}+6$ to both sides gives $x+6=5\sqrt{x}$, and then squaring both sides gives $(x+6)^2=25x$; so $x^2+12x+36=25x\Rightarrow x^2-13x+36=0\Rightarrow(x-9)(x-4)=0\Rightarrow x=9$ or $x=4$.

Pr 1 Simplify the expression $\frac{6x}{\sqrt{x}}+\frac{2x^2}{x^{3/2}}$.

Pr 2 Solve the equation $3\sqrt{x}=x-4$.

Pr 3 Solve the equation $2x+\sqrt{x}=1$.

Pr 4 Solve the following equations:

a) $x^{2/3}=4$. b) $x^{3/2}=8$.

Pr 5 Find all values of $x$ for which the equation $\sqrt{x^2+25}=x+5$ is valid.

Pr 6 Rewrite the expression $\frac{5x^2+3}{\sqrt{x}}$ as a sum of terms with rational exponents.

Pr 7 Solve the equation $x=4\sqrt[3]{x}$.

Pr 8 Solve the equation $\sqrt{x^2-6x}=\sqrt{7x-40}$.

Pr 9 Solve the equation $\sqrt{x+2}+\sqrt{x-2}=\sqrt{4x-2}$.

Pr 10 Solve the equation $\sqrt{x+3}-\sqrt{x-2}=\sqrt{6x-11}$.

Pr 11 Solve the equation $\sqrt{x/2+4}=\sqrt[3]{2x+8}$.

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