Roots and Rational Exponents

Sol 1 $\frac{6x}{\sqrt{x}}+\frac{2x^2}{x^{3/2}}=\frac{6x}{x^{1/2}}+ \frac{2x^2}{x^{3/2}}=6x^{1/2}+2x^{1/2}=8x^{1/2}=8\sqrt{x}$

Sol 2a $3\sqrt{x}=x-4$ gives $0=x-3\sqrt{x}-4$, so factoring yields $(\sqrt{x}-4)(\sqrt{x}+1)=0$. Therefore either $\sqrt{x}=4$ or $\sqrt{x}=-1$, but $\sqrt{x}\ne-1$ since $\sqrt{x}$ is never negative; so $\sqrt{x}=4$ and therefore $x=16$.

Sol 2b Squaring both sides of $3\sqrt{x}=x-4$ gives $9x=(x-4)^2$ or $9x=x^2-8x+16$, so $0=x^2-17x+16$ and $(x-16)(x-1)=0$. Therefore $x=16$ or $x=1$, but $x=16$ is the only solution since $x=1$ does not check in the original equation.

Sol 3a $2x+\sqrt{x}=1$ gives $2x+\sqrt{x}-1=0$, so factoring gives $(2\sqrt{x}-1)(\sqrt{x}+1)=0$. Then $\sqrt{x}=1/2$ or $\sqrt{x}=-1$, but $\sqrt{x}\ne-1$ since $\sqrt{x}$ is never negative; so $\sqrt{x}=1/2$ and therefore $x=1/4$.

Sol 3b $2x+\sqrt{x}=1$ gives $\sqrt{x}=1-2x$, so squaring both sides gives $x=(1-2x)^2=1-4x+4x^2$. Then $0=4x^2-5x+1\Rightarrow (4x-1)(x-1)=0\Rightarrow x=1/4$ or $x=1$. The answer $x=1$ does not check in the original equation, though, so $x=1/4$ is the only answer.

Sol 4

a) Cubing both sides of $x^{2/3}=4$ gives $x^2=4^3=64$, so taking the square root of both sides gives $x=\pm8$.

b) Squaring both sides of $x^{3/2}=8$ gives $x^3=8^2=64$, so taking the cube root of both sides gives $x=4$.

Sol 5 Squaring both sides of $\sqrt{x^2+25}=x+5$ gives $x^2+25=(x+5)^2=x^2+10x+25$, so subtracting $x^2+25$ from both sides gives $0=10x$ and so $x=0$. Therefore this equation is valid only for $x=0$.

Sol 6 $\frac{5x^2+3}{\sqrt{x}}=\frac{5x^2+3}{x^{1/2}}= \frac{5x^2}{x^{1/2}}+\frac{3}{x^{1/2}}=5x^{3/2}+3x^{-1/2}$.

Sol 7 Cubing both sides of $x=4\sqrt[3]{x}$ gives $x^3=64x$, so $x^3-64x=0\Rightarrow x(x^2-64)=0\Rightarrow x(x-8)(x+8)=0\Rightarrow x=0$ or $x=8$ or $x=-8$.

Sol 8 $\sqrt{x^2-6x}=\sqrt{7x-40}\Rightarrow x^2-6x=7x-40\Rightarrow x^2-13x+40=0\Rightarrow (x-8)(x-5)=0\Rightarrow x=8$ or $x=5$. However, $x=5$ does not check in the original equation, so $x=8$ is the only solution.

Sol 9 Squaring both sides of $\sqrt{x+2}+\sqrt{x-2}=\sqrt{4x-2}$ gives $x+2+2\sqrt{x+2}\sqrt{x-2}+x-2=4x-2$, so $2\sqrt{x+2}\sqrt{x-2}=2x-2$ and therefore $\sqrt{x+2}\sqrt{x-2}=x-1$ or $\sqrt{x^2-4}=x-1$. Squaring both sides of this equation yields $x^2-4=(x-1)^2=x^2-2x+1$, so $2x=5$ and $x=5/2$. Since this answer checks in the original equation, it is the only solution.

Sol 10 Squaring both sides of $\sqrt{x+3}-\sqrt{x-2}=\sqrt{6x-11}$ gives $x+3-2\sqrt{x+3}\sqrt{x-2}+x-2=6x-11$, so $-2\sqrt{x+3}\sqrt{x-2}=4x-12$ and therefore $\sqrt{x+3}\sqrt{x-2}=6-2x$. Squaring both sides of this equation gives $(x+3)(x-2)=(6-2x)^2$, so $x^2+x-6=36-24x+4x^2$ and therefore $0=3x^2-25x+42=0$. Factoring gives $(3x-7)(x-6)=0$, so $x=7/3$ or $x=6$. However, $x=6$ does not check in the original equation, so $x=7/3$ is the only solution.

Sol 11 Raising both sides of the equation $\sqrt{x/2+4}=\sqrt[3]{2x+8}$ to the 6th power gives $((x/2+4)^{1/2})^6=((2x+8)^{1/3})^6$ or $(x/2+4)^3=(2x+8)^2$.

Multiplying out both sides gives $(x/2)^3+3(x/2)^{2}(4)+3(x/2)(4)^2+4^3=4x^2+32x+64$, so $x^{3}/8+3x^2+24x+64=4x^2+32x+64$ and therefore $x^{3}/8-x^2-8x=0$ and $x^3-8x^2-64x=0$.

Then $x(x^2-8x-64)=0$, so either $x=0$ or $x^2-8x-64=0$. Completing the square in the last equation gives $x^2-8x+16=64+16$or $(x-4)^2=80$, so $x-4=\pm\sqrt{80}$ and $x=4\pm4\sqrt{5}$.

However, $x=4-4\sqrt{5}$ does not check in the original equation (since $2x+8<0$ when $x=4-4\sqrt{5}$ and therefore $\sqrt{x/2+4}>0$ while $\sqrt[3]{2x+8}<0$). Therefore $x=0$ and $x=4+4\sqrt{5}$ are the only solutions.

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