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How do you solve this: $x-1\dfrac{1}{2}\sqrt{x} = 4\dfrac{1}{2}$

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Set $\sqrt{x} = y$. This gives us $x = y^2$. Hence, $$x - 1 \dfrac12 \sqrt{x} = 4 \dfrac12 \implies x - \dfrac32 \sqrt{x} = \dfrac92 \implies y^2 - \dfrac32 y - \dfrac92=0 \implies 2y^2 - 3y - 9 = 0$$ Solve the quadratic for $y$, keeping mind that $y > 0$.

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For the equation to make sense, we know that $x\geq 0$. Then $x=|x|=(\sqrt{x})^2$. Completing the square by adding $\frac9{16}$ to both sides, we have $$\left(\sqrt{x}-\frac34\right)^2=\frac9{2}+\frac9{16}=\frac{81}{16}=\left(\frac94\right)^2,$$ so $$\sqrt{x}-\frac34=\pm\frac94.$$ Since $\sqrt{x}\geq 0$, then the $-\frac94$ doesn't make sense (check that), so $$\sqrt{x}-\frac34=\frac94,$$ so $$\sqrt{x}=\frac{12}4=3,$$ and so $$x=9.$$

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$$ x-1\tfrac{1}{2}\cdot\sqrt{x} - 4\tfrac{1}{2}=\left(\sqrt{x}-3\right)\cdot\left(\sqrt{x}+1\tfrac12\right) $$

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Isolate the $\sqrt x$ and then square both sides of the equality sign to form a quadratic equation. Then you solve. Don't forget that $x \ge 0$ for $\sqrt x$ to exist.

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