Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

share|improve this question

4 Answers 4

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 $y > 0$.

share|improve this answer

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 $$x=9.$$

share|improve this answer

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

share|improve this answer

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.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.