Q. 8.110 - Intermediate Algebra

Question

$$\begin{gathered} \mathrm{Simplify}: \\ \frac{\sqrt{q}-\sqrt{10}}{\sqrt{q}-\sqrt{10}} \end{gathered}$$

Step-by-Step Solution

Answer

The simplified value is $\frac{{(\sqrt{\mathrm{q}}-\sqrt{10})}^2}{\mathrm{q}-10}$

Step 1. Given information

The given equation is $\frac{\sqrt{q}-\sqrt{10}}{\sqrt{q}-\sqrt{10}}$

Step 2. Multiply the numerator and denominator by the conjugate of the denominator.

$$\begin{gathered} \frac{\sqrt{q}-\sqrt{10}}{\sqrt{q}+\sqrt{10}}\times \frac{\sqrt{q}-\sqrt{10}}{\sqrt{q}-\sqrt{10}} \\ \mathrm{Multiply} \mathrm{the} \mathrm{conjugates} \mathrm{in} \mathrm{the} \mathrm{denominator} \\ \frac{(\sqrt{\mathrm{q}}-\sqrt{10})(\sqrt{\mathrm{q}}-\sqrt{10})}{{\left(\sqrt{\mathrm{q}}\right)}^2-{\left(\sqrt{10}\right)}^2} \\ \mathrm{Simplify} \mathrm{the} \mathrm{denominator} \\ \frac{{(\sqrt{\mathrm{q}}-\sqrt{10})}^2}{\mathrm{q}-10} \\ \mathrm{We} \mathrm{do} \mathrm{not} \mathrm{square} \mathrm{the} \mathrm{numerator}. \mathrm{Leaving} \mathrm{it} \mathrm{in} \mathrm{factored} \mathrm{form}, \mathrm{we} \mathrm{can} \mathrm{see} \mathrm{there} \mathrm{are} \mathrm{no} \mathrm{common} \mathrm{factors} \mathrm{to} \mathrm{remove} \mathrm{from} \mathrm{the} \mathrm{numerator} \mathrm{and} \mathrm{denominator}. \end{gathered}$$