## Textbooks & Solution Manuals

Find the Source, Textbook, Solution Manual that you are looking for in 1 click.

## Tip our Team

Our Website is free to use.
To help us grow, you can support our team with a Small Tip.

## Holooly Tables

All the data tables that you may search for.

## Holooly Help Desk

Need Help? We got you covered.

## Holooly Arabia

For Arabic Users, find a teacher/tutor in your City or country in the Middle East.

Products

## Textbooks & Solution Manuals

Find the Source, Textbook, Solution Manual that you are looking for in 1 click.

## Holooly Arabia

For Arabic Users, find a teacher/tutor in your City or country in the Middle East.

## Holooly Help Desk

Need Help? We got you covered.

## Q. 1.6.5

Solving an Equation Containing Two Radicals

Solve  $\sqrt{2x + 3} – \sqrt{x + 1} = 1$.

## Verified Solution

$\sqrt{2x + 3} – \sqrt{x + 1}= 1$

$\fbox{Isolate one of the radicals on one side of the equation.}$

Step 1   $\sqrt{2x + 3} = 1 +\sqrt{x + 1}$                         Isolate $\sqrt{2x + 3}$.

Step 2  $(\sqrt{2x + 3})² = (1 + \sqrt{x + 1})²$                    Square each side.

$2x + 3 = 1 + 2\sqrt{x + 1} +(x + 1)$          Be careful:

↑                                         (a + b)² = a² + 2ab + b²

$\fbox{Don’t forget this term when squaring.}$

Step 1           $x + 1 = 2\sqrt{x + 1}$                         Isolate the remaining radical.

Step 2           $(x + 1)² =(2\sqrt{x + 1})²$                    Square again.

x² + 2x + 1 = 4(x + 1)←(ab)² = a²b²    Apply the exponents.

x² + 2x + 1 = 4x + 4                             Distributive property

Step 3            x² – 2x – 3 = 0                                 Write in standard form.

(x – 3)(x + 1) = 0                                            Factor.

x – 3 = 0   or     x + 1 = 0                               Zero-factor property

x = 3         or     x = -1                                     Proposed solutions

Step 4

CHECK  $\sqrt{2x + 3} – \sqrt{x + 1} = 1$         Original equation

$\left. \begin{matrix}\sqrt{2(3)+3}-\sqrt{3+1} ≟ 1&\text{Let} x =3.\\ \sqrt{9}-\sqrt{4}≟ 1\\ 3-2≟ 1\\1=1 ✓& \text{True}\end{matrix}\right|\begin{matrix}\sqrt{2(-1)+3}-\sqrt{-1+1} ≟ 1&\text{Let} x = -1.\\\sqrt{1}-\sqrt{0}≟ 1\\ 1-0≟ 1\\1=1 ✓& \text{ True}\end{matrix}$

Both 3 and -1 are solutions of the original equation, so { -1, 3} is the solution set.