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Solving Equations: (lesson 4 of 4)

## Solving Equations with Radicals

A "radical" equation is an equation in which there is a variable inside the radical sign

### Four steps to solve equations with radicals

Step 1: Isolate the radicals to left side of the equal sign.

Step 2: Square each side of the equation

Step 3: Solve the resulting equation

Step 4: Check all solutions

### Equations with one radical

Example 1: Solve $\sqrt {2x + 3} - x = 0$

Solution

Step 1: Isolate the radicals to left side of the equal sign.

$$\sqrt {2x + 3} - x = 0$$ $$\sqrt {2x + 3} = x$$

Step 2: Square each side of the equation

$${\left( {\sqrt {2x + 3} } \right)^2} = {(x)^2}$$ $$2x + 2 = {x^2}$$ $${x^2} - 2x - 3 = 0$$

Step 3: Solve the resulting equation

$${x^2} - 2x - 3 = 0$$ $$a = 1, b = - 2, c = - 3$$ $${x_{1,2}} = \frac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}}$$ $${x_{1,2}} = \frac{{ - ( - 2) \pm \sqrt {{{( - 2)}^2} - 4 \cdot 1 \cdot ( - 3)} }}{{2 \cdot 1}}$$ $${x_{1,2}} = \frac{{2 \pm \sqrt {4 + 12} }}{2}$$ $${x_{1,2}} = \frac{{2 \pm 4}}{2}$$ $${x_1} = \frac{{2 + 4}}{2},{x_2} = \frac{{2 - 4}}{2}$$ $${x_1} = 3,{x_2} = - 1$$

Step 4: Check all solutions

Let's check to see if x1 = 3 is solution:

$$\sqrt {2x + 3} - x = 0$$ $$\sqrt {2 \cdot 3 + 3} - 3 = 0$$ $$\sqrt 9 - 3 = 0$$ $$3 - 3 = 0$$ $$0 = 0, OK$$

Let's check to see if x2 = -1 is solution:

$$\sqrt {2x + 3} - x = 0$$ $$\sqrt {2 \cdot ( - 1) + 3} - ( - 1) = 0$$ $$\sqrt { - 2 + 3} + 1 = 0$$ $$\sqrt 1 + 1 = 0$$ $$2 = 0,NOTOK$$

So, the original equation had a single solution x = 3 .

Exercise 1: Solve equations

Level 1

 $$\color{blue}{\sqrt {2x - 1} = x}$$ $x = 1$ $x = -1$ $x = 2$ $x = -2$

Level 2

 $$\color{blue}{\sqrt {x + 6} = x}$$ $x = 3, x = - 2$ $x = - 3, x = 2$ $x = - 3$ $x = 2$

Example 2: Solve $\sqrt {4x + 3} + 2x - 1 = 0$

Solution

$$\sqrt {4x + 3} + 2x - 1 = 0$$ $${(\sqrt {4x - 3} )^2} = {(1 - 2x)^2}$$ $$4x - 3 = 1 - 4x + 4{x^2}$$ $$4x - 3 - 1 + 4x - 4{x^2} = 0$$ $$- 4{x^2} + 8x - 4 = 0/:( - 4)$$ $${x^2} - 2x + 1 = 0$$ $${(x - 1)^2} = 0$$ $$x = 1$$

Let's check it to see if x = 1 is a solution to the original equation.

$$\sqrt {4x - 3} + 2x - 1 = 0$$ $$\sqrt {4 \cdot 1 - 3} + 2 \cdot 1 - 1 = 0$$ $$\sqrt 1 + 2 - 1 = 0$$ $$2 = 0$$

So, the original equation had no solutions.

Exercise 2: Solve equations

Level 1

 $$\color{blue}{\sqrt {3x + 1} - x - 1 = 0}$$ ${x_1} = 0, {x_2} = - 1$ $x = - 1$ ${x_1} = 0, {x_2} = 1$ $x = 0$

Level 2

 $$\color{blue}{2x + 3 - \sqrt {20x + 9} = 0}$$ ${x_1} = 0, {x_2} = - 2$ ${x_1} = 0, {x_2} = 2$ $x = 0$ $x = - 2$

### Equations with two radicals

Example 3: Solve $\sqrt {3x + 4} - \sqrt {2x + 1} = 1$

Solution

First thing to do is get one of the square roots by itself.

$$\sqrt {3x + 4} - \sqrt {2x + 1} = 1$$ $$\sqrt {3x + 4} = 1 + \sqrt {2x + 1}$$ $${(\sqrt {3x + 4} )^2} = {(1 + \sqrt {2x + 1} )^2}$$ $$3x + 4 = {1^2} + 2 \cdot 1 \cdot \sqrt {2x + 1} + {(\sqrt {2x + 1} )^2}$$ $$3x + 4 = 1 + 2\sqrt {2x + 1} + 2x + 1$$

We have managed to eliminate one of square roots!! We will continue to work this problem as we did in the previous examples.

$$3x + 4 = 1 + 2\sqrt {2x + 1} + 2x + 1$$ $$3x + 4 = 2 + 2x + 2\sqrt {2x + 1}$$ $$3x + 4 - 2 - 2x = 2\sqrt {2x + 1}$$ $${(x + 2)^2} = {(2\sqrt {2x + 1} )^2}$$ $${x^2} + 2 \cdot 2 \cdot x + {2^2} = {2^2}(2x + 1)$$ $${x^2} + 4x + 4 = 8x + 4$$ $${x^2} - 4x = 0$$ $$x(x - 4) = 0$$ $${x_1} = 0$$ $${x_2} = 4$$

Let's check both possible solutions. We will start with x = 0.

$$\sqrt {3x + 4} - \sqrt {2x + 1} = 1$$ $$\sqrt {3 \cdot 0 + 4} - \sqrt {2 \cdot 0 + 1} = 1$$ $$\sqrt 4 - \sqrt 1 = 1$$ $$1 = 1 \Rightarrow OK$$

Now let's check x = 4.

$$\sqrt {3x + 4} - \sqrt {2x + 1} = 1$$ $$\sqrt {3 \cdot 4 + 4} - \sqrt {2 \cdot 4 + 1} = 1$$ $$\sqrt {16} - \sqrt 9 = 1$$ $$1 = 1 \Rightarrow OK$$

Exercise 3: Solve equations

Level 1

 $$\color{blue}{\sqrt {x + 1} - \sqrt x = 1}$$ $x = 0, x = 1$ $x = 0$ $x = 1$ $x = - 1$

Level 2

 $$\color{blue}{\sqrt {x + 1} + \sqrt {2x + 3} = 5}$$ $x = 143$ $x = - 143$ $x = - 3$ $x = 3$