Solving square root equations and extraneous solutions

ac = 70  What two factors of 70 add to 37?

1, 70 no

2, 35 yes

Rewrite:   $7x^2+2x\ +\ 35x\ +10$  

Group:  $\left(7x^2+2x\right)+\left(35x+10\right)$  

Factor: x(7x + 2) + 5(7x + 2)

Factor: (7x + 2)(x + 5)

ac = -45 two factors of -45 that add to 12

1,45 no

-3, 15 yes 

 $5x^2-3x+15x-9$  

 $\left(5x^2-3x\right)+\left(15x-9\right)$  

 $x\left(5x-3\right)+3\left(5x-3\right)$  

(5x - 3)(x + 3)

1. Isolate the radical on one side of the equation (if necessary)

2. Square both sides of the equation

3. Solve the new equation

4. Check your solution(s). When you raise an equation to an EVEN power, you may create extraneous solutions that do NOT solve the original equation.

The radical is isolated so move on to squaring both sides.

 $\left(\sqrt{5x-4}\right)^2=\left(x-2\right)^2$  

5x - 4 = (x - 2)(x - 2)

 $5x-4=x^2-2x-2x+4$  

 $5x-4=x^2-4x+4$  

 $0\ =\ x^2-9x+8$  

0 = (x - 1)(x - 8)

x - 1 = 0 and x - 8 = 0

x = 1 and x = 8

CHECK EACH SOLUTION

 $\sqrt{5\left(1\right)-4}=1\ -\ 2$  

 $\sqrt{1}=-1$  

We are only looking for the principle (positive) square root so this is not true

 $1\ne-1$  

x = 1 is NOT a solution

 $\sqrt{5\left(8\right)-4}=8\ -\ 2$  

 $\sqrt{36}=6$  

We are only looking for the principle (positive) square root so this is true

 $6=6$  

x = 8 is a solution

 $z^2+6z+3=\left(3z+1\right)\left(3z+1\right)$  

 $z^2+6z+3=9z^2+6z+1$  

 $3=8z^2+1$  

 $2=8z^2$  

 $\frac{1}{4}=z^2$  

 $z\ =\ \pm\frac{1}{2}$  

Check  $x=\frac{1}{2}$  :  $\sqrt{\left(\frac{1}{2}\right)^2+6\left(\frac{1}{2}\right)+3}=3\left(\frac{1}{2}\right)+1$  

 $\sqrt{\frac{1}{4}+3+3}=\frac{3}{2}+1$  

 $\sqrt{\frac{25}{4}}=\frac{5}{2}$  

 $\frac{5}{2}=\frac{5}{2}$  x = 1/2 is a solution

Check  $x=-\frac{1}{2}$  :  $\sqrt{\left(-\frac{1}{2}\right)^2+6\left(-\frac{1}{2}\right)+3}=3\left(-\frac{1}{2}\right)+1$  

 $\sqrt{\frac{1}{4}-3+3}=-\frac{3}{2}+1$  

 $\sqrt{\frac{1}{4}}=-\frac{1}{2}$  

 $\frac{1}{2}\ne-\frac{1}{2}$  x =- 1/2 is NOT a solution

Here you need to isolate the radical first

 $\sqrt{7y^2+15y}=5+3y$  

Now solve as before:   $\left(\sqrt{7y^2+15y}\right)^2=\left(5+3y\right)^2$  

 $7y^2+15y=\left(5+3y\right)\left(5+3y\right)$  

 $7y^2+15y=25\ +30y+9y^2$  

 $0=2y^2+15y+25$  

 $0=2y^2+10y+5y+25$  

 $0=2y\left(y+5\right)+5\left(y+5\right)$  

0 = (y + 5)(2y + 5)

y + 5 = 0 or 2y + 5 = 0

y = -5 or y = -5/2

Check y = -5 $\sqrt{7\left(-5\right)^2+15\left(-5\right)}-2\left(-5\right)=5\ +\ -5$  

 $\sqrt{7\left(25\right)-75}+10=0$  

 $\sqrt{175-75}+10=0$  

 $\sqrt{100}+10=0$  

 $10+10=0$  

 $20\ne0$  x = -5 is NOT a solution

check  $x\ =-\frac{5}{2};\sqrt{7\left(-\frac{5}{2}\right)^2+15\left(-\frac{5}{2}\right)}-2\left(-\frac{5}{2}\right)=5-\frac{5}{2}$  

 $\sqrt{\frac{175}{4}-\frac{75}{2}}+5=\frac{5}{2}$  

 $\sqrt{\frac{25}{4}}+5=\frac{5}{2}$  

 $\frac{5}{2}+5=\frac{5}{2}$  

 $\frac{15}{2}\ne\frac{5}{2};\ x\ =\ -\frac{5}{2}$  is not a solution.  There are no solutions

 $x+5=\sqrt{2x^2+3x+7}$  

 $\left(x+5\right)^2=\left(\sqrt{2x^2+3x+7}\right)^2$  

 $\left(x+5\right)\left(x+5\right)=2x^2+3x+7$  

 $x^2+10x+25=2x^2+3x+7$  

 $0=x^2-7x-18$  

0 = (x - 9)(x + 2)

x - 9 = 0 or x + 2 = 0

x = 9 or x = -2

Check x = 9; $9+2=\sqrt{2\left(9\right)^2+3\left(9\right)+7}-3$  

 $11=\sqrt{196}-3$  

11 = 14 - 3

11 = 11; x = 9 is a solution

check x = -2; $-2+2=\sqrt{2\left(-2\right)^2+3\left(-2\right)+7}-3$  

 $0=\sqrt{9}-3$  

0 = 3 - 3

0 = 0; x = -2 is a solution

Only happen when you raise each side of the equation to an EVEN power

You do NOT need to check for extraneous solutions when you cube both sides of the equation

Start like you are going to solve for z and square both sides of the equation

 $\left(\sqrt{\frac{3}{2}z+10}\right)^2=\left(cz+8\right)^2$  

 $\frac{3}{2}z+10=\left(cz+8\right)^2$  

Plug in z = 4

 $\frac{3}{2}\left(4\right)+10=\left(4c+8\right)^2$  

 $16=\left(4c+8\right)^2$  

Take the square root of both sides of the equation:   $\sqrt{16}=\sqrt{\left(4c+8\right)^2}$  

 $\pm4=4c+8$  

This gives us 4c + 8 = 4 and 4c + 8 = -4

The issue is when 4c + 8 = -4 because we only want the priniple (positive) root

4c = -12

When c = -3 we get the extraneous solution z = 4

1. Square root equation intro

2. Square root equation

3. Extraneous solutions of equations

Due tomorrow 11:59 pm

In class on Wednesday we will cover: Cube root equations