Solving square root equations and extraneous solutions

Solving square root equations and extraneous solutions

Multiple Choice

Factor

 $7x^2+37x+10$  

• (7x + 2)(x + 5)

• (7x - 10)(x + 1)

• (7x - 2)(x + 5)

• (7x + 5)(x + 2)

 $7x^2+37x+10;\ a\ =\ 7,\ b\ =\ 37,\ c\ =\ 10$  

• 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)

Multiple Choice

Factor

 $5x^2+12x-9$  

• (5x -3)(x - 3)

• Not factorable

• (5x - 3)(x + 3)

 $5x^2+12x-9;\ a\ =\ 5,\ b\ =\ 12,\ c\ =\ -9$  

• 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)

To solve:

• 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.

Solve

 $\sqrt{5x-4}=x-2$  

• 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

check x = 1

•  $\sqrt{5\left(1\right)-4}=1\ -\ 2$  

•  $\sqrt{1}=-1$  

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

•  $1\ne-1$  

• x = 1 is NOT a solution

check x = 8

•  $\sqrt{5\left(8\right)-4}=8\ -\ 2$  

•  $\sqrt{36}=6$  

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

•  $6=6$  

• x = 8 is a solution

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

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

•  $z^2+6z+3=9z^2+6z+1$  

•  $3=8z^2+1$  

•  $2=8z^2$  

•  $\frac{1}{4}=z^2$  

•  $z\ =\ \pm\frac{1}{2}$  

• 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

Solve

 $\sqrt{7y^2+15y}-2y=5+y$  

• 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

Multiple Select

Solve

 $x+2=\sqrt{2x^2+3x+7}-3$  

• -6

• -2

• 3

• 9

• no solution

 $x+2=\sqrt{2x^2+3x+7}-3$  

•  $x+5=\sqrt{2x^2+3x+7}$  

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

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

•  $x^2+10x+25=2x^2+3x+7$  

•  $0=x^2-7x-18$  

• 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

Extraneous solutions

• 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

Which value for the constant c makes z = 4 an extraneous solution?

 $\sqrt{\frac{3}{2}z+10}=cz+8$  

• 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

Homework in Khan Academy

• 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