Week 12 – Pre-calculus 11

This week in Math 11 we learned how to solve absolute value equations algebraically. When solving with an absolute value and or a square root you may come across extraneous solutions which are solutions that are not true solutions of the equation given. Algebraically we can find out whether a solution is extraneous by verifying and inserting the “x” value back into the original equation to see if the equation is true or to see whether both sides equal each other.

3 = |2x + 5|

  • first thing we do is separate the equation
  • because they original equation uses absolute value we times everything by a negative to make it to make it positive.

3 = 2x + 5

3 = -(2x + 5)

  • now that we have our equations we solve each of them

3 = 2x +5

3 – 5 = 2x

-2 = 2x

\frac{-2}{2} = \frac{2x}{2}

x = -1

Now for the second equation:

3 = -(2x +5)

3= -2x – 5

3 + 5 = -2x

8 = -2x

\frac{8}{-2} = \frac{-2x}{-2}

x = -4

Now we have to make sure these solutions are not extraneous by verifying:

  • to do this we insert each solution back into the original equation

x = -1      x = -4

 

X = -1: Verify

3 = |2x + 5|

3 = |2(-1) + 5|

3 = |-2 + 5|

3 = |3|

3 = 3

X = -4: Verify

3 = |2x + 5|

3 = |2(-4) + 5|

3 = |-8 + 5|

3 = |-3|

3 = 3

We see that both solutions make the equation true because both sides are equal.

 

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