Week 5 – precalculus 11

Identifying Extraneous Roots:

Extraneous roots are found when a variable discovered after solving an equation is not true if used in the original equation. In other words, it’s a root found by squaring both sides, it is not a root in the equation you started with. To show this, we could use an equation like $\sqrt {x-1} = \sqrt {2x+2}$.

First we will square both sides $(\sqrt {x-1})^2 = (\sqrt {2x+2})^2$

This will remove the radical sign

x – 1 = 2x + 2

Then we take away x from the left side and from the right side. We take away 2 and put it on the left side.

-3 = x

If we plug x back in as -3 it will not work. $\sqrt {-3-1} = \sqrt {2(-3)+2}$

Solve, $\sqrt {-4} = \sqrt {-4}$

A radicand of a square root cannot be negative.

Another example of this would be: $\sqrt {3x-1}+ 5 = 2$

We will take away five from both sides $\sqrt {3x-1} = -3$

Then we square both sides

3x – 1 = 9

We will now add one to both sides

3x = 10

Divide both sides by 3 $x = \frac {10}{3}$

If we plug this in to the original equation (After taking away 5 from both sides) this happens, $\sqrt {3(\frac {10}{3})-1} = -3$

This becomes $\sqrt {10-1} = -3$

then, $\sqrt {9} = -3$

3 does not equal -3

Therefore, this root is not accurate when used with the original equation. This makes it a root that is not real, or extraneous.