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.

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