# How do you solve #sqrt[x+5]=3x-7# and find any extraneous solutions?

##### 2 Answers

#### Explanation:

Simplifying the equation would involve first getting rid of the square root:

Simplifying further, we can expand out the square on the right side of the equation:

Since this is a quadratic equation, we want to put everything on one side and then find the zeros of the expression:

Using the quadratic formula (since we cannot factor the expression),

Simplifying gives:

Using a calculator to find the value of this expression,

Plugging the first value back into the equation, we get a valid equation:

However, when we plug the second value back into the equation, we have this:

Since the right side is negative, the second value that we got for

all values of

#### Explanation:

For the solution to remain within the set of numbers called Real the square root must be of a value that is not negative.

Thus all values of

Given:

Square both sides

Subtract

To comply with convention write as:

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

Compare to

In this case:

However, as per smartspot2, substituting 1.485 back into the original question proves this value to fail if you use the 'Principle Square Root'. Thus by his approach there is only 1 value that is correct: