Exercise 29 Page 637

We have a radical equation and are asked which of the given solutions, if any, are extraneous for the equation. ccc Equation & & Solutions x=sqrt(28-3x) & & x=4, x=-7To answer this question, we will substitute both solutions into the equation one at a time and check whether or not we obtain a true statement. Let's start by substituting x=4.

x=sqrt(28-3x)

Substitute

x= 4

4? =sqrt(28-3( 4))

▼

Evaluate

Multiply

Multiply

4? =sqrt(28-12)

SubTerm

Subtract term

4? =sqrt(16)

CalcRoot

Calculate root

4=4 ✓

Since we obtained a true statement, we can conclude that x=4 is not an extraneous solution. Let's repeat the process, this time substituting x=-7 into the equation.

x=sqrt(28-3x)

Substitute

x= -7

-7? =sqrt(28-3( -7))

▼

Evaluate

MultNegNegOnePar

- a(- b)=a* b

-7? =sqrt(28+21)

AddTerms

Add terms

-7? =sqrt(49)

CalcRoot

Calculate root

-7≠ 7 *

Since we did not obtain a true statement, x=-7 is an extraneous solution to the given equation.