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Pearson Algebra 2 Common Core, 2011

PA

Pearson Algebra 2 Common Core, 2011 View details

End-of-Course Assessment

Finish the assignment

Exercise 20 Page 966

What is the inverse operation of square root?

G

Practice makes perfect

To solve equations with a variable expression inside a radical, we first want to make sure the radical is isolated. Then we can raise both sides of the equation to a power equal to the index of the radical. Let's try to solve our equation by using this method!

sqrt(x-2)-7 = - 4

LHS+7=RHS+7

sqrt(x-2)=3

LHS^2=RHS^2

(sqrt(x-2))^2 = 9

( sqrt(a) )^2 = a

x-2 =9

LHS+2=RHS+2

x=11

Finally, we will check our solution by substituting the value into the original equation.

sqrt(x-2)-7 ? = - 4

x= 11

sqrt(11-2)-7 ? = - 4

▼

Simplify left-hand side

Subtract term

sqrt(9)-7 ? = - 4

Calculate root

3-7 ? = - 4

Subtract term

- 4=- 4 ✓

We can see that our solution, x=11, is correct. This corresponds to option G.

Subchapter links

Exercises p.964-971