Question

Given the equation √8r+1 = 5, solve for x and identify if it is an extraneous solution.

Answer 1

The value of x is 3 and solution is extraneous

Solving equations

Given the equation below;

√8x+1 = 5

Square both sides to have;

8x+1 = 5^2

8x + 1 = 25

Subtract 1 from both sides

8x + 1 = 25 -1

8x = 24

x = 24/8

x = 3

Hence the value of x is 3

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Answer 2

Answer:

r = 3, solution is not extraneous

Step-by-step explanation:

I assume you mean

√(8r + 1) = 5

Also, why is the variable in the equation r,

and all solution choices have x = ...

Solving the equation:

√(8r + 1) = 5

Square both sides.

8r + 1 = 5²

8r + 1 = 25

Subtract 1 from both sides.

8r = 24

Divide both sides by 8.

r = 3

Check r = 3 in the original equation.

√(8r + 1) = 5

√(8 × 3 + 1) = 5

√(24 + 1) = 5

√(25) = 5

5 = 5

r = 3 works in the original equation, so the solution is r = 3.

r = 3 is not extraneous.