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Kryger [21]
3 years ago
10

Given the equation square root of quantity 2x plus 8 end quantity equals 6, solve for x and identify if it is an extraneous solu

tion.
Mathematics
2 answers:
irina [24]3 years ago
5 0

Answer:

No extraneous solution

Step-by-step explanation:

The given equation is

\sqrt{2x+8}=6

Taking square on both sides.

(\sqrt{2x+8})^2=(6)^2

2x+8=36

Subtract 8 from both sides.

2x+8-8=36-8

2x=28

Divide both sides by 2.

x=14

The solution of given equation is 14.

The solutions of an equation are known as extraneous solutions if they are invalid.

Substitute x=14 in the given equation.

\sqrt{2(14)+8}=6

\sqrt{36}=6

6=6

LHS=RHS, so x=14 is a valid solution.

Therefore, the given equation have no extraneous solution.

Alex_Xolod [135]3 years ago
3 0

Answer:

Solution: x = 6

Step-by-step explanation:

Given equation is:

\sqrt{2x+8}=6

In order to solve the equation both sides will be squared

(\sqrt{2x+8})^{2} = (6)^2 \\2x+8 = 36\\2x = 36-8\\2x = 28\\x = 14

Verifying the solution

\sqrt{2(14)+8} = 6\\ \sqrt{28+8} = 6\\\sqrt{36} = 6\\6 = 6

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