Question

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

Answer 1

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.

Answer 2

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

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