How do you solve $sqrt[7x+2]-2=7x$ and find any extraneous solutions?

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Answer 1 · 2018-06-15T04:49:25

$x = -1/7$ or $x = -2/7$

Move the constant term

$sqrt(7x+2) = 7x + 2$

Square both sides

$(sqrt(7x+2)) ^ 2 = (7x + 2) ^ 2$

$7x+2 = 49x^2 + 28x + 4$

Move everything over

$0 = 49x ^2 +21x+2$

$x_(1,2) = (-b ± sqrt(b ^2 - 4ac)) / (2a)$

So

$x_(1, 2) = (-21 ± sqrt(21 ^2 - 4(49)(2))) / (2(49))$

Solve to get

$x = -1 / 7" "$ or $" "x = -2 / 7$

To check for extraneous roots, plug the solutions back into the original equation.

$sqrt(7(-1/7)+2) = 7(-1/7) + 2$

$1 = 1 ->$ the solution $(-1/7)$ is correct!

Repeat for the other solution

$sqrt(7(-2/7)+2) = 7(-2/7) + 2$

$0 = 0 ->$ the solution $(-2/7)$ is also correct!

Extraneous solutions appear when your solutions don't satisfy the original equation.

How do you solve sqrt[7x+2]-2=7xsqrt[7x+2]-2=7x and find any extraneous solutions?