How do you solve $sqrt(3x + 1) = -10$ ?

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Answer 1 · 2015-08-24T01:41:16

$x=33$

$sqrt(3x+1)=-10$

Square both sides of the equation.

( $sqrt(3x+1))^2=(-10)^2$

$3x+1=100$

Subtract $1$ from both sides.

$3x=100-1$ $=$

$3x=99$

Divide both sides by $3$ .

$x=99/3$ $=$

$x=33$

Check the answer by substituting $33$ for $x$ .

$sqrt(3x+1)=-10$

$sqrt(3*33+1)=-10$

$sqrt100=-10$

$(-10)^2=100$

$sqrt100=+-10$ , so $33$ is a valid number for $x$ .

How do you solve sqrt(3x + 1) = -10sqrt(3x + 1) = -10?