How do you solve and check for extraneous solutions in $sqrt(x+7) = x + 1$ ?

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Answer 1 · 2015-08-10T10:40:58

$x = 2$

Any solution that you will find must satisfy two conditions

Overall, the solution(s) to this equation must satisfy the condition $x>=-1$ .

Start by squaring both sides of the equation to get rid of the radical term

$(sqrt(x+7))^2 = (x+1)^2$

$x+7 = x^2 + 2x + 1$

Rearrange this equation into classic quadratic form

$x^2 + x -6 = 0$

Use the quadratic formula to find the two solutions to this equation

$x_(1,2) = (-1 +- sqrt(1^2 - 4 * 1 * (-6)))/(2 * 1)$

$x_(1,2) = (-1 +- sqrt(25))/2$

$x_(1,2) = (-1 +- 5)/2 = {(x_1 = (-1 -5)/2 = -3), (x_2 = (-1 + 5)/2 = 2) :}$

Only one of these two solutions, $x=2$ , satisfies the condition $x>=-1$ , which means that $x=-3$ will be an extraneous solution.

Therefore, the solution to this equation is $x=color(green)(2)$ .

Check to see if this is the case

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

$sqrt(9) = 3$

$3 = 3$ $color(green)(sqrt())$

How do you solve and check for extraneous solutions in sqrt(x+7) = x + 1sqrt(x+7) = x + 1?