How do you solve $x = 3 + \sqrt{9 - x}$ $x=3+sqrt( 9-x)$ ?

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How do you solve $x = 3 + \sqrt{9 - x}$ $x=3+sqrt( 9-x)$ ?

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Answer 1 · 2016-03-05T01:56:29

$x = 5 or 0$ $x = 5 or 0$

Explanation:

$x = 3 + \sqrt{9 - x}$ $x = 3 + sqrt(9 - x)$
$x - 3 = \sqrt{9 - x}$ $x - 3 = sqrt(9 - x)$

Squaring both sides: ${(x - 3)}^{2} = 9 - x$ $(x - 3)^2 = 9 - x$
Expanding square term: $x^{2} - 6 x + 9 = 9 - x$ $x^2 - 6x + 9 = 9 - x$
$x^{2} - 5 x = 0$ $x^2 - 5x = 0$
$x (x - 5) = 0$ $x(x - 5) = 0$

$x = 5 or 0$ $x = 5 or 0$

Answer 2 · 2016-03-05T01:57:48

You must square both sides of the equation to get rid of the square root.

Explanation:

First, make sure the radical is isolated.

$x - 3 = \sqrt{9 - x}$ $x - 3 = sqrt(9 - x)$

${(x - 3)}^{2} = {(\sqrt{9 - x})}^{2}$ $(x - 3)^2 = (sqrt(9 - x))^2$

$x^{2} - 6 x + 9 = 9 - x$ $x^2 - 6x + 9 = 9 - x$

$x^{2} - 5 x = 0$ $x^2 - 5x = 0$

$x (x - 5) = 0$ $x(x- 5) = 0$

$x = 0 and 5$ $x = 0 and 5$

When solving radical equations, always make sure to check the solution(s) in the original equation. Often, extraneous solutions can arise. These are solutions that do not satisfy the equation.

Upon checking, you realize that $x = 5$ $x = 5$ is the solution.

Hopefully this helps!

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How do you solve $x = 3 + \sqrt{9 - x}$ $x=3+sqrt( 9-x)$ ?

2 Answers

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