How do you solve $10 - x = \sqrt{3 x + 24}$ $10 - x =sqrt(3x + 24)$ ?

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Answer 1 · 2015-08-10T10:33:43

$x = 4$ $x = 4$

Right from the start, you know that any solution you find for this equation must satisfy two conditions

This means that you must for an $x$ $x$ that satisfies the overall condition $x \in [- 8, 10]$ $x in [-8, 10]$ .

Since the radical is already isolated on one side of the equation, square both sides to get rid of the square root

${(\sqrt{3 x + 24})}^{2} = {(10 - x)}^{2}$ $(sqrt(3x + 24))^2 = (10-x)^2$

$3 x + 24 = 100 - 20 x + x^{2}$ $3x + 24 = 100 - 20x + x^2$

Rearrange this equation into classic quadratic form

$x^{2} - 23 x + 76 = 0$ $x^2 -23x +76 = 0$

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

$x_{1, 2} = \frac{- (- 23) \pm \sqrt{{(- 23)}^{2} - 4 \cdot 1 \cdot 76}}{2 \cdot 1}$ $x_(1,2) = (-(-23) +- sqrt( (-23)^2 - 4 * 1 * 76))/(2 * 1)$

$x_{1, 2} = \frac{23 \pm \sqrt{225}}{2}$ $x_(1,2) = (23 +- sqrt(225))/2$

$x_(1,2) = (23 +- 15)/2 = {(x_1 = (23 + 15)/2 = 19), (x_2 = (23 - 15)/2 = 4) :}$

Now, since $x=19$ $cancel(in) [-8, 10]$ , this is not a valid solution solution.

As a result, the only solution to this equation is $x=color(green)(4)$ .

Check to see if this is the case

$10 - 4 = sqrt(3 * (4) + 24)$

$6 = sqrt(36)$

$6 = 6$ $color(green)(sqrt())$

How do you solve 10 - x =sqrt(3x + 24)10−x=√3x+2410 - x =sqrt(3x + 24)?