Question

Sqrt(x+4)=2-x help me pleaseeeeeee

Answer 1

Answer:

x = 0

Step-by-step explanation:

$\sqrt{x+4}$ = 2 - x ( square both sides to clear the radical )

x + 4 = (2 - x)² ← expand using FOIL

x + 4 = 4 - 4x + x² ( subtract x + 4 from both sides )

0 = x² - 5x ← factor out x from each term

0 = x(x - 5)

equate each factor to zero and solve for x

x = 0

x - 5 = 0 ⇒ x = 5

As a check

substitute these values into the equation and if both sides are equal then they are the solutions

x = 0

left side = $\sqrt{0+4}$ = $\sqrt{4}$ = 2

right side = 2 - 0 = 2

then x = 0 is a solution

x = 5

left side = $\sqrt{5+4}$ = $\sqrt{9}$ = 3

right side = 2 - 5 = - 3 ≠ 3

then x = 5 is an extraneous solution

Answer 2

Answer: $x=\Large\boxed{0}$

Step-by-step explanation:

Given expression

$\sqrt{x+4} =2-x$

Square both sides of the equation

$\sqrt{x+4}^2 =(2-x)^2$

$x+4=4-4x+x^2$

Subtract x on both sides

$x+4-x=4-4x+x^2-x$

$4=x^2-5x+4$

Subtract 4 on both sides

$4-4=x^2-5x+4-4$

$0=x^2-5x$

Factorize the quadratic expression

$0=x(x-5)$

$x=\Large\boxed{0} \text{ or }x=5~(reject)\text{ }$

Check the answer

$\text{When x = 0:}$

$\sqrt{(0)+4} =2-(0)$

$\sqrt{4}=2$

$2=2$   $\boxed{TRUE}$

$\text{When x = 5:}$

$\sqrt{(5)+4} =2-(5)$

$\sqrt{9}=-3$

$3\neq -3$  $\boxed{FALSE}$

Hope this helps!! :)

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