Question

Solving Radical Equations

Square root x+7 = x - 5

Answer 1

The domain:
The radicand must be greater than or equal to 0.

`x+7 \geq 0 \\ x \geq -7`
The value of the square root must be greater than or equal to 0.

`x-5 \geq 0 \\ x \geq 0`
Therefore x≥5.


`√(x+7)=x-5 \\ (√(x+7))^2=(x-5)^2 \\ x+7=x^2-10x+25 \\ 0=x^2-11x+18 \\ 0=x^2-9x-2x+18 \\ 0=x(x-9)-2(x-9) \\ 0=(x-2)(x-9) \\ x-2=0 \ \lor \ x-9=0 \\ x=2 \ \lor \ x=9`
2<5 so it's not a correct solution.
9≥5 so it's a correct solution.

The answer:
x=9

Answer 2

so the square root of the quatity(x+7)=x-5

so you square both sides and get rid of the square root
x+7=(x-5)^2
(x-5)^2=x^2-10x+25
x+7=x^2-10x+25
subtract 7 from both sides
x=x^2-10x+18
subtract x from both sides
0=x^2-11x+18

so if xy=0 we can assume that x or/and y =0

factor out x^2-11+18
(find what two numbers multiply to get 18 and add to get -11)
-2 times -9=18
-2+(-9)=-11

(x-2)(x-9)=0
set them to zero
x-2=0
x=2

x-9=0
x=9

there are two answers -2 and -9