If square root of x^2-8x+16 = x-4, then x=?

Question

It says in the book that the answer to my question is x>=4 but I don't get it. There are choices in my question and one of the choices is '4 only' and I think this is the right answer because x=4, right? I hope you can give me a student average explanation.

Answer 1

Because we are taking the square root of x² - 8x + 16, it can't be negative, so...

 

(x - 4)² ≥ 0

x - 4 ≥ 0

x ≥ 4

 

 

Answer 2

The equation says √(x² - 8x + 16) = x-4

 

If we square both sides then we get

 

x²-8x+16 = (x-4)²

 

Now if we expand the right side of the equation we get

 

x²-8x+16 = (x-4)(x-4)

x²-8x+16 = x²-8x+16

 

So no matter what x is, the right side will be equal to the left side.

 

However, if we look at the original equation on the left side, √(x²-8x+16) , the value under the square root symbol has to be greater than or equal to zero (otherwise we get an imaginary number and you might not be working with those yet)

 

So in order for √(x²-8x+16) to be greater than or equal to zero, x has to be greater than or equal to 4. If x is less than 4 you get a negative number. If x is greater than 4, you'll get the same answer on both sides no matter what x is.