Question

PLEASE HELP ASAP PLEASEEEE show calculations: √(x^2-8x+16) if x≥4

Answer 1

Answer:

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)

Step-by-step explanation:

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

Answer 2

$\qquad\qquad\huge\underline{{\sf Answer}}$

Let's solve ~

$\qquad \sf \dashrightarrow \: \sqrt{ {x}^{2} - 8x + 16}$

$\qquad \sf \dashrightarrow \: \sqrt{ {x}^{2} - 4x - 4x + 16}$

$\qquad \sf \dashrightarrow \: \sqrt{ {x}^{} (x - 4) - 4(x - 4)}$

$\qquad \sf \dashrightarrow \: \sqrt{ (x - 4)(x - 4)}$

$\qquad \sf \dashrightarrow \: \sqrt{ (x - 4) {}^{2} }$

$\qquad \sf \dashrightarrow \: x - 4$

Now, it's given that ~

$\qquad \sf \dashrightarrow \:x \geqslant 4$

plug the value of x as 4 :

$\qquad \sf \dashrightarrow \:4 - 4$

$\qquad \sf \dashrightarrow \:0$

So, we can infer that :

Value of the required expression lies in :

$\qquad \sf \dashrightarrow \: \sqrt{ {x}^{2} - 8x + 16} \geqslant 0$

it's value is always greater than pr equal to 0, satisfying the given condition ~