Math Equation

[Tavrobel]

(3x - 8) = 4 + 2(x - 4)1/2 + x - 4

This is where you squared right? I understand where you got that first 4, but i'm unsure about the 2 and the x-4.

In comparison to the example problem at this step. I don't know how to use subscripts like you so I apologize.

2x-1 = x-5+6 the square root of x-5 + 9


Just to mention in the hinto I'm shown there is no paretheses around the square root of x-5 is it still assumed to multiply?

< sup >exponent< / sup > except, without the spaces.

So, let's break up (2 + (x - 4)^1/2)²
(2 + (x - 4)^1/2) * (2 + (x - 4)^1/2)

Also, I screwed up, since the coefficient should be 4, not 2.

FOIL:
First terms
(2 + (x - 4)^1/2) * (2 + (x - 4)^1/2)
Outer terms
(2 + (x - 4)^1/2) * (2 + (x - 4)^1/2)
Inner terms
(2 + (x - 4)^1/2) * (2 + (x - 4)^1/2)
Last terms
(2 + (x - 4)^1/2) * (2 + (x - 4)^1/2)

2*2 + 2*(x - 4)^1/2 + 2*(x - 4)^1/2 + (x - 4)
4 + 4(x - 4)^1/2 + x - 4
4(x - 4)^1/2 + x
3x - 8 = 4(x - 4)^1/2 + x
2x - 8 = 4(x - 4)^1/2
x/2 - 2 = (x - 4)^1/2

Square both sides:
(x/2 - 2)² = ((x - 4)^1/2)²
x²/4 - 2x + 4 = x - 4
x²/4 - 3x + 8 = 0

Multiply everything by 4 to get rid of the denominator:
x² - 12x + 32 = 0

Different answer. (x - 8)(x - 4) = 0; therefore, x = 8, 4.

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(2x - 1) = (x - 5) + 6(x - 5)^1/2 + 9?