Question

PLEASE HELP QUICK!! WILL OFFER 100 POINTS TO THE FIRST ONE THAT ANSWERS

Answer 1

Given

$x+1 = \sqrt{7x+15}$

We have to set the restraint

$x+1\geq 0 \iff x \geq -1$

because a square root is non-negative, and thus it can't equal a negative number. With this in mind, we can square both sides:

$(x+1)^2=7x+15 \iff x^2+2x+1=7x+15 \iff x^2-5x-14 = 0$

The solutions to this equation are 7 and -2. Recalling that we can only accept solutions greater than or equal to -1, 7 is a feasible solution, while -2 is extraneous.

Similarly, we have

$x-3 = \sqrt{x-1}+4 \iff x-7=\sqrt{x-1}$

So, we have to impose

$x-7\geq 0 \iff x \geq 7$

Squaring both sides, we have

$(x-7)^2=x-1 \iff x^2-14x+49=x-1 \iff x^2-15x+50 = 0$

The solutions to this equation are 5 and 10. Since we only accept solutions greater than or equal to 7, 10 is a feasible solution, while 5 is extraneous.