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Niel De Wet · Answer 1 · 2021-01-31T08:47:48+0000

Given

x+1 = √(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

$x-3 = √(x-1)+4 \iff x-7=√(x-1)$

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.

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