Question

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

2. Use the equation x+1=√(7x+15) to answer these questions.
What is the solution to the equation?
What is the extraneous solution? Why?
Answer:


Use the equation x-3=√(x-1)+4 to answer these questions.
What is the solution to the equation?
What is the extraneous solution? Why?

Answer:

Answer 1

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.