Question

solution set of sqrt(7x+1)-sqrt(x-5)=6

Answer 1

have you done these types of problems before? it normally involves squaring both sides once, then moving some terms over, then squaring a second time.

Answer 2

Ive always struggled with these solution sets

Answer 3

like

\[(\sqrt{7x + 1} - \sqrt{x - 5})^{2} = 36\]

the left hand side becomes:

\[7x + 1 - 2\sqrt{x - 5}\sqrt{7x + 1} + x - 5 = 8x - 2 \sqrt{x - 5}\sqrt{7x + 1} - 4\]

Answer 4

simplified a bit becomes:

\[8x - 2 \sqrt{(x - 5)(7x + 1)} - 4 = 36\]

Answer 5

this was the first time we squared both sides. next we move the 8x and the -4 to the other side by subtracting 8x and adding 4:

\[-2 \sqrt{(x -5)(7x + 1)} = 40 - 8x\]

Answer 6

then we square both sides a second time

Answer 7

\[4(x-5)(7x + 1) = (40 - 8x)^{2}\]

Answer 8

expanding the right side:

\[4(x - 5)(7x + 1) = 64x^{2} - 640x + 1600\]

dividing everything by 4:

\[(x - 5)(7x + 1) = 16x^{2} - 160x + 400\]

expanding the left side:

\[7x^{2} - 34x - 5 = 16x^{2} - 160x + 400\]

combining like terms and rearranging:

\[-9x^{2} + 126x - 405 = 0\]

solve for x

Answer 9

then verify the solutions you get are permitted in the domain of the original equation (because when we square, sometimes we introduce values that are not part of the original domain).

Answer 10

how do you solve for x when you have to x's?

Answer 11

quadratic formula works

Answer 12

you could also rewrite it as

\[9x^{2} - 126x + 405 = 0\]

\[9x^{2} - 126x = -405\]

if you dont want to use quadratic formula, you can also complete the sqaure

Answer 13

5,9

Answer 14

yes those are correct