Question

Approximate the solution to 1/sqrt{x^2+} = 1/x+5

Answer 1

Possibilities:
(a) x = -2.8320
(b) x = -1.2062
(c) x = -2.4000
(d) x = -1.2000
(e) x = 0.3846

Answer 2

o it should be 1/sqrt{x^2+1}

Answer 3

my bad

Answer 4

is that \[\frac{1}{\sqrt{x^{2}+1}} = \frac{1}{x+5} \]

Answer 5

if so just cross multiply (or invert) and square both sides then solve for x to get

\[x = -12/5 =-2\frac{2}{5} = -2.4\]

Answer 6

could you write out the steps please?

Answer 7

I listed the steps

Answer 8

ok so i get x^2+1=sqrt{x+5} then?

Answer 9

Almost. You would get sqrt(x^2 + 1) = x + 5

Answer 10

ok i got that but then u said square both sides so that would get rid of the square on one side and place it on the other

Answer 11

Not exactly. When you square both sides that means to multiply it by itself or raise it to the power of 2 meaning

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

Answer 12

yes i know that but when u raise the side with the radical to the power of 2, they cancel out

Answer 13

correct, but then the term on the other side is squared (not put under a radical as u had said)

Answer 14

ok