Question

sqrt(x+2)<4-x ? solution is -2<=x<2 how?

Answer 1

what did you get? I assume there's no absolute value involved?

Answer 2

anyhow... .so \(\bf \sqrt{x+2} < 4-x\qquad \textit{raise to }{\color{red}{ 2}}
\\ \quad \\
(\sqrt{x+2})^2<(4-x)^2\implies x+2 < (4-x)^2
\\ \quad \\
\implies x+2 < (4-x)(4-x)\implies x+2 < 16-4x-4x+x^2
\\ \quad \\
x+2<16-8x+x^2\implies 0 < 14-9x+x^2\implies 0 < x^2-9x+14\)

now if you factor the right-side.... see what you get

though it's not going to give you \(\bf -2\le x < 2\)

Answer 3

I know, first I did it the same as you wrote but then in solutions it's −2≤x<2. I checked on wolfram and −2≤x<2 is correct. I have no idea how they got that.

Answer 4

hmm

Answer 5

I'm thinking the "7" answer drops off

Answer 6

lemme do some checking

Answer 7

domain of \(\sqrt{x+2}\) is \(x\geq 2\)

Answer 8

ooops i meant
\[x\geq -2\]

Answer 9

in any case whatever you get for the polynomial inequality, you are restrained by \(x\geq -2\) throughout the whole problem

Answer 10

yeap, indeed

Answer 11

@Castiel as you can see, by what satellite73 just pointed out
your solutions are correct is 0 < (x-7) (x-2)

which makes it 7 < x and 2 < x

however, the 7 answers drops off due to DOMAIN RESTRICTIONS on the radical of the original equation

"x" cannot be less than -2, for if it does, the radicand turns negative and the radical will give an imaginary value

Answer 12

Thank you

Answer 13

yw