Question

3x^2+1 divided by x+1 less than or equal to 2 solve for each inequality

Answer 1

i got to \[3^{2}-2x-1\div(x+1)\le0\] but then im stuck

Answer 2

*\[3x ^{2}\]

Answer 3

3x^2 +1 / x+1 < 2

Answer 4

multiply both sides by x+1 so you get 3x^2+1 < 2x+2

Answer 5

3x^2 -2x -1 < 0 and now factor (3x+1)(x-1)<0 take each inequality individually

Answer 6

\[\ \ \ \ \ \ \ \frac{3x^2 + 1}{x+1} < 2\]\[\implies \frac{(3x^2 + 1) - (2x + 2)}{x+1} < 0\]\[\implies \frac{3x^2 - 2x - 1}{x+1} < 0\]\[\implies \frac{(3x + 1)(x-1)}{x+1} < 0\]
\[\implies x \in (-\infty, -1) \cup (-\frac{1}{3},1)\]

Answer 7

Jonny, you lost a root when you multiplied both sides by x+1

Answer 8

thanks that didn't look right to me

Answer 9

Well not really a root. Since you cannot divide by 0, but certainly a point at which the sign changes for the expression.