Inequality Problem: 2 – x2 ≥ x2 – 3x - 4

[svital]

Inequality Problem: 2 – x2 ≥ x2 – 3x - 4

2 – x2 ≥ x2 – 3x - 4

I have been working this for an hour, and keep getting -2 as an answer. But I don't think it's right. I don't think I am taking the right steps for solving. Can anyone help?

Thanks,
Sondra

 

[skeeter]

I'm guessing you mean ...

\(\displaystyle \L 2 - x^2 \geq x^2 - 3x - 4\)

get all terms on one side ...

\(\displaystyle \L 0 \geq 2x^2 - 3x - 6\)

this quadratic will not factor, and you need to find where it equals 0. using the quadratic formula ...

\(\displaystyle \L x = \frac{3 \pm \sqrt{57}}{4}\)

the inequality will be true for values of x between and including the two zeros ...

\(\displaystyle \L \frac{3 - \sqrt{57}}{4} \leq x \leq \frac{3 + \sqrt{57}}{4}\)