Inequalties?

Question

How can you solve these questions? I will post a few. Please try to solve at least one or one someone hasn't solved. Please also don't tell me to do my homework, I am doing it and don't understand thanks.

1) x^4+9x^2-10 <= 0
2) x^3+x^2-6x > 0
3) x^4+x^3-2x^2-2x => 0
Just tell me if #3 equals x=2,-3,-1

Thank u.

(x+3)x^2(x-2)^3 => 0

Answer 1

The roots the person above me solved for this question are right but it is a Quartic function so it should have 4 roots in total.(2 roots are imaginary)

1) First find possible factors of 10 both positive and negative and see which one works wwith the equation.
If you do that you get (x-1) as one of the factors.
2) Use the synthetic division with the root x=1
3) You will get a cubic; solve in the same way to get a quadratic.
4) your final answer will be (x-1)(x+1)(x^2+10)
(x^2+10) are your imaginary roots.

Answer 2

1) Is a quadratic in disguise, if you let y=x^2, it reduces to

y^2+9y-10

which has roots y=1,-10; therefore the original has roots x=1,-1, 10^(1/2)i, -10^(1/2)i.
Then the only points where x^4+9x^2-10 crosses the x-axis are -1 and 1, and is negative between them (the coefficient of x^4 is positive), so the answer is [-1,1].

Answer 3

3) let try to factorize this by grouping
x^3(x+1) - 2x(x+1) = 0
x(x^2-2)(x+1)=0
x(x-sqrt(2))(x+sqrt(2))(x+1)=0

so the roots are -sqrt(2), -1, 0, and sqrt(2)
if you can plot the curve, y >= 0 when x is [-oo, -sqrt(2)], [-1,0], [sqrt(2), oo]