Would I set both of them equal to 0 and solve using the quadratic formula?(Thinking) Those that help me will be thanked!(Wait)

Solve each inequality:

1. 4x^2 - 5x - 6 <= 0

2. 2x^2 + 9x + 4 < 0

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- Aug 9th 2009, 06:13 AMbobbyboy1111How would I set these two inequalities up to solve?Would I set both of them equal to 0 and solve using the quadratic formula?(Thinking) Those that help me will be thanked!(Wait)

Solve each inequality:

1. 4x^2 - 5x - 6 <= 0

2. 2x^2 + 9x + 4 < 0

- Aug 9th 2009, 06:27 AMalexmahone
- Aug 9th 2009, 06:48 AMAladdin
Solving Quadratic Inequalities: Examples

Hope this helps (Rofl) - Aug 9th 2009, 07:05 AMbobbyboy1111
- Aug 9th 2009, 07:14 AMZarathustra
LHS = Left Hand Side

You need to decompose the polynomials on the LHS into a product of factors which when multiplied give the original form.

It should be clear that $\displaystyle (x-2)(4x-3) = 0$ for $\displaystyle x=2$ and $\displaystyle x = -\frac{3}{4}$. It can easily be checked that for values between -3/4 and 2 it is negative.

See also http://www.wolframalpha.com/input/?i=4+x^2+-+5+x+-+6+<%3D+0 - Aug 9th 2009, 07:15 AMAladdin
Dont you know how to factorise quadratice equations ? & did you viewed the link I offered .

- Aug 9th 2009, 10:05 AMbobbyboy1111
- Aug 9th 2009, 01:08 PMstapel
The examples, like "x2 + 2x – 8 < 0" and "–x2 + 6x – 9 > 0" are quadratic, they have "zero" on the other side of the inequality, and they involve factoring, just as you've been advised to proceed here. How are those examples "truly" different from what you are doing? (Wondering)