# Thread: How would I set these two inequalities up to solve?

1. ## How would I set these two inequalities up to solve?

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

Solve each inequality:

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

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

2. Originally Posted by bobbyboy1111
Would I set both of them equal to 0 and solve using the quadratic formula? Those that help me will be thanked!

Solve each inequality:

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

2. 2x^2 + 9x + 4 < 0
Factorize the LHS of both the inequalities by splitting the middle term.

Spoiler:
1. $4x^2 - 5x - 6=4x^2-8x+3x-6=4x(x-2)+3(x-2)=(x-2)(4x+3)$

2. $2x^2 + 9x + 4=2x^2+8x+x+4=2x(x+4)+(x+4)=(x+4)(2x+1)$

3. Originally Posted by alexmahone
Factorize the LHS of both the inequalities by splitting the middle term.

Spoiler:
1. $4x^2 - 5x - 6=4x^2-8x+3x-6=4x(x-2)+3(x-2)=(x-2)(4x+3)$

2. $2x^2 + 9x + 4=2x^2+8x+x+4=2x(x+4)+(x+4)=(x+4)(2x+1)$

Whoa, what does "LHS" mean, how do I split the middle term, and how do I factorize?

4. Originally Posted by bobbyboy1111
Whoa, what is the LHS mean, how do I split the middle term, and how do I factorize?
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 $(x-2)(4x-3) = 0$ for $x=2$ and $x = -\frac{3}{4}$. It can easily be checked that for values between -3/4 and 2 it is negative.

5. Dont you know how to factorise quadratice equations ? & did you viewed the link I offered .

Dont you know how to factorise quadratice equations ? & did you viewed the link I offered .
Yes, I did, but there aren't any truly similar examples.

7. 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?