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. $\displaystyle 4x^2 - 5x - 6=4x^2-8x+3x-6=4x(x-2)+3(x-2)=(x-2)(4x+3)$

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

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

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

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

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 $\displaystyle (x-2)(4x-3) = 0$ 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.