How do you solve #4a ^ { 2} + 4a < 3( a + 1)#?

1 Answer
Alexi De Avila Cadena
Dec 6, 2017

Use the distributive property, make it equal to zero and find #a# with the quadratic formula.

Explanation:

For the purpose of this explanation;
Let #a# be #x#
Put all #x#'s on one side.
#4x^2+4x-3(x+1)<0#
Distribute
#4x^2+4x-3x-3#
Combine like terms
#4x^2+x-3#
Use #x=(-b(+-)sqrt(b^2-4ac))/(2a)#
And get #x>3/4# and #x< -1#

Related questions
Impact of this question
903 views around the world
You can reuse this answer
Creative Commons License