How do you solve #4x^2 + 4x + 1 > 0#?

2 Answers
Lyndhir
Jan 4, 2016

#x!= -1/2 #

Explanation:

Firstly, We have to solve the related second degree equation:

#4x^2+4x+1=0#

We could use the well-known formule:

#(-b +- sqrt(b^2 -4ac))/(2a)#

So we have: #x_1=x_2= - 1/2#

having a double root from the related equation, the solution must be: #x!= -1/2 #

Topscooter
Jan 4, 2016

You need to take a look at the number of real roots this polynomial has.

Explanation:

In order to know where this polynomial is positive and negative, we need its roots. We will of course use the quadratic formula in order to find them.

The quadratic formula gives you the expression of the roots of a trinomial #ax^2 + bx + c#, which is #(-b+-sqrtDelta)/(2a)# where #Delta = b^2 -4ac#. So let's evaluate #Delta#.

#Delta = 16 - 4*4 = 0# so this polynomial has 1 real root only, which means that it will always be positive except at its roots (because #a > 0#).

This root is #(-4)/8 = -1/2#. So #4x^2 + 4x + 1 > 0 iff x != -1/2#. Here is the graph so you can see it.

graph{4x^2 + 4x + 1 [-2.234, 2.092, -0.276, 1.887]}

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