how to factor 5x^2+x-1=0

Question

How to find the solution set by factoring and using the zero product rule or by quadractic formula

Answer 1

This trinomial is not factorable. One needs to use the quadratic formula to find the roots in decimal or radical form. One cannot use the zero product property.

Answer 2

Well, the quadratic formula states that the solution(s) to a trinomial equation in the form ax²+bx-c=0 is

x=(-b ± √(b²-4ac)) / 2a

The first step to finding the solution set using the quadratic formula is to identify the coefficients a, b, and c in the trinomial that was given.  Make sure that the trinomial equals 0!  Otherwise, your coefficients might be different and your answer will be wrong.  Once you have identified a, b, and c from the equation, the next steps are to plug a, b, and c into the quadratic formula stated above and then simplify.

So, for 5x²+x-1=0, a=5, b=1, and c=-1.

x = (-b ± √(b²- 4ac)) / (2a)

x = (-1 ± √(1²- 4*5*-1)) / (2*5)

x = (-1 ± √(1- -20)) / (10)

x = (-1 ± √(1+20)) / (10)

x = (-1 ± √(21)) / (10)

Since 21 is not a perfect square, the easiest way to write this as a decimal would be to plug each answer into a calculator:

Answer #1: x = (-1 + √(21)) / (10) = 0.358 (rounded to the nearest thousandth) 

Answer #2: x = (-1 - √(21)) / (10) = -0.558 (rounded to the nearest thousandth)

Check the answers by plugging them back into the original equation (5x²+x-1=0).  If they work, you're all done!