Anthony P. answered • 03/05/13
Experienced tutor in earth sciences and basic math to trigonometry
I assume the question is to solve for this trinomial quadratic equation. It's already in standard form, Ax2 + By + C = 0, so that's good. The next step is to try and factor the trinomial into two binomials.
Ex. ( )( ) = 0; however, this trinomial is not factorable so we must use the quadratic formula.
x (or p here) = [ -B +/- sqrt(B2 - 4AC)]/2A, where A, B are the coefficients of the variables in the trinomial, and C is the constant in the trinomial (in standard form).
A = 6, B = -2, C = -3
Substitute into the formula and evaluate.
p = [ -(-3) +/- sqrt((-2)2 - 4(6)(-3))]/2(6)
p = [9 +/- sqrt (4 - (-72))]/12
p = [9 +/- sqrt(76)]/12
p = 9/12 +/- 8.72/12
p = 0.75 + 0.73 or 0.75 - 0.73
p = 1.48 or 0.02
Giovanna C.
Ileana is right. But I would leave the solution with the radicals. In this way, you do not use decimal numbers, and the solution isn't approximate. That means: p1=[1+sqrt(19)]/6 and p2=[1-sqrt(19)]/6.
Giovanna C.
06/29/13
Ileana F.
Hi Anthony, I believe you used '-3' instead of '-2' for your first 'B' when you plugged in the numbers in the quadratic formula solution... I think it should be: p=[-(-2)+/-sqrt((-2)(-2)-4(6)(-3))]/2(6) p=(2+/-sqrt(76))/12 p=(2/12)+/-(8.72/12) p=0.17+/-0.73 p=0.90 or -0.5603/09/13