Help with precalculus work.

Question

I've been stuck on this problem for a while. Please help. >__<

The equation 2x^2+16x+3=0 has two solutions A and B where A < B and A=_______ and B=_______

Give your answers to 3 decimal places or as exact expressions.

Answer 1

2x^2 + 16x + 3 cannot be factored so we need to use the quadratic equation.

The solutions will be real but not rational.

x = [-b +-sqrt(b^2 - 4ac)]/2a

a = 2, b = 16, c = 3

x = [-16 +-sqrt(16^2 - 4(2)(3))]2(2)

x = [-16 +-sqrt(256 - 24)/4

x = [-16 +-sqrt232]/4

x = [-16 +-2sqrt58]/4

x = [-8 +-sqrt58]/2

This is the exact solution.

For a rational approximation

x = (-8 +-7.616)/2

x = .384/2 = .192

x = -15.616/2 = -7.808