Question

What method(s) would you choose to solve the equation? Explain your reasoning.
2x^2 +4x - 3 = 0

Answer 1

I would use the quadratic formula to solve this as it can solve any quadratic equation. Did you need to see how to solve this using the quadratic formula?

Answer 2

quadratic formula because it does not factor

Answer 3

yes;

Answer 4

With 2x^2 + 4x - 3 = 0, it's in the form ax^2+bx+c=0 where a = 2, b = 4 and c = -3


Plug these values into the quadratic formula to solve for x


\[\Large x = \frac{-b\pm\sqrt{b^2-4ac}}{2a}\]


\[\Large x = \frac{-(4)\pm\sqrt{(4)^2-4(2)(-3)}}{2(2)}\]


\[\Large x = \frac{-4\pm\sqrt{16-(-24)}}{4}\]


\[\Large x = \frac{-4\pm\sqrt{40}}{4}\]


\[\Large x = \frac{-4+\sqrt{40}}{4} \ \text{or} \ x = \frac{-4-\sqrt{40}}{4}\]


\[\Large x = \frac{-4+2\sqrt{10}}{4} \ \text{or} \ x = \frac{-4-2\sqrt{10}}{4}\]


\[\Large x = \frac{2(-2+\sqrt{10})}{2*2} \ \text{or} \ x = \frac{2(-2-\sqrt{10})}{2*2}\]


\[\Large x = \frac{-2+\sqrt{10}}{2} \ \text{or} \ x = \frac{-2-\sqrt{10}}{2}\]


So the solutions are
\[\Large x = \frac{-2+\sqrt{10}}{2} \ \text{or} \ x = \frac{-2-\sqrt{10}}{2}\]