I'm looking at old past papers and found this question:
"Solve the quadratic equation $3x^2 + 4x – 5$ giving your answer in the form $\frac{a}{b\pm\sqrt{19}}$, where $a$ and $b$ are integers."
I've never seen a quadratic solution in this form, with the surd root on the bottom. Does anybody have any hints on how to rearrange into this format?
Edit: For clarity, I've got $x = \frac{-2 \pm \sqrt{19}}{3}$, I just can't figure out how to convert that answer into the form they want.
Best Answer
In general, the quadratic polynomial $ax^2+bx+c$ has roots $\dfrac{-b+\sqrt{b^2-4ac}}{2a}$ and $\dfrac{-b-\sqrt{b^2-4ac}}{2a}$. If you multiply the first root by $\dfrac{-b-\sqrt{b^2-4ac}}{-b-\sqrt{b^2-4ac}}=1$, you get $\dfrac{b^2-(b^2-4ac)}{-2ab-2a\sqrt{b^2-4ac}}=\dfrac{-2c}{b+\sqrt{b^2-4ac}}$. Doing the same with the other root, you for the desired expressions.
Note: I'm assuming you can actually do the division because the roots are non zero, which is what happens in your problem.