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.
On
Another approach would be to apply Viete's relations: the sum of the roots of $ \ 3x^2 + 4x - 5 \ = \ 0 \ $ is $$ \frac43 \ \ = \ \ - \left(\frac{a}{b + \sqrt{19}} \ + \ \frac{a}{b - \sqrt{19}} \right) \ \ = \ \ -\frac{2ab}{b^2 - 19} $$ and the product of the roots is $$ \frac{-5}{3} \ \ = \ \ \frac{a}{b + \sqrt{19}} \ · \ \frac{a}{b - \sqrt{19}} \ \ = \ \ \frac{a^2}{b^2 - 19} \ \ . $$
The ratio of these equations produces $$ \frac{4/3}{-5/3} \ \ = \ \ -\frac45 \ \ = \ \ \frac{-2ab}{a^2} \ \ = \ \ -\frac{2b}{a} \ \ \Rightarrow \ \ a \ = \ \frac52 · b \ \ . $$
Inserting this into, say, the first equation will lead to $ \ 19b^2 \ = \ 76 \ \ ; $ using the second equation yields a similar result. It will suffice to use the positive value for $ \ b \ \ , $ as using the negative value just gives an equivalent pair of roots in $ \ \frac{a}{b \ \pm \ \sqrt{19}} \ \ . $
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.