# How to convert a quadratic solution to an unusual format

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.

• Use $(a+b)(a-b)=a^2-b^2$. Dec 1 '21 at 13:03
• First use the quadratic formula and then un - rationalise. Dec 1 '21 at 13:04
• Multiply your solution by $1=(-2 \mp \sqrt{19})/(-2 \mp \sqrt{19})$. Dec 1 '21 at 13:11
• That alternative form is useful in many cases to compute the numerical result without loss of accuracy due to subtractive cancellation in the term $-b \pm \sqrt{b^2 - 4ac}$ for one of the two roots. Dec 1 '21 at 13:14
• Ohhh, I just had to multiply the top and bottom by $(-2\pm\sqrt{19})$ and it all cancelled down to $\frac{-5}{2\pm\sqrt{19}}$. Thank you!
– KD97
Dec 1 '21 at 13:17

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.
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}} \ \ .$$