# When to simplify a quadratic equation?

I had the following quadratic equation: $$38x^2 - 140x - 250 = 0$$ And before starting to solve it, I simplified it by dividing all terms by $2$: $$19x^2 -70x - 125 = 0$$ But when I solved it I got: $$x= \frac{35\pm5\sqrt{163}}{19}$$ which is not the exact solution for the original equation. So, I returned to that first equation and solved it. The roots are: $x=-\frac{25}{19}, 5$

Now my question is: How would I know when to simplify a quadratic equation before solving it? Is there something that shows me that I'm right to simplify or not?

• You must have made an error because both those quadratics have the same roots, multiplying or dividing through by the same number doesn't change the roots – Triatticus Jun 15 '16 at 13:46
• @Triatticus: I checked with Wolfram alpha – Rafiq Jun 15 '16 at 13:47
• It should be $x=\frac{35\pm 60}{19}$, not $x=\frac{35\pm 5\sqrt{163}}{19}$. – user236182 Jun 15 '16 at 13:47
• your square root is wrong... $\sqrt{70^2+4.19.125}$ calculate again – Kushal Bhuyan Jun 15 '16 at 13:47
• I did too and got the same answers – Triatticus Jun 15 '16 at 13:48

You can always simplify a quadratic equation (by dividing out common numeric factors) before solving it. Simplifying like this is optional and is always a good idea because it generally gives you easier numbers to work with.

The error is not because you simplified, it's because there is a mistake somewhere in your work after that. I can't tell you where exactly without seeing your work, but here's how it should go:

\begin{align*} x &= \frac{70 \pm \sqrt{(-70)^2 - 4(19)(-125)}}{2(19)}\\[0.3cm] &= \frac{70 \pm \sqrt{4900 + 9500}}{38}\\[0.3cm] &= \frac{70 \pm \sqrt{14400}}{38} \\[0.3cm] &= \frac{70 \pm 120}{38} \end{align*}

When simplified you get $x = 5$ and $x = -25/19$.

When to simplifying a quadratic is up to you. Remember, multiplying both sides by a number retains the equality, the same thing with adding, subtracting and dividing.

You could've left it the way it is, without simplifying, and you would have gotten the solution as it doesn't affect the roots.

It is for last equation :$$x=\frac { 35\pm \sqrt { { 35 }^{ 2 }+125\cdot 19 } }{ 19 } =\frac { 35\pm \sqrt { 3600 } }{ 19 } =\frac { 35\pm 60 }{ 19 }$$

But using $\frac{\sqrt{b^2-4ac}}{2a}$ in the reduced equation gives the same roots .