# Found an example for solving via quadratic formula in a book where I am wondering if this is correct

As a refresher, I was skimming through a free Calculus online textbook "MOOCULUS massive open online calculus" (https://mooculus.osu.edu/handouts) and stumbled upon the following example solving a quadratic equation via the quadratic formula: Is the step where they get rid of the -10 in the denominator correct? I am wondering because sqrt(81)/10 is != sqrt(8.1) for example.

• They divided $20$ by $100$, and got $0.2$, which they wrote as $.2$. You may not have noticed the decimal point. It is correct. – André Nicolas Mar 3 '14 at 6:14
• @AndréNicolas why can you just say that $\dfrac{\sqrt{a+b}}{c}=\dfrac{\sqrt{a}}{c}+\dfrac{\sqrt{b}}{c}$? That is what I see in the solution – TrueDefault Mar 3 '14 at 6:19
• @JChau, square both sides, and see for yourself! (You can get rid of all the $c$ denominators first.) – Sammy Black Mar 3 '14 at 6:21
• @SammyBlack I have been taught that $\sqrt{a+b} \neq \sqrt{a}+\sqrt{b}$. So by saying $\dfrac{\sqrt{a+b}}{c}=\dfrac{\sqrt{a}}{c}+\dfrac{\sqrt{b}}{c}$, it implies that $\dfrac{\sqrt{a+b}}{c}=\dfrac{\sqrt{a}+\sqrt{b}}{c}$, which means that $\sqrt{a+b}=\sqrt{a}+\sqrt{b}$. That is not true. So, if André is right, where is my mistake? – TrueDefault Mar 3 '14 at 6:27
• You are correct that square roots don't distribute over sums, but I don't see how anything that @AndréNicolas wrote suggests that. – Sammy Black Mar 3 '14 at 6:31

It is correct. $$t=\dfrac{-30\pm \sqrt{30^2-4(-5)(60-h)}}{2(-5)}$$ $$t=\dfrac{-30\pm \sqrt{900+20(60-h)}}{-10}$$ $$t=\dfrac{-30\pm \sqrt{900+1200-20h}}{-10}$$ $$t=\dfrac{-30\pm \sqrt{2100-20h}}{-10}$$ $$t=\dfrac{-30}{-10}\pm \dfrac{\sqrt{2100-20h}}{-10}$$ $$t=3 \mp \dfrac{\sqrt{2100-20h}}{10}$$ $$t=3 \mp \dfrac{\sqrt{2100-20h}}{\sqrt{100}}$$ $$t= 3 \mp \sqrt{\dfrac{2100-20h}{100}}$$ $$\displaystyle \large \boxed{t= 3\mp \sqrt{21-0.2h}}$$ I hope this post helped you.