# Faster way to find zeros of quadratic equations. [closed]

I use 'splitting the middle term' method to find zeros of a quadratic equation. Sometimes it takes a lot of time to split it. As for example: \begin{align} & x^2+5x-1476=0 \\ & x^2+41x-36x-1476=0 \\ & x(x+41)-36(x+41)=0 \\ & (x+41)(x-36)=0 \\ & x=-41, \, 36 \end{align}

Is there any trick to split the middle term, or do you have any faster way to find the zeros.

## closed as off-topic by Shailesh, Claude Leibovici, Xander Henderson, GNUSupporter 8964民主女神 地下教會, NamasteMay 6 '18 at 15:25

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• The zeroes of $x^2+px+q=0$ are $\frac{-p\pm\sqrt{p^2-4q}}2$ in constant time – Hagen von Eitzen Apr 28 '18 at 10:20
• So you need to see that $1476$ factors into $41$ and $36$? – user251257 Apr 28 '18 at 10:20
• Yes, any trick to get it faster. – Gufran Mozahir Apr 28 '18 at 10:25
• @GufranMozahir I don't mean it rudely, but that is a trick in mine eyes at best and a less useful method. – user251257 Apr 28 '18 at 10:34
• No, I didn't take it rudely instead I am thankful to you that you gave your time on my question. – Gufran Mozahir Apr 30 '18 at 14:11

I am not sure whether any of the previous answers really address the question, which is to have a trick to solve the equation $x^2+5x-1476=0$. Of course, the explicit formula for the roots gives the answer, but I would no qualify this as a trick, but as a generic solution for this type of problem.

There is however a trick, if you assume that, as most of the elementary questions of this type, the equation has integer solutions. In this case, you can write $x^2+5x-1476= (x+a)(x-b)$ for some integers $a$ and $b$. It follows $ab = 1476$ and $a-b = 5$. Now comes the trick: the prime decomposition of $1476$ is $41 \times 2^2 \times 3^2$ (easy to find), which let you few possibilities for $a$. Just try the first one, $a = 41$, then $b = 2^2 \times 3^2 = 36$. Bingo! one gets $a-b = 5$, so we are done.

• Thankyou J.-E. Pin this is indeed a good hack. – Gufran Mozahir May 2 '18 at 17:43

The best way in my opinion is using the quadratic formula,

Given a quadratic $ax^2+bx+c =0$

$x=\dfrac{-b\pm\sqrt{b^2-4ac}}{2a}$

You could also try completing the square,in order to get the solution

When you split a polynomial as $$x^2+px+q=x^2+ax+bx+ab=(x+a)(x+b)$$

you readily see that the roots are $-a$ and $-b$. Thus, such splitting is equivalent to finding roots.

By the way, there's an explicit formula for the roots:

$$\frac{-p\pm \sqrt{p^2-4q}}{2}$$

• Thankyou lisyarus. But finding the roots as in my question: √5929 is also lengthy. – Gufran Mozahir Apr 28 '18 at 10:39
• Method 1: Use a calculator. Method 2: Find a number close to $5929$ whose square root you know. E.g. $\sqrt{4900}=70$ or $\sqrt{6400}=80$. Also consider that if $n^2=5929$, then $n$ must end in a $3$ or $7$ if it is a whole number. Since you know its between $70$ and $80$, you can check $73^2=5329$ by hand and $77^2=5929$ which you seek. – Rhys Hughes Apr 28 '18 at 16:27
• Precisely explained, thankyou again. This helped me alot. – Gufran Mozahir May 2 '18 at 18:04

The only way what i know is the substitution $$x=t-\frac{b}{2a}$$ then we get $$t^2+\frac{b^2}{4a^2}-\frac{bt}{a}+\frac{bt}{a}-\frac{b^2}{2a^2}+\frac{c}{a}=0$$ and we get $$t^2+\frac{4ac-b^2}{4a^2}=0$$ to solve. And you get no middle term.

As others have mentioned, use the Quadratic Formula: $$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$$ Here $$x=\frac{-5\pm\sqrt{25-4(1)(-1476)}}{2}$$ $$x=\frac{-5\pm\sqrt{5929}}{2}$$ $$x=\frac{-5\pm 77}{2}$$ $$x=36, x=41$$ from changing between the $\pm$ between a $+$ and a $-$

• Thankyou Rhys Hughes – Gufran Mozahir May 2 '18 at 18:07
• No worries Gufran. – Rhys Hughes May 2 '18 at 18:08

I am too lazy to memorize the $p$-$q$-formula, I like completing the square, thus applying $a^2 + 2ab + b^2 = (a+b)^2$: \begin{align} 0 &= x^2+5x-1476 \\ &= x^2+5x + \biggl(\frac{5}{2}\biggr)^2 - \biggl(\frac{5}{2}\biggr)^2 - 1476 \\ &= \biggl(x + \frac{5}{2}\biggr)^2 - \biggl(\frac{5929}{4}\biggr) \iff \\ \pm \sqrt{\frac{5929}{4}} &= x + \frac{5}{2} \\ \pm \frac{77}{2} &= x + \frac{5}{2} \iff \\ x &= \frac{-5 \pm 77}{2} \end{align}

• Thankyou mvw, it helps.. – Gufran Mozahir May 2 '18 at 18:06