Faster way to find zeros of quadratic equations.

Question

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.

Answer 1

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.

Answer 2

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

Answer 3

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:

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

Answer 4

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.

Answer 5

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}

Answer 6

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 $-$