Questions about a trick for solving quadratic equations quickly

Question

I just watched a video that teaches a trick to solve some quadratic equations faster:

Suppose we have $3x^2-152x+100=0$ It takes a lot of time to solve it by finding discriminant because we have to calculate $152^2$ and so on. we divide $3x^2$ by $3$ and multiply $100$ by $3$ and we get: $x^2-152x+300=0$ we can solve it easily by factoring $(x-150)(x-2)=0$ then we divide the roots by $3$ so the roots of original quadratic are $\frac{150}3$ and $\frac23$

It is the first time I see this trick. so is it a known method?

And How we can prove this method works mathematically?

Answer 1

Nice trick, I never saw it.

$$ax^2+bx+c=0\iff a^2x^2+abx+ac=0\iff (ax)^2+b(ax)+ac=0.$$

So you solve for $ax$ and divide by $a$.

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You can combine with another trick that I like, when $b$ is even or a factor $2$ can be pulled out (this is a frequent situation):

$$x^2+2bx+c=0\iff x=-b\pm\sqrt{b^2-c}.$$

Answer 2

Let the original equation be $ax^2+bx+c=0$.

Let us divide degree $2$ term by $a$ and multiply constant term by $a$. Then we get $x^2+bx+ac=0$.

The roots of this are $$x={{-b±\sqrt{b^2-4ac}}\over 2}$$ Dividing the roots by $a$, we get $${x\over a}={{-b±\sqrt{b^2-4ac}}\over 2a}$$ which are the roots of the original equation. Hence proved.

Answer 3

Sure, your method can be proved to be correct.

Consider the quadratic - $ax^{2} +bx+c=0$ where $a$ is assumed to be non-zero. It's roots are given by the quadratic formula- \begin{equation*} \frac{-b\pm \sqrt{b^{2} -4ac}}{2a} \end{equation*} And by your method, the quadratic changes to $x^{2}+bx+ac=0$. It's roots are - \begin{equation*} \frac{-b\pm \sqrt{b^{2} -4ac}}{2} \end{equation*} You propose to divide the roots by $a$, so that just means they are the roots of the original equation.

Nice trick, but is it really feasible?

Answer 4

I have known a cousin of this trick for long, not sure of where from though. Essentially like in factoring method, you multiply $3\times100$ and split the product $300$ into two so that the sum is $-152$, like so $$3x^2-150x - 2x +100 = 3x(x-50)-2(x-50)=(3x-2)(x-50)$$

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In general the logic here is that $$(px+q)(rx+s) = prx^2+(ps+qr)x+qs$$ so if there is a nice factoring of the quadratic, the middle term coefficient should be a sum of two whose product is $pqrs$. One then considers the possible ways to split it back.

Answer 5

This trick is correct. But, unfortunately I cannot call it "nice". I will try to show you the reason for this.

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First alternatively, you can easily prove this trick using Vieta's formula.

$$ax^2+bx+c=0$$

$$x_1+x_2=-\frac ba, ~ x_1x_2=\frac ca$$

Then,

$$y^2+by+ac=0$$

$$y_1+y_2=-b, ~ y_1y_2=ac$$

Putting $$x_1=\frac {y_1}{a},~ x_2=\frac{y_2}{a}$$

You get,

$$x_1+x_2=-\frac ba, ~ x_1x_2=\frac ca$$

which is correct.

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I did not find this trick useful specifically for this equation for the following reason:

Observation

The implementation of this trick is based on the following observation:

$$150+2=152, ~ 150\times 2=300.$$

Isn't it?

But, this trick already works in the original equation:

$$\begin{align}3x^2-152x+100=0\end{align}$$

$$\iff 3x^2-150x-2x+100=0.$$

It is likewise easy to observe:

$$\frac{150}{3}=50, ~ \frac{100}{2}=50$$

This fact is equivalent to observe that

$$150+2=152, ~ 150\times 2=300.$$

Finally we have,

$$\begin{align}3x^2-152x+100=0 &\iff 3x^2-150x-2x+100 \\ &\iff 3x(x-50)-2(x-50) \\ &\iff (3x-2)(x-50).\end{align}$$

Which one is easy? Make up your own mind.

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Also, don't forget about the "half-discriminant" method.

If $$ax^2+2kx+c=0, a≠0$$ then

$$x_{1,2}=\dfrac{-k±\sqrt{k^2-ac}}{a}.$$