Factoring a quadratic equation using two different methods. Why does this work?

Question

I have the quadratic equation: $$4x^2 - 49$$

Using

$$\frac{b\pm\sqrt{b^2-4ac}}{2a}$$

I can factor the equation to

$$(x-3.5)(x+3.5)$$.

However, if I use

$$\frac{b\pm\sqrt{b^2-4ac}}{2}$$

I get 14, which I can then use the grouping method $$4x^2 +14x - 14x -49$$ to get an equivalent result of:

$$(2x-7)(2x+7)$$

Why does removing the $$a$$ from the denominator in the quadratic formula work to give me a number (this case 14), which I can then use to factor out the original quadratic equation?

Also, are there any benefits in solving the equation using the latter method? (the result to looks better to me)

Answer 1

You got lucky actually. It just so happens that in general, a problem of the form $ax^2-b$ can be factored via grouping if $a$ and $b$ are perfect squares, and it further happens that the value needed to perform the grouping process was given. In particular, doing this will apparently result in the correct grouping coefficients if $a=4$.

Answer 2

This is too long for a comment and I don't know what is the "grouping method" but note that the quadratic formula

$$ x_{1,2} = \frac{b \pm \sqrt{b^2 - 4ac}}{2a} $$

gives you the roots $x_{1,2}$ of $ax^2 + bx + c = 0$. When you factor the expression, it is not enough to know the roots but you also need to know $a$. Thus, using the "first" method, you discovered that the roots are $\pm 3.5$ but this does not mean you can factor $4x^2 - 49$ as $(x - 3.5)(x - 3.5)$!

If we open the brackets, we see

$$ (x - 3.5)(x + 3.5) = x^2 - (3.5)^2 = x^2 - \left( \frac{7}{2} \right)^2 = x^2 - \frac{49}{4} $$

so we don't get the same expression. But if we multiply by $a = 4$ both sides, we get the correct factorization

$$ 4x^2 - 49 = 4(x - 3.5)(x + 3.5) = 2(x-3.5)2(x+3.5) = (2x - 7)(2x + 7). $$

Answer 3

Back to completing the square:

\begin{align*} ax^2+bx+c &= a\left( x+\frac{b}{2a} \right)^2- \left( \frac{b^2-4ac}{4a} \right) \\ &= a\left[ \left( x+\frac{b}{2a} \right)^2- \left( \frac{b^2-4ac}{4a^2} \right) \right] \\ &= a\left(x+\frac{b+\sqrt{b^2-4ac}}{2a} \right) \left(x+\frac{b-\sqrt{b^2-4ac}}{2a} \right) \end{align*}

Answer 4

If $ax^2 + bx + c = 0;a \ne 0$ has solutions $x_1,x_2 = \frac{b \pm \sqrt{b^2 - 4ac}}{2a}$ and $v_1, v_2 = \frac{b \pm \sqrt{b^2 - 4ac}}{2} = ax_1, ax_2$

then $(x -x_1)(x - x_2)=x^2 + \frac bax + \frac ca = 0 \iff a(x- x_1)(x-x_2) = ax^2 + bx + c = 0 \iff (x - \frac {v_1}a)(x - \frac{v_2}a) =(x -x_1)(x - x_2)= 0 \iff (ax - v_1)(ax-v_2) = a^2(x - x_1)(x-x_2) = a^2x^2 + bax + ca = 0$

And $(\sqrt{a}x - \sqrt{a}x_1)(\sqrt{a}x - \sqrt{a}x_2) = (\sqrt{a}x - \frac{v_1}{\sqrt{a}})(\sqrt{a}x - \frac{v_2}{\sqrt{a}}) = a^2 + bx + c = 0 \iff etc.$.

If I knew what the "grouping method" was supposed to be (I've never heard of it) I'd be able to answer specifically but the above should be ... exhaustive and exhausting.

Answer 5

Perhaps this simplification is worth investigating? $$\begin{align} ax^2+bx+c&=\frac{1}{a}(a^2x^2+abx+ac)\\ &=\frac{1}{a}((ax)^2+b(ax)+ac)\\ &=\frac{1}{a}(t^2+bt+ac)\\ \end{align}$$ Where $t = ax$

Answer 6

No you didn't get lucky what you found is to two methods for factoring quadratic polymomials that are a consequence of the $\color{red}{\text{the ac method}}$.

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$\mathbf{THEOREM}$. Suppose that $a\ne 0,b$ and $c$ are relatively prime integers and $ax^2 + bx + c$ factors over the set of rational numbers. Then there exists integers $u$ and $v$ such that

$x^2 + bx + ac = (x-u)(x-v)$

$ax^2 + bx + c$ can be factored by grouping as $ax^2 + bx + c = (ax^2 - ux) + (-vx + c)$

The roots of $ax^2 + bx + c$ are $\dfrac ua$ and $\dfrac va$.

$ax^2 + bx + c$ can be factored by simplifying $\dfrac{(ax-u)}{\gcd(a,u)}\;\dfrac{(ax-v)}{\gcd(a,v)}$

$\gcd(a,u)\gcd(a,v) = a$

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$\mathbf{EXAMPLE}$. $ax^2 + bx + c = 10x^2 -3x - 18$.

$x^2 + bx + ac = x^2 - 3x - 180 = (x - 15)(x + 12) = (x-u)(x-v)$

\begin{align} ax^2 + bx + c &= (ax^2 - ux) + (-vx + c) \\ &= (10x^2 - 15x) + (12x - 18) \\ &= 5x(2x - 3) + 6(2x - 3) \\ &= (5x + 6)(2x - 3) \end{align}

\begin{align} ax^2 + bx + c &= \dfrac{(ax-u)}{\gcd(a,u)}\dfrac{(ax-v)}{\gcd(a,v)} \\ &= \dfrac{(10x-15)}{5} \dfrac{(10x+12)}{2}\\ &= (2x-3)(5x+6) \\ \end{align}

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$\mathbf{PROOF}$.

We assume that $\gcd(a,b,c) = 1$ and that $ax^2 + bx + c$ has two real rational roots.

The roots of $x^2 + bx + ac$ are $u,v = \dfrac{-b \pm \sqrt{b^2-4ac}}{2}$. So the roots of $ax^2 + bx + c$ are $\dfrac{-b \pm \sqrt{b^2-4ac}}{2a}=\dfrac ua, \dfrac va$.

The next part is ugly, but I couldn't think of any other way to present it.

We need to reduce the fractions $\dfrac ua$ and $\dfrac va$. Let $s = \gcd(a,u)$ and let $t = \gcd(a,v)$

Then $\dfrac ua = \dfrac{u'}{s'}$ and $\dfrac va = \dfrac{v'}{t'}$ where $u' = \dfrac us,\; s' = \dfrac as,\; v' = \dfrac vt,\; t' = \dfrac at$. Then we must have $$ax^2 + bx + c = a\left( x - \dfrac{u'}{s'} \right) \left( x - \dfrac{v'}{t'} \right) = ax^2 - a\left( \dfrac{u'}{s'} + \dfrac{v'}{t'}\right)x +\dfrac{au'v'}{s't'}. $$

Because $c = \dfrac{au'v'}{s't'}$ is an integer, then $s't'$ must be a divisor of $a$. Lets say $a=s't'a'$. So \begin{align} ax^2 + bx + c &= s't'a'x^2 - s't'a'\left( \dfrac{u'}{s'} + \dfrac{v'}{t'}\right)x +\dfrac{s't'a'u'v'}{s't'} \\ &= s't'a'x^2 - a'(t'u' +s'v')x +a'u'v'\\ \end{align}

Comparing coefficients, we must have \begin{align} a &= s't'a' \\ b &= -a'( t'u' + s'v') \\ c &= a'u'v' \end{align} It follows that $a' \mid \gcd(a,b,c) = 1$. So $a'=1$ Hence \begin{align} a &= s't' \\ b &= -t'u' - s'v' \\ c &= u'v' \\ u &= t'u' \\ v &= s'v' \\ \end{align}

As a consequence, we find that $s = \dfrac{a}{s'} = t'$ and $t = \dfrac{a}{t'} = s'$

We end up with \begin{align} s &= \gcd(a,u) \\ t &= \gcd(a,v) \\ a &= st \\ b &= -u - v \\ c &= u'v' \\ u &= su' \\ v &= tv' \\ \end{align}

If you replace $b$ with $-u-v$, then you find

\begin{align} ax^2 + bx + c &= ax^2 - (u+v)x + c\\ &= stx^2 - (su'+tv')x + u'v'\\ &= (stx^2 - su'x) - (tv'x - u'v') \\ &= sx(tx - u') - v'(tx - u') \\ &= (sx - v')(tx - u') \\ \end{align}

Which agrees with the roots being $\dfrac{u'}{t} = \dfrac ua$ and $\dfrac{v'}{s} = \dfrac va$.

We also note that $\dfrac{(ax-u)}{\gcd(a,u)}\;\dfrac{(ax-v)}{\gcd(a,v)} = \dfrac{(stx-su')}{s}\;\dfrac{(stx-tv')}{t} = (tx-u')(sx-v')$