# Why doesn't quadratic formula lead to a the correct factored form of the original equation?

Applying the quadratic formula to $2x^2-3x+1$ we have

\begin{eqnarray*} a&=&2 \\ b&=&-3 \\ c&=&1 \end{eqnarray*} which gives me two roots: \begin{eqnarray*} x_1&=&1 \\ x_2&=&\tfrac{1}{2} \end{eqnarray*}

Therefore you can re-write the original quadratic as:

$$(x-1)\left(x-\tfrac{1}{2}\right)$$

However, if you actually multiply this out then you get:

$x^2-\frac{3}{2}x + \frac{1}{2}$ which is not the same as $2x^2-3x+1$.

So why the discrepancy? Aren't you supposed to get the original equation?

• Because $(x-x_1)(x-x_2)$ will always give you a quadratic with leading coefficient $1$. Note that if you multiply what you got by $2$, you get back your quadratic. Commented Sep 1, 2015 at 19:49
• You can also rewrite the quadratic as $$a(x-x_1)(x-x_2)$$ Commented Sep 1, 2015 at 22:45

Since you are applying quadratic formula to factorize, I am assuming that you made a typo in the degree of x in the first term. It should be $2x^2$ and not $2x^3$

Also,

$f(x)=ax^2+bx+c=a(x-x_1)(x-x_2)$ is the correct factorization of the trinomial.

$f(x)=0$ does have roots $x_1$ and $x_2$, but the multiplicand '$a$' $=2$ is missing in your reconstruction of the trinomial as a product of its factors.

• To obtain $x_1$, type x_1 when you are in math mode. Commented Sep 2, 2015 at 15:05

The quadratic formula gives the solution to the equation $$ax^2+bx+c=0.$$ But the formula is derived by completing the square, and to do so the coefficient of $x^2$ must be 1. Therefore, the first step is to divide by $a$ (it is assumed without loss of generality that $a>0$). This gives $$x^2+\tfrac{b}{a}x+\tfrac{c}{a}=0.$$ Now notice that $$x^2+\tfrac{b}{a}x+\tfrac{c}{a}=\left(x-\left(\frac{-b+\sqrt{b^2-4ac}}{2a}\right)\right)\left(x-\left(\frac{-b-\sqrt{b^2-4ac}}{2a}\right)\right).$$

Long story short, the quadratic formula gives the solutions of $ax^2+bx+c=0$ by finding the solutions to the $equivalent$ equation $x^2+\tfrac{b}{a}x+\tfrac{c}{a}=0.$

How can $x=\frac{1}{2}$ and $x=1$ both be solutions to $x^2 - \tfrac{3}{2}x+\tfrac{1}{2} = 0$ and $2x^2-3x+1=0$?

The simple fact is that $2x^2-3x+1 \equiv 2\!\left(x^2 - \tfrac{3}{2}x+\tfrac{1}{2}\right)$, and so

$2x^2-3x+1 = 0$ if, and only if, $x^2 - \tfrac{3}{2}x+\tfrac{1}{2}=0$. They have the same solutions.

By a similar argument: $4x^2-6x+2 \equiv 4\!\left(x^2 - \tfrac{3}{2}x+\tfrac{1}{2}\right)$, and so

$4x^2-6x+2 = 0$ if, and only if, $x^2 - \tfrac{3}{2}x+\tfrac{1}{2}=0$. They have the same solutions.

In general, for any non-zero number $k$, we have

$ax^2+bx+c=0$ if, and only if, $(ka)x^2+(kb)x+(kc)=0$. Why not try to prove this?!

Even more generally, for any non-zero number $k$: $\mathrm{f}(x)=0$ if, and only if, $k\cdot \mathrm{f}(x)=0$.

Your linear factor is $(2x-1)$, i.e., $$2x^2-3x+1=(x-1)(2x-1)\neq (x-1)(x-1/2).$$

• You mean $2x^2-3x-1$. Commented Sep 1, 2015 at 22:00