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I have a very simple problem which must have a simple answer and I was wondering if anyone can point out my error.

I have the following quadratic equation to factor:

$2x^2+5x+1$

Which is of the form:

$Ax^2+Bx+C$

All I want to now do is factor this into the form:

$(x+\alpha)(x+\beta)$

So using the quadratic formula

$x = \frac{-B\pm\sqrt{B^2-4AC}}{2A}$

I get

$x = \frac{-5 \pm \sqrt{17}}{4}$

So I would think then that

$(x - \frac{-5 + \sqrt{17}}{4})(x - \frac{-5 - \sqrt{17}}{4}) = 2x^2+5x+1 $

But it doesn't seem to work. Any x that I choose results in the answer being off by a factor of A (in this case A=2).

What silly thing am I missing?

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  • $\begingroup$ The left hand side of your equation is the factorization of $$x^2 + \frac{5}{2}x + \frac{1}{2}$$ rather than the factorization of $2x^2 + 5x + 1$. $\endgroup$ – N. F. Taussig Mar 4 '15 at 18:37
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We can obtain the correct factorization by completing the square. \begin{align*} 2x^2 + 5x + 1 & = 2\left(x^2 + \frac{5}{2}x\right) + 1\\ & = 2\left(x^2 + \frac{5}{2}x + \frac{25}{16}\right) - \frac{25}{8} + 1\\ & = 2\left(x + \frac{5}{4}\right)^2 - \frac{17}{8}\\ & = 2\left[\left(x + \frac{5}{4}\right)^2 - \frac{17}{16}\right]\\ & = 2\left[x + \frac{5}{4} + \frac{\sqrt{17}}{4}\right]\left[x + \frac{5}{4} - \frac{\sqrt{17}}{4}\right]\\ & = 2\left[x + \frac{5 + \sqrt{17}}{4}\right]\left[x - \frac{5 + \sqrt{17}}{4}\right] \end{align*} However, when we solve the equation $2x^2 + 5x + 1 = 0$ by completing the square, we first transform it into the monic equation $$x^2 + \frac{5}{2}x + \frac{1}{2} = 0$$ whose roots are the same as those of the original equation. When we derive the Quadratic Formula, we complete the square on the equation $$ax^2 + bx + c = 0$$ where $a \neq 0$. The first step in the derivation is to divide $ax^2 + bx + c = 0$ by $a$ to transform it into the monic equation $$x^2 + \frac{b}{a}x + \frac{c}{a} = 0$$ whose roots are equivalent to those of the equation $ax^2 + bx + c = 0$.

If we factor the monic equation, we obtain $$\left(x - \frac{-b + \sqrt{b^2 - 4ac}}{2a}\right)\left(x - \frac{-b + \sqrt{b^2 - 4ac}}{2a}\right) = 0$$ The sum of the roots is $-\dfrac{b}{a}$ and the product of the roots is $\dfrac{c}{a}$.

To obtain the original equation from the monic equation, we must multiply the monic equation by $a$. Thus, the factorization of the original equation is \begin{align*} ax^2 + bx + c & = a\left(x^2 + \frac{b}{a}x + \frac{c}{a}\right)\\ & = a\left(x - \frac{-b + \sqrt{b^2 - 4ac}}{2a}\right)\left(x - \frac{-b + \sqrt{b^2 - 4ac}}{2a}\right) \end{align*}

What you found by using the sum and product of the roots was the factorization of the monic equation $$x^2 + \frac{5}{2}x + \frac{1}{2} = 0$$ To obtain the original equation, you must multiply your factorization by $a = 2$.

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When you use the quadratic formula to find the roots of a quadratic, you search for solutions $x$ to the quadratic equation $$Ax^2 + Bx + C = 0$$

Note that this is equivalent to the quadratic equation $(A\neq 0)$ $$x^2 + \frac BA\cdot x+ \frac CA = 0$$

Both the quadratics $Ax^2 + Bx + C$ and $x^2 + \frac BAx +\frac CA$ share the same roots.

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$Ax^2+Bx+C$ factorize as : $A(x-\alpha)(x-\beta)$ wiht $\alpha + \beta= \dfrac{-B}{A}$ and $\alpha \beta=\dfrac{C}{A}$.

obviously if $A \ne 0$, and $\alpha$ and $\beta$ are roots of the the two equations:

$$ A(x-\alpha)(x-\beta)=0 \iff Ax^2+Bx+C=0 $$ and

$$ (x-\alpha)(x-\beta)=0 \iff x^2+\dfrac{B}{A}x+\dfrac{C}{A}=0 $$

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