How do we factor quadratic equations?

Mar 27, 2015

A quadratic expression is completely factorizable if and only if its discriminant is positive. Given a quadratic expression of the form $a {x}^{2} + b x + c$, the discriminant $\setminus \Delta$ is defined as ${b}^{2} - 4 a c$.

If the discriminant is negative, the solving formula
${x}_{1 , 2} = \setminus \frac{- b \setminus \pm \setminus \sqrt{\setminus \Delta}}{2 a}$ doesn't work with real numbers, because it would involve the square root of a negative number.

If the discriminant equals zero, the solving formula reduces to
${x}_{1 , 2} = \setminus \frac{- b}{2 a}$, i.e. ${x}_{1} = {x}_{2}$. Once the solutions are found, we can write $p \left(x\right) = \left(x - {x}_{1}\right) \left(x - {x}_{2}\right)$, where $p \left(x\right)$ is the quadratic form and ${x}_{1}$,${x}_{2}$ are the solutions. Since in this case ${x}_{1} = {x}_{2}$, we have that $\left(x - {x}_{1}\right) \left(x - {x}_{2}\right) = {\left(x - {x}_{1}\right)}^{2}$, and this would be the factorization of the quadratic.

If the discriminant is positive the same formulas hold, and this time $p \left(x\right) = \left(x - {x}_{1}\right) \left(x - {x}_{2}\right)$ is the final representation of the quadratic, since it cannot be further simplified, because the terms $\left(x - {x}_{1}\right)$ and $\left(x - {x}_{2}\right)$ are linear.

Mar 27, 2015

A quadratic equation is simply another way of solving a problem if the solution cannot be factored logically.

Let’s say we have the equation ${x}^{2} + 2 x - 3$ for example. This equation could be solved logically using the factors of the first and last terms.

To begin, we can state the factors of the first term, ${x}^{2}$. Imagine there’s an invisible 1 in front of the ${x}^{2}$, therefore the factors are 1, because only $1 \cdot 1 , \mathmr{and} - 1 \cdot - 1$ will multiply to get one. Then we can analyze the third term, $- 3$. The factors of $- 3$ are either $1 \cdot - 3 , \mathmr{and} - 1 \cdot 3$.

Now we can check and see if any of the factors can combine in order to get a $+ 2$, the middle term (don’t worry about the x’s, those will carry over). Recall $1 = - 1 , 1$, and $- 3 = 1 , - 1 , 3 , - 3$

From our factors we can use a -1 and a 3 to get +2. Therefore,
$\left(x + 3\right) \left(x - 1\right) = 0$ is our derived factorization. Then plug in the values to make the statement true, -3 or 1 will both result in an answer of 0 and our the possible values for x.

However , when the logical factorization seen above is not possible, we can plug our numbers into the quadratic equation .

$a {x}^{2} + b x + c$ is the standard way we view an equation. Using the values from the equation above, $a = 1 , b = 2 , \mathmr{and} c = - 3$.

After our a, b, and c values are found we can plug them into the actual quadratic equation.

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

Note : This equation may look intimidating, but as long as you follow factoring rules, you should have no problem. It’s totally normal to come out with an answer containing square roots.