MHB Method for Checking the Solutions to a Quadratic

[Ackbach]

So you start out with the standard form of a quadratic, $ax^{2}+bx+c=0$, and we know that the solutions are from the quadratic formula:
$$x= \frac{-b \pm \sqrt{b^{2}-4ac}}{2a}.$$
Let's assume for the sake of this thread that $a,b,c$ are all real.

Now, your mantra when solving equations should be that the solution is not correct unless it's checked. How could you check these solutions? Obviously, by plugging them into the original equation. However, that's going to be a fair amount of algebra, which might be just as error-prone as solving the original. There are several short-cuts which should reduce the amount of time you need to check your solutions.

We begin with the discriminant, $b^{2}-4ac$. The nature of the solutions will depend greatly on the sign of the discriminant. Here are the cases:

1. $b^{2}-4ac>0$ - you will have two distinct real roots.
2. $b^{2}-4ac=0$ - you will have a single repeated real root.
3. $b^{2}-4ac<0$ - you will have two complex conjugate pair roots.

So the solutions you obtain should check in with whichever case you have. Note that if you have complex coefficients, all bets are off with checking the discriminant.

Next, there are two quick checks you can do which are almost certain to catch any errors in your computations, and are much easier to do that plugging into the original equation. Let's begin with adding the two solutions provided by the quadratic formula together, and see what we get:
$$\frac{-b + \sqrt{b^{2}-4ac}}{2a}+\frac{-b - \sqrt{b^{2}-4ac}}{2a}
= \frac{-b + \sqrt{b^{2}-4ac}-b - \sqrt{b^{2}-4ac}}{2a}=-\frac{2b}{2a}=- \frac{b}{a}.$$
So the sum of the two solutions should give you $-b/a$. Note that this check works even if the solutions are complex conjugate pairs. Theoretically, this check should work even with complex coefficients!

Next, let's try multiplying out the two solutions. We get a difference-of-squares pattern:
$$\frac{-b + \sqrt{b^{2}-4ac}}{2a} \times \frac{-b - \sqrt{b^{2}-4ac}}{2a}
= \frac{b^{2}-(b^{2}-4ac)}{4a^{2}}= \frac{4ac}{4a^{2}}= \frac{c}{a}.$$
So the product of the two solutions should give you $c/a$. Again, this should work for any quadratic, even ones with complex coefficients.

To review, then, there are three checks you should have when you're solving a quadratic:

1. The sign of the discriminant should match up with the kinds of solutions you found.
2. The sum of the solutions should be $-b/a$.
3. The product of the solutions should be $c/a$.

If you're dealing with a quadratic where the coefficient of $x^{2}$ is just $1$, then you get the further simplification that the sum of the solutions must be $-b$, and the product must be $c$.

Comments and questions should be posted here:

http://mathhelpboards.com/commentary-threads-53/commentary-method-checking-solutions-quadratic-4208.html

 

[Prove It]

temp

Moderator edit: This commentary topic pertains to the following tutorial:

http://mathhelpboards.com/math-notes-49/method-checking-solutions-quadratic-4101.html

And for anyone wondering how we get the Quadratic Formula, it's just completing the square on a standard quadratic equation [math]\displaystyle a\,x^2 + b\,x + c = 0[/math].

[math]\displaystyle \begin{align*} a\,x^2 + b\,x + c &= 0 \\ a \left( x^2 + \frac{b}{a}\,x + \frac{c}{a} \right) &= 0 \\ x^2 + \frac{b}{a}\,x + \frac{c}{a} &= 0 \\ x^2 + \frac{b}{a}\,x + \left( \frac{b}{2a} \right) ^2 - \left( \frac{b}{2a} \right) ^2 + \frac{c}{a} &= 0 \\ \left( x + \frac{b}{2a} \right) ^2 - \frac{b^2}{4a^2} + \frac{4ac}{4a^2} &= 0 \\ \left( x + \frac{b}{2a} \right) ^2 &= \frac{b^2 - 4ac}{4a^2} \\ x + \frac{b}{2a} &= \frac{\pm \sqrt{b^2 - 4ac}}{2a} \\ x &= \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \end{align*}[/math]

 

[Ackbach]

Re: Method for Checking the Solutions to a Quadratic

And for anyone wondering how we get the Quadratic Formula, it's just completing the square on a standard quadratic equation [math]\displaystyle a\,x^2 + b\,x + c = 0[/math].

[math]\displaystyle \begin{align*} a\,x^2 + b\,x + c &= 0 \\ a \left( x^2 + \frac{b}{a}\,x + \frac{c}{a} \right) &= 0 \\ x^2 + \frac{b}{a}\,x + \frac{c}{a} &= 0 \\ x^2 + \frac{b}{a}\,x + \left( \frac{b}{2a} \right) ^2 - \left( \frac{b}{2a} \right) ^2 + \frac{c}{a} &= 0 \\ \left( x + \frac{b}{2a} \right) ^2 - \frac{b^2}{4a^2} + \frac{4ac}{4a^2} &= 0 \\ \left( x + \frac{b}{2a} \right) ^2 &= \frac{b^2 - 4ac}{4a^2} \\ x + \frac{b}{2a} &= \frac{\pm \sqrt{b^2 - 4ac}}{2a} \\ x &= \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \end{align*}[/math]

Right, or you can do a http://www.mathhelpboards.com/f2/two-methods-deriving-quadratic-formula-i-not-taught-school-2629/.