Quadratic Equations, Part 3

So we are in the midst of solving quadratic equations of the form:

\[
{ax}^{2}\hspace{0.33em}{+}\hspace{0.33em}{bx}\hspace{0.33em}{+}\hspace{0.33em}{c}\hspace{0.33em}{=}\hspace{0.33em}{0}
\]

where a,b, and c are some numbers. So far, we have been looking at equations like this that can be factored so that we can use the Null Factor Law. But most quadratics cannot be factored by hand and the solutions to the equations are frequently decimal numbers. So what to do? Well fortunately mathematicians have heard your concerns and way back in the year 628, an Indian mathematician Brahmagupta came up with a formula called the quadratic formula. This formula will solve any quadratic equation.

So given a quadratic equation, if you identify the three coefficients a,b, and c, the solution can be obtained by using the following formula:

\[
{x}\hspace{0.33em}{=}\hspace{0.33em}\frac{{-}{b}\hspace{0.33em}\pm\hspace{0.33em}\sqrt{{b}^{2}\hspace{0.33em}{-}\hspace{0.33em}{4}{ac}}}{2a}
\]

This is a very powerful equation. It is not difficult to prove but I’d rather you research that yourself and for now, just accept  my word for it truthfulness. Let’s use it in some examples.

\[
{2}{x}^{2}\hspace{0.33em}{+}\hspace{0.33em}{5}{x}\hspace{0.33em}{+}\hspace{0.33em}{3}\hspace{0.33em}{=}\hspace{0.33em}{0}
\]

Here a = 2, b = 5, and c = 3. Let’s put these numbers in the quadratic formula. Notice that there is a symbol ± in the formula. That means that there are generally two solutions: one using the + symbol and the other using the – symbol. So the solutions to this equation are:

\[
{x}\hspace{0.33em}{=}\hspace{0.33em}\frac{{-}{5}\hspace{0.33em}\pm\hspace{0.33em}\sqrt{{5}^{2}\hspace{0.33em}{-}\hspace{0.33em}{4}{(}{2}{)(}{3}{)}}}{2(2)}\hspace{0.33em}{=}\hspace{0.33em}\frac{{-}{5}\hspace{0.33em}\pm\hspace{0.33em}{1}}{4}\hspace{0.33em}{=}\hspace{0.33em}{-}{1}{,}\hspace{0.33em}{-}{1}{.}{5}
\]

So there are two solutions to this quadratic equation. Let’s do another one:

\[
{3}{x}^{2}\hspace{0.33em}{-}\hspace{0.33em}{4}{x}\hspace{0.33em}{-}\hspace{0.33em}{7}\hspace{0.33em}{=}\hspace{0.33em}{0}
\]

Here a = 3, b = -4, and c = -7. It’s important to include the signs in the a,b, and and include them in the quadratic formula.

So the solutions are:

\[
{x}\hspace{0.33em}{=}\hspace{0.33em}\frac{{-}{(}{-}{4}{)}\hspace{0.33em}\pm\hspace{0.33em}\sqrt{{(}{-}{4}{)}^{2}\hspace{0.33em}{-}\hspace{0.33em}{4}{(}{3}{)(}{-}{7}{)}}}{2(3)}\hspace{0.33em}{=}\hspace{0.33em}\frac{{4}\hspace{0.33em}\pm\hspace{0.33em}{10}}{6}\hspace{0.33em}{=}\hspace{0.33em}{2}{.}{33}{,}\hspace{0.33em}{-}{1}
\]

Now look at the \[
{b}^{2}\hspace{0.33em}{-}\hspace{0.33em}{4}{ac}
\] part of the formula. If this is a positive number, like in the last two examples, then you can take the square root and get two solutions: one by using the + symbol and the other by using the – symbol. If this is zero, then you only get one solution as adding or subtracting a zero doesn’t give two solutions, just one. And if the expression is negative, well there is no solution in the real world because you can’t take the square root of a negative number. However, scientists and engineers work in a complex world where you can take the square root of a negative number, and this has a physical meaning. Maybe some day I will talk about these special numbers.

Quadratic Factoring, Part 2

Last time I introduced the method to factor some quadratic equations so that you can use the Null Factor Law to solve the equation. The example used was

\[
{x}^{2}\hspace{0.33em}{-}\hspace{0.33em}{2}{x}\hspace{0.33em}{-}\hspace{0.33em}{35}\hspace{0.33em}{=}\hspace{0.33em}{0}
\]

which when factored became

\[
{(}{x}\hspace{0.33em}{-}\hspace{0.33em}{7}{)(}{x}\hspace{0.33em}{+}\hspace{0.33em}{5}{)}\hspace{0.33em}{=}\hspace{0.33em}{0}
\]

where you can readily see the two solutions x = 7 and -5. Let’s do another example.

\[
{x}^{2}\hspace{0.33em}{+}\hspace{0.33em}{2}{x}\hspace{0.33em}{-}\hspace{0.33em}{24}\hspace{0.33em}{=}\hspace{0.33em}{0}
\]

So we know the factors will look like \[
{(}{x}\hspace{0.33em}{+}\hspace{0.33em}{a}{)(}{x}\hspace{0.33em}{+}\hspace{0.33em}{b}{)}
\] and we need to find the a and b so that the factored form is equivalent to the left side of the equation. a and b must be numbers that multiply to equal 24 and add or subtract to equal +2, the coefficient in front of the x. 8 and 3 do not work as they do not add or subtract to equal 2. However, 6 and 4 look like contenders.

So the signs of 6 and 4 must be such that they add to equal +2 but multiply to equal -24. Looks like +6 and -4 work so

\[
{x}^{2}\hspace{0.33em}{+}\hspace{0.33em}{2}{x}\hspace{0.33em}{-}{24}\hspace{0.33em}{=}\hspace{0.33em}{(}{x}\hspace{0.33em}{+}\hspace{0.33em}{6}{)(}{x}\hspace{0.33em}{-}\hspace{0.33em}{4}{)}\hspace{0.33em}{=}\hspace{0.33em}{0}
\]

where the two solution can be found by equating each factor to 0 which gives the two solutions x = 4 and -6.

Some quadratic equations may not be factorable but there is another method to solve these. I will show this method in my next post.

Now the generic quadratic equation looks like

\[
{a}{x}^{2}\hspace{0.33em}{+}\hspace{0.33em}{b}{x}\hspace{0.33em}{+}\hspace{0.33em}{c}\hspace{0.33em}{=}\hspace{0.33em}{0}
\]

where ab, and c are specific numbers. I have been merciful so far in that I have only looked at equations where a = 1. If it is some other value, it’s a little more difficult to factor, but it can be done. However, I will use the quadratic formula to work with these in my next post.

Quadratic Factoring

In my last post, I showed how to solve

\[
{(}{x}\hspace{0.33em}{-}\hspace{0.33em}{7}{)(}{x}\hspace{0.33em}{+}\hspace{0.33em}{5}{)}\hspace{0.33em}{=}\hspace{0.33em}{0}
\]

Now this is really a quadratic equation in disguise. When I covered the Distributive Property, you saw how to distribute a factor within a set of brackets:

\[
{x}{(}{x}\hspace{0.33em}{+}\hspace{0.33em}{5}{)}\hspace{0.33em}{=}\hspace{0.33em}{x}\hspace{0.33em}\times\hspace{0.33em}{x}\hspace{0.33em}{+}\hspace{0.33em}{5}\hspace{0.33em}\times\hspace{0.33em}{x}\hspace{0.33em}{=}\hspace{0.33em}{x}^{2}\hspace{0.33em}{+}\hspace{0.33em}{5}{x}
\]

This can also be done in reverse by un-distributing the x to get the expression back into factored form so that you can take advantage of the Null Factor Law. What I didn’t cover, was how to take an expression like \[
{(}{x}\hspace{0.33em}{-}\hspace{0.33em}{7}{)(}{x}\hspace{0.33em}{+}\hspace{0.33em}{5}{)}
\] and un-factor it. To do this, you can distribute each term in the first set of brackets with each term in the second set:

\[
\begin{array}{l}
{{(}{x}\hspace{0.33em}{-}\hspace{0.33em}{7}{)}{(}{x}\hspace{0.33em}{+}\hspace{0.33em}{5}{)}\hspace{0.33em}{=}\hspace{0.33em}{x}{(}{x}{+}{5}{)}\hspace{0.33em}{-}\hspace{0.33em}{7}{(}{x}\hspace{0.33em}{+}\hspace{0.33em}{5}{)}\hspace{0.33em}{=}\hspace{0.33em}{x}^{2}\hspace{0.33em}{+}\hspace{0.33em}{5}{x}\hspace{0.33em}{-}{7}{x}\hspace{0.33em}{-}\hspace{0.33em}{35}}\\
{{=}\hspace{0.33em}{x}^{2}\hspace{0.33em}{-}\hspace{0.33em}{2}{x}\hspace{0.33em}{-}\hspace{0.33em}{35}}
\end{array}
\]

So this is now recognisable as a quadratic expression. However, if I originally had the equation

\[
{x}^{2}\hspace{0.33em}{-}\hspace{0.33em}{2}{x}\hspace{0.33em}{-}\hspace{0.33em}{35}\hspace{0.33em}{=}\hspace{0.33em}{0}
\]

we could not use the Null Factor Law. So what can you do if given this equation? There is a way to solve this using something called the Quadratic Formula, but that will be covered later. Here I will show how to factor this equation back to the original form we started with so that we can use the Null Factor Law.

So the goal is to take \[
{x}^{2}\hspace{0.33em}{-}\hspace{0.33em}{2}{x}\hspace{0.33em}{-}\hspace{0.33em}{35}
\] and get it in the form

(something)(something else). If you look at how we unfactorised this, you can see that I can start with

\[
{(}{x}\hspace{0.33em}{+}\hspace{0.33em}{a}{)(}{x}\hspace{0.33em}{+}\hspace{0.33em}{b}{)}
\], where we need to find the a and the b so that the expressions are equivalent. I know this is the way to start as the two x‘s are needed to get \[
{x}^{2}
\] when they are multiplied together. Now to find the a and the b, including the sign of each, you need to look at all the possible factors of the known number in the quadratic, in this case -35, that add up to the coefficient of the middle term, in this case -2. Again, this is suggested if you look at how we expanded \[
{(}{x}\hspace{0.33em}{-}\hspace{0.33em}{7}{)}{(}{x}\hspace{0.33em}{+}\hspace{0.33em}{5}{)}
\]. The last number in the expansion (-35) is generated by multiplying the -7 and the +5. These two numbers also multiply the x‘s which are eventually added together to get the middle term, in this case -2x.

So to find the a and the b in \[
{(}{x}\hspace{0.33em}{+}\hspace{0.33em}{a}{)(}{x}\hspace{0.33em}{+}\hspace{0.33em}{b}{)}
\], let’s look at the possible factors of 35 (ignoring the sign for the moment): 35 and 1 or 7 and 5. 35 and 1 do indeed multiply to equal 35 but their addition or subtraction together do not equal 2. That leaves 7 and 5 which do satisfy both requirements: 7 × 5 = 35, 7 – 5 = 2. Now all that remains is to determine the signs. Since their multiplication has to equal -35, one of the numbers needs to be negative. And since the coefficient of the middle term is negative, the large number 7 needs to be negative as well. So in this case a = -7 and b = + 5 so that the factorisation we are looking for is \[
{(}{x}\hspace{0.33em}{+}\hspace{0.33em}{a}{)(}{x}\hspace{0.33em}{+}\hspace{0.33em}{b}{)}
\]. Now we can use the Null Factor Law to solve the equation, as we had done previously.

Not all quadratics can be factored like this, but this is a good skill to develop with practise. I will do several more examples in my next post.

Null Factor Law

I’d like to return to algebra and discuss the Null Factor Law which is useful in solving certain equations:

If two or more factors multiplied together equal zero, then the solutions can be found be equating each factor separately to zero.

This makes sense if you think of two numbers multiplied together equal 0:

\[
{ab}\hspace{0.33em}{=}\hspace{0.33em}{0}
\] can only be true if either a is zero, b is zero, or both are zero. No other non-zero numbers multiplied together can equal zero.

This is true for any algebraic expressions multiplied together. For example:

\[
{(}{x}{-}{7}{)(}{x}{+}{5}{)}\hspace{0.33em}{=}\hspace{0.33em}{0}
\]

Can only be true if \[
{(}{x}{-}{7}{)\hspace{0.33em}=}\hspace{0.33em}{0}
\] or if \[
{(}{x}{+}{5}{)\hspace{0.33em}=}\hspace{0.33em}{0}
\]

Without formal algebra, you can see the two solutions to this equation are then x = 7 or -5.

Now this one was easy, but sometimes you are given an equation that is not a factored one:

\[
{x}^{2}\hspace{0.33em}{+}\hspace{0.33em}{5}{x}\hspace{0.33em}{=}\hspace{0.33em}{0}
\]

At first glance, it looks like the null factor law doesn’t apply here. But I did a post on the distributive property. Please review that if needed, but notice that there is a common factor of x in each of the terms on the left side of the equation. I can un-distribute this x to get:

\[
{x}{(}{x}\hspace{0.33em}{+}\hspace{0.33em}{5}{)}\hspace{0.33em}{=}\hspace{0.33em}{0}
\]

Looks like the null factor law can be used as there are now two factors on the left side. So mentally setting each of these to zero, we get the two solutions x = 0 or -5.

I covered a similar example on my post about quadratic equations. You can click on the tags on the right or below (depending on the device you are viewing this on) to directly go to previous posts on the listed topics.

I will be covering more complex examples in my next post.

Quadratic equations

I think we are ready for more interesting equations to solve. Given the knowledge from the past posts, you have the tools to solve:

\[
{x}^{2}\hspace{0.33em}{+}\hspace{0.33em}{7}{x}\hspace{0.33em}{+}\hspace{0.33em}{5}\hspace{0.33em}{=}\hspace{0.33em}{18}{x}\hspace{0.33em}{+}\hspace{0.33em}{5}
\]

 

This equation is an example of things called quadratic equations. A quadratic equation is one that only has three kinds of terms: \[
{\mathrm{a}}{x}^{2}{,}\hspace{0.33em}{\mathrm{b}}{x}{,}\hspace{0.33em}{\mathrm{and}}\hspace{0.33em}{\mathrm{c}}
\] where the a, b, and c things are known numbers. I will refine this definition a bit later. Now let’s get back to our equation.

Remember, our goal is to do legal things to both sides of the equation to eventually get x = a known number. So you may notice that there is a +5 on both side of the equation. Well a good first step would be to subtract 5 from both sides and this would eliminate them:

\[
{x}^{2}\hspace{0.33em}{+}\hspace{0.33em}{7}{x}\hspace{0.33em}{+}\hspace{0.33em}{5}\hspace{0.33em}{-}\hspace{0.33em}{5}\hspace{0.33em}{=}\hspace{0.33em}{18}{x}\hspace{0.33em}{+}\hspace{0.33em}{5}\hspace{0.33em}{-}\hspace{0.33em}{5}\hspace{0.33em}\hspace{0.33em}\Longrightarrow\hspace{0.33em}\hspace{0.33em}{x}^{2}\hspace{0.33em}{+}\hspace{0.33em}{7}{x}\hspace{0.33em}{=}\hspace{0.33em}{18}{x}
\]

 

Now I’ve decided on the next step because I know what the answer is. Please bear with me and all will be revealed. So let’s now subtract 18x from both sides. Now remember yesterday I said there is a mental shortcut to doing this. The result can be obtained by just moving the 18x to the left side and change its sign. But showing the full detail:

\[
{x}^{2}\hspace{0.33em}{+}\hspace{0.33em}{7}{x}\hspace{0.33em}{-}\hspace{0.33em}{18}{x}\hspace{0.33em}{=}\hspace{0.33em}{18}{x}\hspace{0.33em}{-}\hspace{0.33em}{18}{x}\hspace{0.33em}\hspace{0.33em}\Longrightarrow\hspace{0.33em}\hspace{0.33em}{x}^{2}\hspace{0.33em}{-}\hspace{0.33em}{11}{x}\hspace{0.33em}{=}\hspace{0.33em}{0}
\]

 

Notice that on the left side, I combined the like terms 7x and -18x and that the right side is now zero. In fact, this equation is now in the form:

\[
{\mathrm{a}}{x}^{2}\hspace{0.33em}{+}\hspace{0.33em}{\mathrm{b}}{x}\hspace{0.33em}{+}\hspace{0.33em}{\mathrm{c}}\hspace{0.33em}{=}\hspace{0.33em}{0}
\]

where in this case, a = 1, b = -11 and c = 0. This is the formal definition of a quadratic equation. The only restriction on a, b, or c is that “a” cannot be zero.

Now most quadratic equations can only be solved with something called the quadratic formula. We’ll eventually cover that, but this one can be solved using what we know now.

Now we will not be using square roots for this one. The next step will be to “un-distribute” the x from the terms on the left side:

\[
{x}^{2}\hspace{0.33em}{-}\hspace{0.33em}{11}{x}\hspace{0.33em}{=}\hspace{0.33em}{0}\hspace{0.33em}\hspace{0.33em}\Longrightarrow\hspace{0.33em}\hspace{0.33em}{x}{(}{x}\hspace{0.33em}{-}\hspace{0.33em}{11}{)}\hspace{0.33em}{=}\hspace{0.33em}{0}
\]

 

So now we have two things multiplied together that equal zero. How can two numbers multiplied together equal zero? Well only if one of them is zero. So this sets up two possible solutions: either x = 0 or x – 11 = 0. If you go back to the original equation, you will see that substituting 0 for x will give the true equation 5 = 5. I will leave solving x – 11 = 0 to you, but the answer is x = 11. This is another solution to our equation.

A quadratic equation will have up to two solutions, and in this case, it does have two solutions, 0 and 11. Again, if the equation comes from a real world problem, only one of the answer may make sense so you would discard the other solution.