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.

Posted on Categories Algebra, Pre-VCE