This guide provides useful resources for a wide variety of math topics. It is targeted at students enrolled in a math course or any other Centennial course that requires math knowledge and skills.

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An equation is a **quadratic equation** if it can be written in the form:

\(ax^2+bx+c=0\)

where \(a, b, c\) are known numbers, \(a \ne 0\), and \(x\) is some unknown variable.

__Examples__

- \(4w^2-10=0\) is a quadratic equation (it is in the form \(ax^2+bx+c=0\) where \(b=0\) and instead of \(x\) as our unknown variable we have \(w\)).
- \(3(2x^2+1)=-5x\) is a quadratic equation (we can first expand the left-hand side into \(6x^2+3=-5x\), then rearrange it into \(6x^2+5x+3=0\)).
- \(7x-4\) is
__not__a quadratic equation (because \(a=0\)).

A **root** or **solution **of a quadratic equation is a value that satisfies the equation. For example, \(x = 1\) is a solution to \(x^2-2x+1=0\) since \((1)^2-2(1)+1=0\). *Is it possible to find another solution?*

We can use the factoring techniques we've learned to solve quadratic equations. You can go to the 'Factoring' page to review the various methods of factoring you can use.

To solve quadratic equations using factoring, follow these steps:

- Rewrite/rearrange the quadratic equation so that one side is equal to 0 (This is
__very important__ - Factor the other side of the equation (i.e. the non-zero side).
- Set each of the factors equal to 0 and solve these linear equations separately for roots.

Example: Solve \(2x^2-6=x\) by factoring.

**Solution:**

1. First, we need to rearrange this equation so that one side is equal to 0. We do this by moving \(x\) from the right to the left side by subtracting it on both sides:

\(2x^2-6-x=\cancel{x}-\cancel{x}\)

\(2x^2-x-6=0\)

2. Now we factor the left side of the equation:

\(2x^2-x-6=0\)

\((x-2)(2x+3)=0\)

3. Set each of the factors equal to zero and solve each equation separately to find the roots of the original quadratic equation:

\(x-2=0\) *or * \(2x+3=0\)

\(x=2\) \(2x=-3\)

\(x=2\) \(x=\frac{-3}{2}\)

So, the solutions to the quadratic equation \(2x^2-6=x\) are \(x=2\) or \(x=\frac{-3}{2}\).

Watch the video below for another example of solving a quadratic equation by factoring using the 'Difference of Squares' factoring technique:

What happens if a quadratic equation has roots that aren't integers? You may find it tricky or near impossible to solve such equations using factoring. In those cases, another way to find the roots of a quadratic equation is using the **quadratic formula:**

\(x=\frac{-b\ \pm\sqrt{b^2-4ac}}{2a}\)

where \(ax^2+bx+c=0\).

Notice that in the formula, there is a symbol with a plus sign and a minus sign: \(\pm\)

What this means is that we actually have to consider two expressions:

\(x=\frac{-b\ + \sqrt{b^2-4ac}}{2a}\) and \(x=\frac{-b\ - \sqrt{b^2-4ac}}{2a}\)

**Tip:** For this formula to work properly, you first need to make sure that your quadratic equation is in the form: \(ax^2+bx+c=0\).

Then, you can plug the \(a, b, c\) values into the formula to calculate for the roots of the quadratic equation!

See the below video for an example of solving a quadratic equation using the quadratic formula:

When approaching word problems, often multiple equations can be extracted from the question. This system of equations then needs to be solved to find the answer.

**Solution:**

Since we're given that "one number is the square of another", if we let \(x\) represent one number, and \(y\) represent the other number, we have the equation representing their relationship:

\(y=x^2\)

Since "their sum is 176", we have the equation:

\(x+y=176\)

Now, we can substitute the first equation into the second to end up with one equation we will solve:

\(x+y=x+x^2=176\)

Notice that we now have a quadratic equation. To determine its solutions, we need to make one side equal to 0, then factor it:

\(x^2+x=176 \Rightarrow x^2+x-176=0\)

Since this equation does not easily factor, we apply the Quadratic Formula to find the solutions:

\(x=\frac{-(1)\pm \sqrt{(1)^2-4(1)(-176)}}{2(1)}\)

\(x=\frac{-1\pm \sqrt{705}}{2}\)

\(x \approx 12.78\) or \(x \approx -13.78\)

Now that we our solutions, we can plug them back into the original equations to find the values for \(y\), as well as check our work to make sure our solutions are valid.

Since \(x+y=176\), we can rearrange this equation and use it to find \(y\):

\(y=176-x\)

For \(x \approx 12.78\), \(y \approx 176-12.78 \approx 163.22\) Checking our work that \(y=x^2\), indeed \(163.22 \approx (12.78)^2\) |
For \(x \approx -13.78\), \(y \approx 176-(-13.78) \approx 189.78\) Checking our work that \(y=x^2\), indeed \(189.78 \approx (-13.78)^2\) |

So, we actually have two pairs of numbers that work in the given statement:

- \(x \approx 12.78, y \approx 163.22\)
- \(x \approx -13.78, y \approx 189.78\)

In our previous examples, you might have noticed that some equations had a different number of roots/solutions - 0 roots, 1 root or 2 roots.

For example, for \(x^2-2x+1=0\), we mentioned that \(x=1\) is a root/solution to this quadratic equation. In fact, it is the __only__ root of this equation (i.e. this quadratic equation only has one root).

To determine the number of roots a quadratic equation has, we can use a part of the quadratic formula called the **discriminant:**

\(D = b^2-4ac\)

There are three possible cases:

- If \(b^2-4ac < 0\)
**no roots.** - If \(b^2-4ac = 0\) , then \(ax^2+bx+c=0\) has one root.
- If \(b^2-4ac > 0\) , then \(ax^2+bx+c=0\) has two (distinct) roots.

**Tip:** Make sure that the quadratic equation you are working with is written in \(ax^2+bx+c=0\) form before calculating its discriminant!

Let's apply this idea to our previous example: \(x^2-2x+1=0\).

To calculate the discriminant, we plug in \(a=1, b=-2, c=1\) into the discriminant formula:

\(D=(-2)^2-4(1)(1)=4-4=0\)

Since \(D=0\), this tells us that \(x^2-2x+1=0\) only has one root.

Time for you to apply these ideas!

- Last Updated: Oct 29, 2024 1:02 PM
- URL: https://libraryguides.centennialcollege.ca/mathhelp
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