Tips For Solving Quadratic Equations

Every algebra student at higher levels needs to learn to solve quadratic equations. These are a type of polynomial equation that includes a power of 2 but none higher, and they have the general form: ​ax​² + ​bx​ + ​c​ = 0. You can solve these by using the quadratic equation formula, by factorizing or by completing the square.

Factorization exploits the fact that the right hand side of the standard quadratic equation equals zero. This means if you can split the equation up into two terms in brackets multiplied by each other, you can work out the solutions by thinking about what would make each bracket equal zero. To give a concrete example:

\(x^2 + 6x + 9 = 0\)

Compare this to the standard form:

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

In the example, ​a​ = 1, ​b​ = 6 and ​c​ = 9. The challenge of factorizing is finding two numbers that add together to give the number in the ​b​ spot and multiply together to get the number in the place for ​c​.

So, representing the numbers by ​d​ and ​e​, you're looking for numbers that satisfy:

\(d + e = b\)

Or in this case, with ​b​ = 6:

\(d + e = 6\)

And

\(d × e = c\)

Or in this case, with ​c​ = 9:

\(d × e = 9\)

Focus on finding numbers that are factors of ​c​, and then add them together to see if they equal ​b​. When you have your numbers, put them in the following format:

\((x + d) (x + e)\)

In the above example, both ​d​ and ​e​ are 3:

\(x^2 + 6x + 9 = (x + 3) (x + 3) = 0\)

If you multiply out the brackets, you'll end up with the original expression again, and this is good practice to check your factorization. You can run through this process (by multiplying the first, inner, outer and then last parts of the brackets in turn — see Resources for more detail) to see it in reverse:

\(\begin{aligned}
(x + 3) (x + 3) &= (x × x) + (3 × x ) + (x × 3) + (3 × 3)\)
\(&= x^2 + 3x + 3x + 9\)
\(&= x^2 + 6x + 9 \
\end{aligned}\)

Factorization effectively runs through this process in reverse, but it can be challenging to work out the right way to factor the quadratic equation, and this method isn't ideal for every quadratic equation for this reason. Often you have to guess at a factorization and then check it.

The problem is now making either of the expressions in brackets come out to equal zero through your choice of value for ​x​. If either bracket equals zero, the whole equation equals zero, and you've found a solution. Look at the last stage [(​x​ + 3) (​x​ + 3) = 0] and you'll see that the only time the brackets come out to zero is if ​x​= −3. In most cases, though, quadratic equations have two solutions.

Factorization is even more challenging if ​a​ isn't equal to one, but focusing on simple cases is better at first.

Completing the square helps you solve quadratic equations that can't be easily factorized. This method can work for any quadratic equation, but some equations suit it more than others. The approach involves making the expression into a perfect square and solving that. A generic perfect square expands like this:

\((x + d)^2 = x^2 + 2dx + d^2\)

To solve a quadratic equation by completing the square, get the expression into the form on the right hand side of the above. First divide the number in the ​b​ position by 2, and then square the result. So for the equation:

\(x^2 + 8x = 0\)

The coefficient ​b​ = 8, so ​b​ ÷ 2 = 4 and (​b​ ÷ 2)² = 16.

Add this to both sides to get:

\(x^2 + 8x + 16 = 16\)

Note that this form matches the perfect square form, with ​d​ = 4, so 2​d​ = 8 and ​d​² = 16. This means that:

\(x^2 + 8x + 16 = (x + 4)^2\)

Insert this into the previous equation to get:

\((x + 4)^2 = 16\)

Now solve the equation for ​x​. Take the square root of both sides to get:

\(x + 4 = \sqrt{16}\)

Subtract 4 from both sides to get:

\(x = \sqrt{16} – 4\)

The root can be positive or negative, and taking the negative root gives:

\(x = -4 – 4 = -8\)

Find the other solution with the positive root:

\(x = 4 – 4 = 0\)

Therefore the only non-zero solution is −8. Check this with the original expression to confirm.

The quadratic equation formula looks more complicated than the other methods, but it's the most reliable method, and you can use it on any quadratic equation. The equation uses the symbols from the standard quadratic equation:

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

And states that:

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

Insert the appropriate numbers into their places and work through the formula to solve, remembering to try both subtracting and adding the square root term and note both answers. For the following example:

\(x^2 + 6x + 5 = 0\)

You have ​a​ = 1, ​b​ = 6 and ​c​ = 5. So the formula gives:

\(\begin{aligned}
x &= \frac{-6 ± \sqrt{6^2 – 4×1×5}}{2×1} \
&= \frac{-6 ± \sqrt{36 – 20}}{2} \
&= \frac{-6 ± \sqrt{16}}{2} \
&= \frac{-6 ± 4}{2}
\end{aligned}\)

Taking the positive sign gives:

\(\begin{aligned}
x &= \frac{-6 + 4}{2} \
&= \frac{-2}{2} \
&= -1
\end{aligned}\)

And taking the negative sign gives:

\(\begin{aligned}
x &= \frac{-6 – 4}{2} \
&= \frac{-10}{2} \
&= -5
\end{aligned}\)

Which are the two solutions for the equation.

Look for a factorization before trying anything else. If you can spot one, this is the quickest and easiest way to solve a quadratic equation. Remember that you're looking for two numbers that sum to the ​b​ coefficient and multiply to give the ​c​ coefficient. For this equation:

\(x^2 + 5x + 6 = 0\)

You can spot that 2 + 3 = 5 and 2 × 3 = 6, so:

\(x^2 + 5x + 6 = (x + 2) (x + 3) = 0\)

And ​x​ = −2 or ​x​ = −3.

If you can't see a factorization, check to see if the ​b​ coefficient is divisible by 2 without resorting to fractions. If it is, completing the square is probably the easiest way to solve the equation.

If neither approach seems suitable, use the formula. This seems like the hardest approach, but if you're in an exam or otherwise pushed for time, it can make the process a lot less stressful and much faster.