Introduction to algebra
This page introduces the concept of quadratic equations and explains some techniques used to solve them.
In brief, it covers the following:
- The definition of a quadratic equation
- How to solve a quadratic equation when b = 0
- How to solve a quadratic equation by factorising
- How to solve a quadratic equation using the quadratic formula
What is a quadratic equation?
A quadratic equation, or second degree equation, is an algebraic equation of the form:
\[ax^2 + bx + c = 0,\]where \(x\) is a variable and \(a\), \(b\) and \(c\) are known numbers, where \(a \neq 0\). These are referred to as coefficients of the equation. NOte that if \(a = 0\) then the equation is linear, not a quadratic.
The most basic quadratic equation occurs when \(a = 1\), \(b = 0\) and \(c = 0\). In this case the equation becomes:
\[x^2 = 0\]From this, you can see that the value of the variable \(x\) must be \(0\), because only \(0^2 = 0\). So, the solution to this quadratic equation is \(0\).
This quadratic equation has only one solution. However, many quadratic equations have two solutions, and some have no real solutions at all. To understand why, it can be helpful to look at the graphs of quadratic equations. These graphs, shown below, are called parabolas. The points where a parabola crosses \(x\) axis (the horizontal axis) are the solutions to the equation, because these are the points where the equation equals \(0\).
The first image below shows a parabola that crosses the \(x\) axis once, so there is only one solution. The second image shows a parabola that crosses the \(x\) axis twice, so there are two solutions. The third image shows a parabola that does not cross the \(x\) axis at all, so there are no real solutions.
(Remember, the point where a graph crosses the \(x\) axis is called the \(x\) intercept.)
There are several ways to solve quadratic equations, depending on the type of equation. This page explains three different techniques.
Solving when \(b = 0\)
Probably the easiest type of quadratic equation to solve is one where \(b = 0\); that is, when the equation has the form:
\[ax^2 + c = 0\]You can solve equations like this by following these steps:
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First, rearrange the equation so that \(ax^2\) is by itself on the left-hand side. Do this by adding or subtracting the constant \(c\) from both sides, just as you would for a linear equation.
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Next, divide both sides of the equation by \(a\) (as long as \(a \neq 0\)), again using the same method you would use for a linear equation.
- Then, take the square root of both sides of the equation. When you do this, remember that there are three possible outcomes:
- there is no real solution, if you are taking the square root of a negative number, because no real nunmber squared gives a negative result;
- there is one solution, \(x = 0\), if you are taking the square root of \(0\); or
- there are two solutions, one positive and one negative, if you are taking the square root of a positive number. For example, \(\sqrt{100} = \pm 10\) (“plus or minus ten”), because both \(10\) and \(-10\) are square roots of \(100\).
- Finally, check your answer(s) by substituting each value of \(x\) back into the original equation. If the result is not correct, go back through your steps to find and fix the mistake.
Example: Solve the quadratic equation \(2x^2 - 72 = 0\)
Solution: Following the steps above, we have:
\[\begin{aligned} && 2x^2 - 72 &= 0 && \\\ \therefore && 2x^2 - 72 + 72 &= 0 + 72 && \textrm{ (undo subtraction by adding) } \\\ \therefore && 2x^2 &= 72 && \textrm{ (simplify) } \\\ \therefore && \frac{2x^2}{2} &= \frac{72}{2} && \textrm{ (divide both sides by 2) } \\\ \therefore && x^2 &= 36 && \textrm{ (simplify) } \\\ \therefore && \sqrt{x^2} &= \sqrt{36} && \textrm{ (take square root of both sides) } \\\ \therefore && x &= \pm 6 && \textrm{ (evaluate - equation is solved!) } \end{aligned}\]Finally, substituting \(\pm 6\) back into the original equation gives \(2(\pm 6)^2 - 72 = 2(36) - 72 = 72 - 72 = 0\) as required. So the solution is correct.
Once you have worked through the example above, have a go at using this technique to find the value of \(x\) in some or all of the following quadratic equations:
Solving by factorising
Here is another way to solve a quadratic equation. This method uses factorising, where you write the equation using two sets of brackets. This method is not always easy. Sometimes it is not even possible. However, if you can see that the equation is easy to factorise, this method can help you solve it quickly. Follow these steps:
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Factorise the quadratic equation. You can learn how to do this in the Factorising section of this module.
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For the quadratic equation to equal zero, at least one of the factors must equal zero. This means you can solve the equation by setting each factor equal to zero. Then solve each one separately, the same way you would solve a linear equation.
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Check that your answer, or answers, are correct. To do this, put the value back into the original equation and work it out. If the answer is not correct, look back through your steps to find your mistake. Then try again.
Example: Solve the quadratic equation \(x^2 + 7x + 12 = 0\)
Solution: Following the steps above, we have:
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\(x^2 + 7x + 12 = 0\) can be factorised as \((x + 3)(x + 4)\)
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Therefore either \(x + 3 = 0\), or \(x + 4 = 0\)
Solving \(x + 3 = 0\) gives \(x = -3\), while solving \(x + 4 = 0\) gives \(x = -4\)
Therefore the two solutions are \(x = -3\) and \(x = -4\) -
Substituting these values back into the original equation gives \((-3)^2 + 7(-3) + 12 = 9 - 21 + 12 = 0\) and \((-4)^2 + 7(-4) + 12 = 16 - 28 + 12 = 0\) as required.
Once you have worked through the example above, have a go at using this technique to find the value of \(x\) in some or all of the following quadratic equations:
Factorising gives \((x + 6)(x + 1) = 0\)
Therefore \(x + 6 = 0\), or \(x + 1 = 0\)
Solving these equations respectively gives two solutions: \(x = -6\) or \(x = -1\)
Factorising gives \((x + 1)(x - 5) = 0\)
Therefore \(x + 1 = 0\), or \(x - 5 = 0\)
Solving these equations respectively gives two solutions: \(x = -1\) or \(x = 5\)
Factorising gives \(x(x + 3) = 0\)
Therefore \(x = 0\), or \(x + 3 = 0\)
Solving gives two solutions: \(x = 0\) or \(x = -3\)
Solving using the quadratic formula
Finally, there is one more way to solve any quadratic equation. This method uses the quadratic formula. This method will not always be the fastest way to find the answer. Because of this, it is a good idea to check first whether the other two methods will work. However, the quadratic formula works in every situation. Sometimes, it is the only method that will work.
The quadratic formula says that for a quadratic equation written as \(ax^2 + bx + c = 0\), you can find the value of \(x\) using this formula:
\[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\]Example: Solve the quadratic equation \(2x^2 + 3x - 10 = 0\)
Solution: Following the steps above, we have:
\[\begin{aligned} x &= \frac{-3 \pm \sqrt{3^2 - 4(2)(-10)}}{2(2)} \\\ &= \frac{-3 \pm \sqrt{9 + 80}}{4} \\\ &= \frac{-3 \pm \sqrt{89}}{4} \\\ &= \frac{-3 \pm 9.43}{4} \\\ &= \frac{6.43}{4} \textrm{ or } \frac{-12.43}{4} \\\ &= 1.61 \textrm{ or } -3.11 \end{aligned}\]Finally, substituting these values back into the original equation gives \(2(1.61)^2 + 3(1.61) - 10 \approx 0\) and \(2(-3.11)^2 + 3(-3.11) - 10 \approx 0\), as required.
Once you have worked through the example above, have a go at solving some or all of the following quadratic equations using one of the three techniques described above:
Factorising gives \((x + 2)^2 = 0\)
Therefore \(x + 2 = 0\)
Solving gives one solution: \(x = -2\)
Factorising gives \((x + 6)(x - 2) = 0\)
Therefore \(x + 6 = 0\) or \(x - 2 = 0\)
Solving these equations respectively gives two solutions: \(x = -6\) or \(x = 2\)
Factorising gives \((x + 13)(x - 4) = 0\)
Therefore \(x + 13 = 0\) or \(x - 4 = 0\)
Solving these equations respectively gives two solutions: \(x = -13\) or \(x = 4\)