8 Quadratics
Quadratics often appear in mathematics, they occur when you have something squared, like \(x^2\). They produce ‘U’ shaped graphs that can be either way up (depending on the sign of the \(x^2\) term), and, a powerful formula is known that we can use to solve them.
A plot of \(y=x^2\) is below:
Quadratics can occur when we expand pairs of brackets, so I’ve included in this section.
8.1 Expanding pairs of brackets
Expanding a pair of brackets is much the same as a single bracket. However there is a little more going on. Consider this example of a mental method to calculate \(25 \times 16\).
\[ \begin{aligned} 25 \times 16 &= (20 + 5) \times (10 + 6) \\ &= \overbrace{20 \times 10 + 20 \times 6}^{20 \times (10 + 6)} + \overbrace{5 \times 10 + 5 \times 6}^{5 \times (10 + 6)}\\ &= 200 + 120 + 50 + 30 \\ &= 400 \end{aligned} \]
With algebra it works in the same way:
\[ \begin{aligned} (a+b)(c+d) &= (a+b) \times (c+d) \\ &= \overbrace{a \times c + a \times d}^{a \times (c+d)} + \overbrace{b \times c + b \times d}^{b \times (c+d)}\\ &= ac + ad + bc + bd \end{aligned} \]
8.2 Factorising pairs of brackets
To factorise a quadratic in the form \(x^2 +bx +c\) into a pair of brackets like \((x+p)(x+q)\), we look to see if there are a pair of numbers \(p\) and \(q\) that add to get \(b\) and multiply to get \(c\).
\[ p+q = b \quad \quad p \times q = c \]
If we can find this pair of numbers we can factorise the quadratic. For example for the quadratic \(x^2 + 8x +12\) we can look at the factors of \(12\) to help us.
\[ \begin{aligned} 12 &= 1 \times 12, &\ \ \ 1+12 &= 13\\ 12 &= 2 \times 6, &\ \ \ 2+6 &= 8\\ 12 &= 3 \times 4, &\ \ \ 3+4 &= 7\\ \end{aligned} \]
Notice how \(2\) and \(6\) multiply to get \(12\) and add to get \(8\). This means we have the correct pair. So we can now factorise the quadratic:
\[ x^2 + 8x +12 = (x+2)(x+6) \]
Here are some practice questions.
8.3 Solving Quadratics
Interestingly three things can happen when we solve a quadratic. There can be:
- two different values that satisfy the equation
- one repeated value
- no real values (only imaginary ones - and yes that is a thing!)
Here are some methods to solve quadratic equations.
8.3.1 Factorisation
We can solve some quadratics by factorisation. Take for example the following equation \(x^2 + 8x = -12\). To solve via factorisation we must first make it equal to zero and then factorise. So we have:
\[ \begin{aligned} x^2 + 8x &= -12 \\ x^2 + 8x +12 &= -12 +12 \\ x^2 + 8x +12 &= 0 \end{aligned} \]
Now, with a little sense of deja vu (see the example in the previous section) we can factorise our quadratic to get \((x+2)(x+6)=0\). Notice that this is one bracket multiplied by another to get the answer zero. When this happens, i.e. when you multiply two numbers and the answer is zero, either the first number is zero or the second one is. This means either \(x+2=0\) or \(x+6=0\). Solving these two mini-equations gives the two solutions: either \(x=-2\) or \(x=-6\).
We can quickly get from the factorised quadratic to the solutions by flipping the signs in the bracket.
Try some questions.
8.3.2 Quadratic Formula
For a quadratic equation of the form \(ax^2 + bx + c = 0\) we can use the quadratic formula to find solutions for \(x\).
\[ x = \frac{-b \pm \sqrt{b^2-4ac}}{2a} \]
We can use the formula on the equation \(x^2 -4x +2 = 0\). In this example the values of \(a\), \(b\) and \(c\) are:
- \(a=1\) since \(x^2\) means \(1 \times x^2\)
- \(b=-4\) notice how the negative sign is owned by the \(x\) coefficient
- \(c=2\) finially we just have \(2\)
Substituting into the quadratic formula we have:
\[ \begin{aligned} x &= \frac{-(-4) \pm \sqrt{(-4)^2-4(1)(2)}}{2(1)} \\ &= \frac{4 \pm \sqrt{16-8}}{2} \\ &= \frac{4 \pm \sqrt{8}}{2} \end{aligned} \]
It is possible to simplify the square roots in this answer to give \(2 \pm \sqrt{2}\). So don’t be surprised if your calculator gives you that answer.
Finally, we must deal with the \(\pm\) symbol. This means do the calculation once using addition, \(+\), and another time using subtraction, \(-\). This will give two possible answers for \(x\), given to \(2\) decimal places.
\[ \begin{aligned} x_1 &= \frac{4 + \sqrt{8}}{2} \\ &= 3.41 \end{aligned} \]
and
\[ \begin{aligned} x_2 &= \frac{4 - \sqrt{8}}{2} \\ &= 0.59 \end{aligned} \]
Notice the use of \(x_1\) and \(x_2\). It is common in maths to use subscript numbers to show different particular values of the same variable. That’s all it’s doing \(x_1\) is just a value for \(x\) named \(x_1\) and \(x_2\) is just a value for \(x\) named \(x_2\).
8.4 Simultaneous equations
We are going to solve this type of equation by substitution i.e. substituting one equation into another.
To solve a pair of simultaneous equations of this type we want to rearrange the linear equation such that it is in terms of \(x\) or \(y\), which we can then substitute into the equation with the quadratic terms. This will result in a quadratic equation in terms of one variable only.
For the equations:
\[ \begin{aligned} 2x+y &\ = 1 &\ (1) \\ 3x^2+3y^2 &\ = 4 &\ (2) \end{aligned} \]
we can rearrange equation \((1)\) to make \(y\) the subject:
\[ \begin{aligned} y = 1-2x &\ &\ (3) \end{aligned} \]
Substituting equation \((3)\) into equation \((2)\) we have:
\[ \begin{aligned} 3x^2+3y^2 &\ = 4 \\ 3x^2+3(1-2x)^2 &\ = 4 \\ 3x^2 + 3 (1-2x)(1-2x) &\ = 4 \\ 3x^2+3(1-4x+4x^2) &\ = 4 \\ 3x^2+3-12x+12x^2 &\ = 4 \\ 15x^2-12x-1 &\ = 0 \end{aligned} \]
There are a few things to be careful of here:
- \((1-2x)^2\) was expanded as a pair of brackets, \((1-2x)(1-2x)\) before being multiplied by \(3\).
- The finial stage was to make the equation equal zero so we can use the quadratic formula.
Now we have an equation we can solve we can use the quadratic formula. To find values of \(x\). This gives two solutions \(x_1 = -0.08\) to 2 decimal places, and, \(x_2 = -0.88\) again to 2 decimal places.
Finally, since our equations for \(x\) and \(y\) we need to find corresponding \(y\) values for each \(x\). The easiest way to do this is to use equation \((3)\). This gives, \(y_1 =1.15\) and \(y_2 = -0.75\). Note, to maintain accuracy you’ll need to put your full values for \(x_1\) and \(x_2\) into equation (3) and then round to \(2\) decimal places afterwards.
This gives two pairs of numbers for our answer. \((x_1,y_1) = (-0.08,1.15)\) and \((x_2,y_2)=(0.88,-0.75)\).
notice our answers look a lot like co-ordinates on a graph. That’s because they are. If you plot the lines \(2x+y = 1\) and \(3x^2+3y^2 = 4\) on the same graph (don’t do this by hand! Use something like desmos) the places where the two lines cross will correspond with our answers.
Here are some practice questions. Don’t forget you can graph them if it helps.