Quadratics

Solving Quadratics

A quadratic expression is any expression with an x squared term, an x term and a constant. For example, 3x^2-4x+7 is a quadratic expression. Note that it doesn’t have to be an x, it could by y or any other letter as long as it is the same throughout. Furthermore, there are three ways in which you can solve quadratics – each method requires setting the quadratic to 0 first. See Example 1.

Factorising

Firstly, the simplest method, provided that it is possible, is factorising.

Example: 

\begin{array}{lll}2x^2-5x-3=0\hspace{5pt}&\Rightarrow&\hspace{5pt}(2x+1)(x-3)=0\hspace{5pt}\\&\Rightarrow&\hspace{5pt} x=-\frac{1}{2}, x=3\end{array}.

Quadratic Formula

If factorising doesn’t work but a quadratic does have roots, the quadratic formula will find them instead. Recall that the discriminant will tell you how many roots a quadratic has.  See Discriminants page.
The quadratic formula says that if ax^2+bx+c=0 then the roots are given by:

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

Example: In the following quadratic, a=1, b=3 and c=-3:

\begin{array}{lll}x^2+3x-3=0\hspace{5pt}&\Rightarrow&\hspace{5pt}x=\frac{-3\pm\sqrt{3^2-4\times 1\times-3}}{2}\hspace{5pt}\\&\Rightarrow&\hspace{5pt}x=\frac{-3 +\sqrt{21}}{2}, \frac{-3-\sqrt{21}}{2}\end{array}

given exactly, i.e. not as a rounded decimal. Rounded to two decimal places using a calculator, the solutions are x=0.79 and x=-3.79. See Example 2.

Completing the Square

Alternatively, another infallible method for finding roots (if a quadratic is solvable) is to complete the square. See Completing the Square page and Example 3.

Example: 

\begin{array}{lll}x^2+6x+5=0\hspace{3pt}\hspace{5pt}&\Rightarrow&\hspace{5pt}(x+3)^2-4=0\hspace{5pt}\\&\Rightarrow&\hspace{5pt} x+3=\pm 2\hspace{3pt}\hspace{5pt}\\&\Rightarrow&\hspace{5pt} x=-1, x=-5\end{array}

Sketching Quadratics

It is worth noting that completing the square is also useful for sketching a quadratic. The reason for this is that, by writing the quadratic in completed square form, we can see the transformations applied to the graph of x^2 (the shape of a quadratic is a known as a parabola). For example, to get y=(x+3)^2+1 we shift the graph of x^2 by 3 (x transformation) and then up by 1 (y transformation).

1. Firstly, find the roots using one of the above methods, roots occur when y=0.

2.  Then, find the y-intercept, this occurs when x=0.

3. Finally, find the coordinates of the vertex by completing the square and applying transformations to y=x^2.

See Completing the Square for more details and check out Example 4.

DESMOS 


DESMOS is a fantastic sketching tool. Click here to try it out. Firstly click the start graphing button and type y=x^2+4x-5 in the bar where the cursor starts. Then try adding more graphs and experimenting with the options. Finally, try exporting your graphs.

Examples

Solve the following quadratic equations:
1. x^2+x=12
2. 2x^2=12x-16

Solution:

  1. Write x^2+x=12 as x^2-x-12=0 and then factorise to (x-4)(x+3)=0. The two factors x-4 and x+3 could both be zero to give a product of 0 and so the solutions are x=4 or x=-3.
  2. First write 2x^2=12x-16  as 2x^2-12x+16=0 and then divide both sides by 2 to give x^2-6x+8=0. Then factorise to (x-4)(x-2)=0 and so the solutions are x=4 and x=2.

Solve 2x^2-5x-6=0, giving your answers to two decimal places.

Solution:

Since the question says find the solutions to two decimal places, it suggests that we should use the quadratic formula. Selecting a=2, b=-5 and c=-6, the solutions are given by

\begin{array}{lll}x&=&\frac{-b\pm\sqrt{b^2-4ac}}{2a}=\frac{5\pm\sqrt{(-5)^2-4\times 2\times -6}}{2\times 2}\\&=&\frac{5\pm\sqrt{73}}{4}\approx3.39, -0.89\end{array}

Solve the equation 2x^2+6x-9=0 by completing the square. Give exact answers, simplified where appropriate.

Solution:

Taking out a factor of 2 gives the equation:

\begin{array}{lll}2\left(x^2+3x-\frac{9}{2}\right)=0&\Rightarrow &\hspace{5pt}2\left(\left(x+\frac{3}{2}\right)^2-\frac{27}{4}\right)=0\\&\Rightarrow &\hspace{5pt}\left(x+\frac{3}{2}\right)^2=\frac{27}{4}\\&\Rightarrow& \hspace{5pt}x=-\frac{3}{2}\pm\sqrt{\frac{27}{4}}\end{array}.

Hence, x=-\frac{3}{2}+\frac{3\sqrt{3}}{2} and x= -\frac{3}{2}-\frac{3\sqrt{3}}{2}.

Videos

https://youtu.be/ruryPLFNJW0

Sketching and finding the discriminant of a quadratic in a completing the square style question.

https://youtu.be/wy_EARDCOU8

A real life quadratics question rooted in the context of profit versus selling price.