Quadratic Simultaneous Equations¶

Given a quadratic equation, and a linear equation, you can use the following method to solve them.

Example¶

\[ x^2 + y^2 = 5 \\ 3x - y = -1 \]

Rearrange the linear equation¶

Before you can start proceeding with solving the equations, you need to rearrange the linear equation to make \(y\) the subject.

\[ \begin{align} 3x - y & = -1 \\ 3x & = -1 + y \\ 3x + 1 & = y \end{align} \]

The subject of the linear equation is \(y\).

Substitute the linear equation¶

In order to use the quadratic formula, the equation needs to equal \(0\), and for it to only contain one variable.

\[ x^2 + (3x + 1)^2 = 5 \]

The linear equation has been substituted into the quadratic equation.

Expand the brackets¶

By this point chances are there are some brackets involved, these need to be squashed.

	\(3x\)	\(+1\)
\(3x\)	\(9x^2\)	\(+3x\)
\(+1\)	\(+3x\)	\(+1\)

\[ \begin{align} x^2 + 9x^2 + 6x + 1 & = 5 \\ 10x^2 + 6x + 1 & = 5 \end{align} \]

The brackets in the equation were expanded, and the equation simplified.

Make the equation equal to 0¶

To be able to use the quadratic formula, the equation needs to equal \(0\). It is worth noting that the equation may well be better to solve by use of factorising or, my favourite, completing the square.

\[ \begin{align} 10x^2 + 6x + 1 & = 5 \\ 10x^2 + 6x - 4 & = 0 \end{align} \]

The equation was made to equal \(0\).

Use the Quadratic Formula¶

I am now ready to use the Quadratic Formula to solve the equation.

\[ \begin{align} x & = \frac{-b \pm\sqrt{b^2 - 4ac}}{2a} \\ & = \frac{-6 \pm\sqrt{6^2 - 4(10)(-4)}}{2(10)} \\ & = \frac{-6 \pm\sqrt{36 + 160}}{20} \\ & = \frac{-6 \pm\sqrt{196}}{20} \end{align} \]

Solve¶

Now having an equation where \(x\) is the subject, the equations can now be solved. The established \(x\) values will need to be factored into the original linear equation, with \(y\) as the subject.

\[ \begin{align} x & = \frac{-6 + \sqrt{196}}{20} \\ & = 0.4 \\ \\ y & = 3x + 1 \\ & = 3(0.4) + 1 \\ & = 1.2 + 1 \\ & = 2.2 \end{align} \]

\[ \begin{align} x & = \frac{-6 - \sqrt{196}}{20} \\ & = -1 \\ \\ y & = 3x + 1 \\ & = 3(-1) + 1 \\ & = -3 + 1 \\ & = -2 \end{align} \]

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