Quadratic Simultaneous Equations (AQA GCSE Further Maths): Revision Note

• If the resulting quadratic has a repeated root then the line is a tangent to the curve

• If the resulting quadratic has no roots then the line does not intersect with the curve – or you have made a mistake!

• When giving your final answer, make sure you indicate which x and y values go together

  • If you don’t make this clear you can lose marks for an otherwise correct answer

• Don't make the common mistake of thinking each squared term in x² + y² = 25 can be square-rooted to give x + y = 5 (they can't, the most you can do is $\sqrt{x^2+y^2}=\pm 5$, but you shouldn't be making x or y the subject of this anyway!)

Solve the equations

x² + y² = 36
x = 2y + 6

Number the equations.

$$\begin{gathered} x^2+y^2=36 \\ x=2y+6 \end{gathered}$$$$\begin{gathered} \left(1\right) \\ \left(2\right) \end{gathered}$$ 

There is one quadratic equation and one linear equation so this must be done by substitution.

Equation (2) is equal to $x$ so this can be eliminated by substituting it into the $x$ part for equation (1).
Substitute $x=2y+6$ into equation (1).

${\left(2y+6\right)}^2+y^2=36$

Expand the brackets, remember that a bracket squared should be treated the same as double brackets.

$$\begin{gathered} \left(2y+6\right)\left(2y+6\right)+y^2=36 \\ 4y^2+6\left(2y\right)+6\left(2y\right)+6^2+y^2=36 \end{gathered}$$

Simplify.

$\begin{array}{rcl}4y^2+12y+12y+36+y^2& =& 36\\ 5y^2+24y+36& =& 36\end{array}$

Rearrange to form a quadratic equation that is equal to zero.

$\begin{array}{rcl}5y^2+24y+36-36& =& 0\\ 5y^2+24y& =& 0\end{array}$

The question does not give a specified degree of accuracy, so this can be factorised.
Take out the common factor of $\begin{array}{rcl}& & y\end{array}$.

$\begin{array}{rcl}y\left(5y+24\right)& =& 0\end{array}$

Solve to find the values of $y$.
Let each factor be equal to 0 and solve.

$\begin{array}{rcl}{y}_1& =& 0\\ 5y_2+24& =& 0\Rightarrow y_2=-\frac{24}{5}=-4.8\end{array}$

Substitute the values of $y$ into one of the equations (the linear equation is easier) to find the values of $x$.

              $x_1=2\left(0\right)+6=6x_2=2\left(-\frac{24}{5}\right)+6=-9.6+6$

${\mathit{x}}_{\mathbf{1}}\mathbf{=}\mathbf{6}\mathbf{,}{\mathit{y}}_{\mathbf{1}}\mathbf{=}\mathbf{0}$ 
${\mathit{x}}_{\mathbf{2}}\mathbf{=}\mathbf{-}\mathbf{3}\mathbf{.}\mathbf{6}\mathbf{,}{\mathit{y}}_{\mathbf{2}}\mathbf{=}\mathbf{-}\mathbf{4}\mathbf{.}\mathbf{8}$