Hi - what is the best way to solve simultaneous equations where one is quadratic and the other linear? For example:

Hi - what is the best way to solve simultaneous equations where one is quadratic and the other linear?

For example:

y=x+3

y=x²+3x

1. Formula line ( linear equation)

$y_{\text{line}}=m\cdot x_{\text{line}}+b$

2. Formula parabola (quadatic equation)

$y_{\text{parabola}}=A\cdot x_{\text{parabola}}^2+B\cdot x_{\text{parabola}} + C$

3. set equal  $y_{\text{line}}=y_{\text{parabola}}=y_{\text{intersection}}$ :

$\begin{array}{|rcll|} \hline m\cdot x_{\text{intersection}}+b &=& A\cdot x_{\text{intersection}}^2+B\cdot x_{\text{intersection}} + C \\\\ Ax_{\text{intersection}}^2+x_{\text{intersection}}(B-m)+C-b &=& 0 \\ x_{\text{intersection}_{1,2}} &=& \dfrac{m-B\pm \sqrt{(m-B)^2-4\cdot A \cdot(C-b)} }{2A} \\ y_{\text{intersection}_{1,2}} &=& m\cdot x_{\text{intersection}_{1,2}} + b \\ \hline \end{array}$

4. Example:

$\begin{array}{|rcll|} \hline y &=& x + 3 \quad & \quad m=1 \quad b = 3 \\ y &=& x^2+3x \quad & \quad A=1 \quad B = 3 \quad C = 0 \\ x_{\text{intersection}_{1,2}} &=& \dfrac{1-3\pm \sqrt{(1-3)^2-4\cdot 1 \cdot(0-3)} }{2\cdot 1} \\ x_{\text{intersection}_{1,2}} &=& \dfrac{-2\pm \sqrt{4+12} }{2} \\ x_{\text{intersection}_{1,2}} &=& \dfrac{-2\pm 4 }{2} \\ x_{\text{intersection}_{1}} &=& \dfrac{-2 + 4 }{2} \\ &=& 1 \\\\ x_{\text{intersection}_{2}} &=& \dfrac{-2 - 4 }{2} \\ &=& -3 \\\\ y_{\text{intersection}_{1}} &=& 1\cdot x_{\text{intersection}_{1}} + 3 \\ &=& 1\cdot 1 + 3 \\ &=& 4 \\\\ y_{\text{intersection}_{2}} &=& 1\cdot x_{\text{intersection}_{2}} + 3 \\ &=& 1\cdot (-3) + 3 \\ &=& 0 \\ \hline \end{array}$

Just set the equations equal......so we have...

x^2 + 3x   = x + 3           subtract  x + 3 from both sides

x^2 + 2x - 3   = 0           factor

(x + 3) ( x - 1)  = 0

Set both factors = 0   and solve for x........so

x = -3           and      x = 1 

And when x = -3, y = -3 + 3  = 0

And when x = 1, y = 1 + 3  = 4

So......the intersection points are   ( -3, 0)   and   ( 1, 4)