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
Thank you
+26367 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:
\(\begin{array}{|rcll|} \hline y &=& x + 3 \quad & \quad m=1 \quad b = 3 \\ y &=& x^2+3x \quad & \quad A=1 \quad B = 3 \quad C = 0 \\ x_{\text{intersection}_{1,2}} &=& \dfrac{1-3\pm \sqrt{(1-3)^2-4\cdot 1 \cdot(0-3)} }{2\cdot 1} \\ x_{\text{intersection}_{1,2}} &=& \dfrac{-2\pm \sqrt{4+12} }{2} \\ x_{\text{intersection}_{1,2}} &=& \dfrac{-2\pm 4 }{2} \\ x_{\text{intersection}_{1}} &=& \dfrac{-2 + 4 }{2} \\ &=& 1 \\\\ x_{\text{intersection}_{2}} &=& \dfrac{-2 - 4 }{2} \\ &=& -3 \\\\ y_{\text{intersection}_{1}} &=& 1\cdot x_{\text{intersection}_{1}} + 3 \\ &=& 1\cdot 1 + 3 \\ &=& 4 \\\\ y_{\text{intersection}_{2}} &=& 1\cdot x_{\text{intersection}_{2}} + 3 \\ &=& 1\cdot (-3) + 3 \\ &=& 0 \\ \hline \end{array}\)
+128406
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)
