+0  
 
0
173
2
avatar

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

Guest Apr 11, 2017
Sort: 

2+0 Answers

 #1
avatar+18777 
+2

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:

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

 

laugh

heureka  Apr 11, 2017
 #2
avatar+79894 
+1

 

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)

 

 

 

cool cool cool

CPhill  Apr 11, 2017

5 Online Users

avatar
We use cookies to personalise content and ads, to provide social media features and to analyse our traffic. We also share information about your use of our site with our social media, advertising and analytics partners.  See details