Deciphering Quadratic Simultaneous Equations

What is a Quadratic Simultaneous Equation?

A quadratic simultaneous equation involves two equations:

1. A linear equation (no powers of $x$ or $y$ greater than 1).
2. A quadratic equation (one variable, typically $x$, has a power of 2).

For example: $$y = 2x + 1 \\ y = x^2 - 3x + 2$$

To solve these, we find the values of $x$ and $y$ that satisfy both equations simultaneously.

 

How to Solve Quadratic Simultaneous Equations

1. Substitution Method

1. Step 1: Rearrange the linear equation to express $y$ (or $x$)
2. Step 2: Substitute this expression into the quadratic equation.
3. Step 3: Solve the resulting quadratic equation.
4. Step 4: Use the $x$-values to find the corresponding $y$-values

 

Examples

Example 1

Solve the simultaneous equations: $$y = 2x + 1 \\ y = x^2 - 3x + 2$$

Solution:

1. Substitute $y = 2x + 1$ into the quadratic equation $y = x^2 - 3x + 2$: $$2x + 1 = x^2 - 3x + 2$$
   

2. Rearrange into standard quadratic form $ax^2 + bx + c = 0$: Subtract $2x + 1$ from both sides: $$0 = x^2 - 3x - 2x + 2 - 1 \\ \text{Simplify: } \\ x^2 - 5x + 1 = 0$$
3. Solve the quadratic equation using the quadratic formula: The quadratic formula is: $$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \\ \text{For } x^2 - 5x + 1 = 0, a = 1, b = -5, \text{ and } c = 1. \text{Substitute these values:} \\  x = \frac{-(-5) \pm \sqrt{(-5)^2 - 4(1)(1)}}{2(1)} \\ \text{Simplify: } \ x = \frac{5 \pm \sqrt{25 - 4}}{2} \\ x = \frac{5 \pm \sqrt{21}}{2} \\ \text{Approximate: } \ x = \frac{5 + \sqrt{21}}{2} \quad \text{or} \quad x = \frac{5 - \sqrt{21}}{2}$$
4. Find $y$ for each $x$: Substitute $x$ back into $y = 2x + 1$:
   • For $x = \frac{5 + \sqrt{21}}{2}$: $$y = 2\left(\frac{5 + \sqrt{21}}{2}\right) + 1 \\  y = 5 + \sqrt{21} + 1 \\ y = 6 + \sqrt{21}$$
   • For $x = \frac{5 - \sqrt{21}}{2}$: $$y = 2\left(\frac{5 - \sqrt{21}}{2}\right) + 1 \\  y = 5 - \sqrt{21} + 1 \\ y = 6 - \sqrt{21}$$

Final Solutions: $$x = \frac{5 + \sqrt{21}}{2}, \, y = 6 + \sqrt{21} \\ x = \frac{5 - \sqrt{21}}{2}, \, y = 6 - \sqrt{21}$$

 

 

 

 

 

Example 2

Solve the simultaneous equations: $$y = 3x - 4 \\ x^2 + y^2 = 25$$

Solution:

1. Substitute $y = 3x - 4$ into $x^2 + y^2 = 25$: $$x^2 + (3x - 4)^2 = 25$$
   

2. Expand $(3x - 4)^2$: $$x^2 + (3x - 4)^2 = 25 \\ \text{Simplify: } \\ 10x^2 - 24x + 16 = 25$$
   
   
3. Rearrange into standard quadratic form: $$10x^2 - 24x - 9 = 0$$
   
4. Solve using the quadratic formula: For $10x^2 - 24x - 9 = 0, a = 10, b = -24, \text{and} c = -9$. Use the quadratic formula: $$x = \frac{-(-24) \pm \sqrt{(-24)^2 - 4(10)(-9)}}{2(10)} \\ \text{Simplify: } \\ x = \frac{24 \pm \sqrt{576 + 360}}{20} \\  x = \frac{24 \pm \sqrt{936}}{20} \\ \text{Approximate: } \\ x = \frac{24 + 30.6}{20} \quad \text{or} \quad x = \frac{24 - 30.6}{20} \\  x = 2.73 \quad \text{or} \quad x = -0.33$$
5. Find $y$ for each $x$: Substitute $x$ back into $y = 3x - 4$: 
   • For $x = 2.73$: $$y = 3(2.73) - 4 = 8.19 - 4 = 4.19$$
   • For $x = -0.33$: $$y = 3(-0.33) - 4 = -0.99 - 4 = -4.99$$

Final Solutions: $$x = 2.73, \, y = 4.19 \\ x = -0.33, \, y = -4.99$$