08Esimeqns

= Simultaneous Equations including Quadratics =

To solve simultaneous equations means
 * Recall **
 * to find the coordinates of the point (or points) where two equations intersect
 * at those coordinates, both equations are true.

If we solve a pair of equations where one is a quadratic and the other is linear, three things could happen:
 * The straight line could intersect the parabola twice ( see line **AB** )


 * The straight line could intersect the parabola once ( see line **CD** )
 * ** CD ** is said to be a ** tangent ** to the parabola because it touches in one place. The gradient of ** CD ** is the same as the gradient of the parabola at the point where it touches.


 * The straight line could not intersect the parabola at all ( see line **EF** )

Therefore, solving equations simultaneously could result in
 * two real solutions
 * one real solutions
 * no real solutions


 * Example 1 **

Solve the following equations simultaneously
 * y = x 2 + 2x + 2 ... ** [1] **
 * y = 7 – 2x ... ... ... ** [2] **


 * Solution: **

{Make ** [1] = [2] ** }

math \\ . \qquad x^2 + 2x + 2 = 7 - 2x \qquad \qquad \big\{ \text{Rearrange to make RHS = 0} \big\} \\. \\ . \qquad x^2 + 4x - 5 = 0 \qquad \qquad \qquad \big\{ \text{Factorise} \big\} \\. \\ . \qquad \big( x - 1 \big) \big( x + 5 \big) = 0 \qquad \qquad \quad \big\{ \text{Use NFL} \big\} \\. \\ . \qquad x - 1 = 0 \quad \text{or} \quad x + 5 = 0 \\. \\ . \qquad \quad x = 1 \;\; \quad \text{or} \quad \;\; x = -5 math

{sub ** x = 1 ** into ** [2] ** }

math . \qquad y = 7 - 2(1) = 5 \qquad \text{Hence one solution is } \; \big( 1, \; 5 \big) math

{sub ** x = –5 ** into ** [2] ** }

math . \qquad y = 7 - 2(-5) = 17 \qquad \text{Hence one solution is } \; \big( -5, \; 17 \big) math


 * Confirm Solutions **

{sub ** (1, 5) ** into ** [1] ** }

math \\ . \qquad y = x^2 + 2x + 2 \\. \\ . \qquad 5 = (1)^2 + 2(1) + 2 \\. \\ . \qquad 5 = 1 + 2 + 2 \\. \\ . \qquad 5 = 5 math

{sub ** (–5, 17) ** into ** [1] ** }

math \\ . \qquad y = x^2 + 2x + 2 \\. \\ . \qquad 17 = (-5)^2 + 2(-5) + 2 \\. \\ . \qquad 17 = 25 - 10 + 2 \\. \\ . \qquad 17 = 17 math

Hence the solutions are:
 * ** (1, 5) **
 * ** (–5, 17) **

{The straight line, y = 7 – 2x, intersects the parabola, y = x 2 + 2x + 2 , in two places}


 * Example 2 **

Solve the following equations simultaneously
 * y = x 2 + x + 4 ... ... ** [1] **
 * y = 2x – 1 ... ... ... . ** [2] **


 * Solution: **

{Make ** [1] = [2] ** }

math \\ . \qquad x^2 + x + 4 = 2x - 1 \qquad \qquad \big\{ \text{Rearrange to make RHS = 0} \big\} \\. \\ . \qquad x^2 - x + 5 = 0 \qquad \qquad \qquad \big\{ \text{Doesn't factorise so use Quadratic Formula} \big\} math . math \\ . \qquad a = 1 \quad b = -1 \quad c = 5 \\. \\ . \qquad x = \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a} \\. \\ . \qquad \;\; = \dfrac{-(-1) \pm \sqrt{(-1)^2 - 4(1)(5)}}{2(1)} \\. \\ . \qquad \;\; = \dfrac{1 \pm \sqrt{1 - 20}}{2} \\. \\ . \qquad \;\; = \dfrac{1 \pm \sqrt{-19}}{2} math . math \Delta < 0 \; \text{ hence there are no solutions} math

{The straight line, y = 2x – 1, does not intersect the parabola, y = x 2 + x + 4 }

.