Quadratic and non-linear simultaneous equationsIB MYP Maths Extended: Revision notes

To solve a pair such as $y=x+2$ and $y=x^2$ by substitution:

1. Make $y$ the subject of the linear equation (or use the equation that already has $y$ alone).
2. Substitute into the quadratic, or set the two expressions for $y$ equal: $x^2=x+2$.
3. Rearrange to $ax^2+bx+c=0$ and solve by factorising or the formula.
4. Put each $x$ back into the linear equation to find $y$. Give answers as coordinate pairs.

Solve $y=3x-4$ and $y=x^2-4x+8$. Set equal: $x^2-4x+8=3x-4$, so $x^2-7x+12=0$ and $(x-3)(x-4)=0$. Then $x=3$ or $x=4$. Use $y=3x-4$: $y=5$ and $y=8$. The solutions are $(3,5)$ and $(4,8)$. Check in the quadratic: $x=4$ gives $16-16+8=8$. Correct.

Sometimes the linear equation is not in the form $y=\dots$. For $x+y=10$ and $xy=21$, rearrange to $y=10-x$, substitute to get $x(10-x)=21$, which is $x^2-10x+21=0$. Then $x=3$ or $x=7$, giving the pairs $(3,7)$ and $(7,3)$. In a context, think about what each solution means. Both pairs may describe the same shape.

After substitution you have $ax^2+bx+c=0$. The discriminant $b^2-4ac$ shows how the line and the curve meet:

• $b^2-4ac>0$: two solutions, so the line cuts the curve twice.
• $b^2-4ac=0$: one solution, so the line touches the curve (a tangent).
• $b^2-4ac<0$: no solutions, so the line and curve do not meet. Example: $y=2x-2$ and $y=x^2-4x+8$ give $x^2-6x+10=0$, with discriminant $36-40=-4$, so there is no intersection.

The solutions of the pair are the coordinates of the points where the graphs cross. Plot the line and the parabola (or use graphing software) and read the intersection points. Graphical answers are approximate. To solve $f(x)=g(x)$, find the $x$-coordinates of the intersections of $y=f(x)$ and $y=g(x)$. In a model, a ball landing on a slope, a profit equal to a target or a break-even point is an intersection. Reject any solution that is impossible, such as a negative length, and say what the answer means in words.