Solving quadratic equationsIB MYP Maths Standard: Revision notes

A quadratic equation can be written in the form $ax^2+bx+c=0$ where $a\neq0$. Its solutions are also called roots. A quadratic can have two, one or no real solutions. Many methods rely on the zero-product rule: if $A\times B=0$ then $A=0$ or $B=0$. This is why we must get the equation equal to $0$ before we factorise.

To solve by factorising: (1) rearrange to $=0$, (2) factorise, (3) set each bracket equal to $0$. Example: $x^2-5x-14=0$. Two numbers that multiply to $-14$ and add to $-5$ are $-7$ and $2$, so $(x-7)(x+2)=0$ and $x=7$ or $x=-2$. If there is no constant term, take out a common factor: $x^2-4x=0$ gives $x(x-4)=0$, so $x=0$ or $x=4$. Also watch for the difference of two squares: $x^2-25=(x-5)(x+5)$. When $a\neq1$, e.g. $2x^2+3x-5=0$, split the middle term: $(2x+5)(x-1)=0$, so $x=-\frac52$ or $x=1$.

Some quadratics do not factorise neatly. For $ax^2+bx+c=0$ the quadratic formula gives the solutions: $x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$ Example: $3x^2-4x-5=0$ has $a=3$, $b=-4$, $c=-5$. Then $x=\frac{4\pm\sqrt{16+60}}{6}=\frac{4\pm\sqrt{76}}{6}$, so $x=2.12$ or $x=-0.786$ (3 s.f.). Use brackets when $b$ or $c$ is negative so that $b^2$ is always positive.

The expression under the square root, $b^2-4ac$, is the discriminant. It tells you how many solutions there are without solving:

• $b^2-4ac>0$: two different solutions
• $b^2-4ac=0$: one solution (a repeated root, $x=-\frac{b}{2a}$)
• $b^2-4ac<0$: no real solutions Example: $2x^2-3x+5=0$ has $9-40=-31<0$, so no real solutions.

The solutions of $ax^2+bx+c=0$ are the $x$-coordinates where the graph of $y=ax^2+bx+c$ crosses the $x$-axis. Use a graphing calculator or graphing software to draw the curve and find these roots. Two crossing points mean two solutions, a curve that just touches the axis means one solution, and a curve that never reaches the axis means none. To solve $x^2-3x=4$ graphically, plot $y=x^2-3x$ and $y=4$ and read the $x$-values where they meet.

To model a situation: define the unknown, write an equation from the information (area, product, height), rearrange to $=0$ and solve. Example: a rectangle has length $x+3$ and width $x$, and area $54$ m$^2$. Then $x(x+3)=54$, so $x^2+3x-54=0$ and $(x+9)(x-6)=0$. Reject $x=-9$ because a length cannot be negative, so the width is $6$ m. Always interpret each answer in context and reject any that are impossible (negative lengths or times) before you write your conclusion.