2.7 Quadratic equations, inequalities and the discriminantIB Maths: Analysis and Approaches SL: Revision notes

Write the equation as $ax^{2} + bx + c = 0$, factorise, then use the fact that if $pq = 0$ then $p = 0$ or $q = 0$.

$2x^{2} - 5x - 3 = 0 \Rightarrow (2x + 1)(x - 3) = 0 \Rightarrow x = -\tfrac{1}{2} \text{ or } x = 3.$ The solutions are also called the roots of the equation or the zeros of the function $f(x) = 2x^{2} - 5x - 3$. They are the $x$-intercepts of its graph.

Always rearrange to $= 0$ first: $x^{2} = 5x$ gives $x(x - 5) = 0$, so $x = 0$ or $x = 5$.

Rewrite $x^{2} + bx$ as $\left(x + \frac{b}{2}\right)^{2} - \left(\frac{b}{2}\right)^{2}$. For example $x^{2} - 6x + 14 = (x - 3)^{2} - 9 + 14 = (x - 3)^{2} + 5.$ To solve $x^{2} - 6x + 4 = 0$: $(x - 3)^{2} - 5 = 0$, so $x - 3 = \pm\sqrt{5}$ and $x = 3 \pm \sqrt{5}$.

Completing the square also gives the vertex $(h, k)$ of $y = a(x - h)^{2} + k$, and shows at once whether the graph can cross the $x$-axis: $(x - 3)^{2} + 5 \ge 5$ for all $x$.

For $ax^{2} + bx + c = 0$ with $a \ne 0$, $x = \frac{-b \pm \sqrt{b^{2} - 4ac}}{2a}.$ It is in the formula booklet and works for every quadratic, including those that do not factorise. For $x^{2} - 4x - 1 = 0$: $x = \frac{4 \pm \sqrt{16 + 4}}{2} = \frac{4 \pm 2\sqrt{5}}{2} = 2 \pm \sqrt{5}$.

On Paper 1 give exact answers (surds); on Paper 2 a GDC may be used and answers are given to 3 significant figures.

The discriminant is $\Delta = b^{2} - 4ac$, the expression under the square root in the formula.

• $\Delta > 0$: two distinct real roots (the graph crosses the $x$-axis twice).
• $\Delta = 0$: two equal real roots (a repeated root; the graph touches the $x$-axis at its vertex).
• $\Delta < 0$: no real roots (the graph does not meet the $x$-axis).

Example: for $3kx^{2} + 2x + k = 0$ (with $k \ne 0$ so that it is a quadratic), $\Delta = 4 - 12k^{2}$. Two distinct roots when $k^{2} < \frac{1}{3}$, i.e. $-\frac{1}{\sqrt{3}} < k < \frac{1}{\sqrt{3}}$, $k \ne 0$; equal roots when $k = \pm\frac{1}{\sqrt{3}}$; no real roots when $k < -\frac{1}{\sqrt{3}}$ or $k > \frac{1}{\sqrt{3}}$.

To solve $f(x) < 0$ or $f(x) > 0$ for a quadratic $f$:

1. Solve $f(x) = 0$ to find the critical values.
2. Use the shape: if $a > 0$ the parabola opens upwards, so $f(x) < 0$ between the roots and $f(x) > 0$ outside them.

$2x^{2} - 5x - 3 < 0 \Rightarrow -\frac{1}{2} < x < 3$, while $2x^{2} - 5x - 3 \ge 0 \Rightarrow x \le -\frac{1}{2}$ or $x \ge 3$.

The same method solves inequalities in a parameter, e.g. $k^{2} - 4k - 12 < 0 \Rightarrow -2 < k < 6$.

Where the line $y = mx + c$ meets a quadratic curve, equate the two expressions to get a quadratic in $x$. Its discriminant counts the intersection points: $\Delta > 0$ two points, $\Delta = 0$ the line is a tangent, $\Delta < 0$ no intersection.

In context, the discriminant finds greatest values: if an area $A$ satisfies $2x^{2} - 40x + A = 0$, then real lengths exist only when $1600 - 8A \ge 0$, so the largest possible area is $200$.