The discriminantEdexcel International A Level Maths: Revision notes

The solutions of $ax^2+bx+c=0$ are given by the quadratic formula $x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}.$ The expression under the square root, $b^2-4ac$, is the discriminant. Its value decides whether the square root exists and whether the two values from $\pm$ are different, so it tells you how many real roots there are without solving the equation. Example: for $x^2-4x+7=0$, $a=1$, $b=-4$, $c=7$, so $b^2-4ac=16-28=-12$.

• $b^2-4ac>0$: two distinct real roots.
• $b^2-4ac=0$: one repeated real root (two equal roots), $x=-\frac{b}{2a}$.
• $b^2-4ac<0$: no real roots, because the square root of a negative number is not real. Example: $x^2+6x+9=0$ has $b^2-4ac=36-36=0$, so it has a repeated root $x=-3$, and indeed $(x+3)^2=0$. Use the discriminant to decide, but give the conclusion in words, for example 'the discriminant is negative so there are no real roots'.

The roots are where the graph of $y=ax^2+bx+c$ meets the $x$-axis.

• $b^2-4ac>0$: the graph crosses the $x$-axis at two points.
• $b^2-4ac=0$: the graph touches the $x$-axis at its vertex.
• $b^2-4ac<0$: the graph does not meet the $x$-axis. When there are no real roots the sign of $a$ decides the position: if $a>0$ the whole graph is above the axis ($y>0$ for all $x$), and if $a<0$ it is below ($y<0$ for all $x$). For $y=-x^2+2x-5$: $b^2-4ac=4-20=-16<0$ and $a<0$, so the graph lies entirely below the $x$-axis.

Exam questions give a condition on the roots and ask for an unknown constant. Translate the words into an (in)equality for $b^2-4ac$:

• equal or repeated roots: $b^2-4ac=0$;
• two distinct real roots: $b^2-4ac>0$;
• no real roots: $b^2-4ac<0$;
• real roots (distinct or equal): $b^2-4ac\ge0$. Example: $x^2+kx+9=0$ has equal roots, so $k^2-36=0$ and $k=\pm6$. If $k$ is positive then $k=6$, and the equation is $(x+3)^2=0$. Example: $mx^2-4x+1=0$ has no real roots when $16-4m<0$, so $m>4$.

Sometimes the condition on $b^2-4ac$ is itself a quadratic inequality in the unknown constant. Solve it by finding the critical values and sketching. Example: $f(x)=kx^2+4x+(k-3)$, $k\neq0$, has two distinct real roots when $16-4k(k-3)>0$, which simplifies to $k^2-3k-4<0$, that is $(k-4)(k+1)<0$. The graph of $k^2-3k-4$ is $\cup$-shaped, so it is negative between its roots: $-1<k<4$ (with $k\neq0$). Check with a value: $k=2$ gives $16-4(2)(-1)=24>0$, but $k=5$ gives $16-4(5)(2)=-24<0$ (no real roots).

To see how a line meets a curve, substitute the line into the curve to get a quadratic in $x$, then use its discriminant.

• $b^2-4ac>0$: the line cuts the curve at two points.
• $b^2-4ac=0$: the line is a tangent, touching the curve at one point.
• $b^2-4ac<0$: the line and curve do not meet. Example: $y=2x+c$ and $y=x^2-3x+4$ give $x^2-5x+(4-c)=0$ with discriminant $25-4(4-c)=4c+9$. For a tangent $c=-\frac94$; the repeated root is $x=-\frac{b}{2a}=\frac52$ and $y=\frac{11}{4}$. If $c>-\frac94$ the line cuts the curve twice.