Quadratic functions and the discriminantEdexcel A-Level Maths: Revision notes

A quadratic function has the form $f(x)=ax^2+bx+c$ with $a\neq0$. Its graph is a parabola: U-shaped if $a>0$ (minimum) and n-shaped if $a<0$ (maximum). Key features to find when sketching:

• $y$-intercept: $(0,c)$.
• Roots (where it crosses the $x$-axis): solve $f(x)=0$.
• Turning point: on the line of symmetry $x=-\frac{b}{2a}$, found by completing the square. Solving $f(x)=0$ can be done by factorising, completing the square, the formula or a calculator.

$ax^2+bx+c=a\left(x+\frac{b}{2a}\right)^2+c-\frac{b^2}{4a}.$The turning point is at $\left(-\frac{b}{2a},\ c-\frac{b^2}{4a}\right)$. Example: $f(x)=2x^2-12x+7=2(x^2-6x)+7=2\left[(x-3)^2-9\right]+7=2(x-3)^2-11$. The minimum value is $-11$ at $x=3$, so the turning point is $(3,-11)$. To solve: $2(x-3)^2=11$, so $x=3\pm\sqrt{\frac{11}{2}}$. The form also proves positivity: $n^2-6n+10=(n-3)^2+1\geqslant1$.

For $ax^2+bx+c=0$ the discriminant is $b^2-4ac$. It is the part under the root in $x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$.

• $b^2-4ac>0$: two distinct real roots (graph crosses the $x$-axis twice).
• $b^2-4ac=0$: one repeated root (graph touches the $x$-axis).
• $b^2-4ac<0$: no real roots (graph does not meet the $x$-axis). Example: $x^2+kx+16=0$ has $k^2-64$. Repeated root when $k=\pm8$, no real roots for $-8<k<8$, two real roots for $k<-8$ or $k>8$.

When a coefficient is unknown, set up an inequality in the unknown and solve it. Example: $x^2+px+(p+3)=0$. The discriminant is $p^2-4(p+3)=p^2-4p-12=(p-6)(p+2)$.

• Real roots: $(p-6)(p+2)\geqslant0$, so $p\leqslant-2$ or $p\geqslant6$.
• Equal roots: $p=6$ (root $x=-3$) or $p=-2$ (root $x=1$). Sketch the quadratic in $p$ to read the inequality: 'greater than' means outside the roots, 'less than' means between them.

Some equations are quadratic in $x^2$, $\cos x$, $2^x$ or $\ln x$. Substitute, solve the quadratic, then solve for $x$.

• $x^4-5x^2+4=0$: let $y=x^2$, so $(y-1)(y-4)=0$ and $x=\pm1,\pm2$.
• $2\cos^2x-\cos x-1=0$: $(2\cos x+1)(\cos x-1)=0$, so $\cos x=-\frac12$ or $1$, giving $x=0^\circ,120^\circ,240^\circ$ in $0\leqslant x<360^\circ$.
• $2^{2x+1}-9\times2^x+4=0$: $2y^2-9y+4=0$ with $y=2^x$, so $y=\frac12$ or $4$, giving $x=-1$ or $2$. Reject impossible values: $2^x>0$ and $-1\leqslant\cos x\leqslant1$, so for example $2^x=-3$ has no solution.