Problems

2013 Paper 2 Q3

D: 1600.0 B: 1500.0

polynomials cubic roots of polynomials curve sketching turning points Vieta's formulas inequality sign analysis

Given that the cubic equation $x^3+3ax^2 + 3bx +c=0$ has three distinct real roots and $c<0$, show with the help of sketches that either exactly one of the roots is positive or all three of the roots are positive.
Given that the equation $x^3 +3ax^2+3bx+c=0$ has three distinct real positive roots show that \begin{equation*} a^2>b>0, \ \ \ \ a<0, \ \ \ \ c<0\,. \tag{$*$} \end{equation*} [Hint: Consider the turning points.]
Given that the equation $x^3 +3ax^2+3bx+c=0$ has three distinct real roots and that \begin{equation*} ab<0, \ \ \ \ c>0\,, \end{equation*} determine, with the help of sketches, the signs of the roots.
Show by means of an explicit example (giving values for $a$, $b$ and $c$) that it is possible for the conditions ($*$) to be satisfied even though the corresponding cubic equation has only one real root.

Solution:

First notice that this cubic has leading first term $1$ and three real roots, so it must have the shape:

With the $x$-axis running somewhere between the dashed lines. Since $c < 0$, the $y$-axis must meet the curve below the $x$-axis, ie somewhere on the blue section of this curve:

Therefore there will be either $1$ (if it meets it in the $\cup$ area) or $3$ (if it meets it on the far left) positive roots.
First notice that if $c > 0$ we cannot have three positive real roots since the function would need to pass $0$ between $0$ and $-\infty$. Secondly, notice both turning points must be larger than zero, ie \begin{align*} && 0 &= 3x^2 + 6ax + 3b \\ \Leftrightarrow && 0 &= (x+a)^2 + b - a^2 \end{align*} has both roots larger than zero, (and it needs to have two roots, so $a^2 > b$ and $-a > 0$, ie $a < 0$. If $b < 0$, then just looking at $x^2+2ax+b$ we can see that it is $<0$ at $0$ and one of the roots will be negative, therefore $c < 0$, $a^2 > b > 0$ and $a < 0$
Since $c > 0$ we can see that at least one root is negative.

ie the $y$-axis passes through an orange section of this curve. What now matters is where the larger turning point is. Considering $x^2 + 2ax + b$, we notice that $ab < 0$ means that $(x-\alpha)(x-\beta)$ we must have $(\alpha + \beta)\alpha \beta > 0$ which isn't possible if both roots are negative. Therefore the $y$-axis passes through the orange $\cap$ and there are $2$ positive real roots.
If we take $a = 1, b = -1, c = 1$ then we have $x^3 + 3x^2-3x+1$. This has turning points when $x^2+2x-1 = 0$, ie $x = -1 \pm \sqrt{2}$ Notice that \begin{align*} && y(-1\pm \sqrt2) &= (-1 \pm \sqrt{2})^3 + 3(-1 \pm \sqrt{2})^2-3(-1 \pm \sqrt{2}) + 1 \\ &&&= (-1\pm \sqrt{2}) \cdot (3 \mp 2\sqrt2) + 3(3 \mp \sqrt2) -3(-1\pm \sqrt2) + 1 \\ &&&= (-7 \pm 5 \sqrt2) + (9 \mp 3\sqrt2) +(3 \mp 3\sqrt2) + 1 \\ &&&= 24 \mp 16\sqrt2 = 8(3 \mp 2 \sqrt2) >0 \end{align*} ie both turning points are above zero and hence only one real root