TMUA 2022 Paper 2 Q17 — Worked Solution

P2·Differentiation, Logic & Quantifiers·7.0Official · 2022 · Q17

proof-errorconversecubic-roots

A student answered the following question:

a

and

b

are non-zero real numbers.Prove that the equation

x^{3} + a x^{2} + b = 0

has three distinct real roots if

27 b (b + \frac{4 a ^{3}}{27}) < 0.

Here is the student's solution:

We differentiate

y = x^{3} + a x^{2} + b

to get

\frac{d y}{d x} = 3 x^{2} + 2 a x = x (3 x + 2 a)

. Solving

\frac{d y}{d x} = 0

shows that the stationary points are at

(0, b)

and

(- \frac{2 a}{3}, b + \frac{4 a ^{3}}{27})

27 b (b + \frac{4 a ^{3}}{27}) < 0

, then

b

and

b + \frac{4 a ^{3}}{27}

must have opposite signs, and so one of the stationary points is above the

x

-axis and one is below.

III

If the cubic has three distinct real roots, then one of the stationary points is above the

x

-axis and one is below.

Hence if

27 b (b + \frac{4 a ^{3}}{27}) < 0

, then the equation has three distinct real roots.

Which one of the following options best describes the student's solution?

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