Name: Number Theory: Diophantine Equation
Uploaded: 2019-07-19
Duration: 9 min 43 s
Description: Once you learn to solve Diophantine equations with a single variable, the next step is to consider equations with two variables. Watch and learn about Diophantine equations.

Question

Suppose

a

and

b

are two positive real numbers such that the roots of the cubic equation

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

are all real. If

α

is a root of this cubic with minimal absolute value prove that

\frac{b}{a} < α \leq \frac{3 b}{2 a}

.

Accepted Answer

Let $α, β, γ$ be the roots of the given cubic $x^{3} - a x + b = 0$ , where $a > 0$ and $b > 0$ . Then

$\begin{matrix} α + β + γ = 0 \\ α β + β γ + γ α = - a \\ α β γ = - b \end{matrix}\} . . . . . . (1)$

By Descartes rule of sign change, we ca conclude that two roots must be positive and one root must be negative.

So, two of $α, β, γ$ are positive and the remaining root is negative. The first equation in $(1)$ implies that the negative root is numerically larger than the other two positive roots. Hence, we may assume that $γ < 0 < α \leq β$ where $|α| \leq |β| \leq |γ|$ .

Now, we have

$b - a α = - α β γ + α (α β + β γ + γ α)$

$\Rightarrow b - a α = α^{2} (β + γ) = - α^{3} < 0$

Since, $a$ is positive, we get $\frac{b}{a} < α$ proving the first inequality.

Again, we have

$3 b - 2 a α = - 3 α β γ + 2 α (α β + β γ + γ α)$

$= - α β γ + 2 α^{2} β + 2 α^{2} γ$

$= α [2 α (β + γ) - β γ]$

$= α [- 2 {(β + γ)}^{2} - β γ]$

(since $α = - (β + γ)$ )

$= - α (2 β^{2} + 5 β γ + 2 γ^{2})$

$= - α (2 β + γ) (β + 2 γ)$

$= - α (β - α) (γ - α)$

Observe that $- α < 0, β \geq α, γ - α < 0$ .

Hence, $3 b - 2 a α$ is non-negative. This proves the second inequality, $α \leq \frac{3 b}{2 a}$ .

Suppose $a$ and $b$ are two positive real numbers such that the roots of the cubic equation $x^{3} - a x + b = 0$ are all real. If $α$ is a root of this cubic with minimal absolute value prove that $\frac{b}{a} < α \leq \frac{3 b}{2 a}$ .

Important Questions on Algebra

Learn and Prepare for any exam you want !

Suppose a and b are two positive real numbers such that the roots of the cubic equation x3-ax+b=0 are all real. If α is a root of this cubic with minimal absolute value prove that ba<α≤3b2a.

Important Questions on Algebra

Learn and Prepare for any exam you want !

Suppose $a$ and $b$ are two positive real numbers such that the roots of the cubic equation $x^{3} - a x + b = 0$ are all real. If $α$ is a root of this cubic with minimal absolute value prove that $\frac{b}{a} < α \leq \frac{3 b}{2 a}$ .