Algebraic Mess!

Algebra Level 5

The number of integral solution(s) the two inequalities

− 5 < 2 x − 1 < 2 2 α β γ ≤ y ≤ α 3 + β 3 + γ 3 \begin{aligned} -5 < & 2x-1 & < 2 \sqrt{2} \\ \alpha \beta \gamma \leq & y& \leq \alpha^3 + \beta^3 + \gamma^3 \\ \end{aligned}

are p p and q q respectively.

Given that

1 α + 1 β + 1 γ = 1 2 1 α 2 + 1 β 2 + 1 γ 2 = 9 4 α + β + γ = 2 \begin{aligned} \dfrac{1}{\alpha} + \dfrac{1}{\beta} + \dfrac{1}{\gamma} & = & \dfrac{1}{2} \\ \dfrac{1}{\alpha^2} + \dfrac{1}{\beta^2} + \dfrac{1}{\gamma^2} & = & \dfrac{9}{4} \\ \alpha + \beta + \gamma & = & 2 \\ \end{aligned}

What is the value of ∣ p − q ∣ ? \lvert p-q\rvert?

This problem is an extension to the note: Solving Linear Inequalities from the set: Inequalities


The answer is 8.

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Priyansh Sangule by Priyansh Sangule , 24, India

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2 solutions

Aditya Raut
Jul 22, 2014

We see that the first equation is equivalent to − 4 < 2 x < 2 2 + 1 -4<2x<2\sqrt{2}+1 and this gives − 2 < x < 2 -2<x< 2 ( \color{#D61F06}{\biggl(} because 1 < 2 2 + 1 2 < 2 ) \color{#D61F06}{1< \dfrac{2\sqrt{2}+1}{2} < 2 \biggr)} . Hence there will be 3 3 integral values of x x possible.

In the seconds equation, by solving the three equations and three variables, we get that unordered ( α , β , γ ) = ( − 1 , 1 , 2 ) \color{#3D99F6}{\text{unordered}(\alpha,\beta,\gamma) = (-1,1,2)} . (This means that there maybe different permutations, but values will be these only).

Thus the second condition, i.e. for y y is transformed to

α β γ ≤ y ≤ α 3 + β 3 + γ 3 ⟹ − 2 ≤ y ≤ 8 \alpha \beta \gamma \leq y \leq \alpha^3+ \beta^3 + \gamma^3 \implies -2\leq y \leq 8

Hence there are 11 11 integer values of y y possible and

∣ p − q ∣ = ∣ 3 − 11 ∣ = 8 \Huge{\color{#20A900}{|p-q|= |3-11| = \boxed{8}}}

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Since you didn't show how you solved the system of equations, I'm going to show how I did it, maybe it can be useful for someone. Let us think in 1 α , 1 β \frac{1}{\alpha}, \frac{1}{\beta} and 1 γ \frac{1}{ \gamma} as roots of a cubic polynomial: p ( x ) = x 3 + A x 2 + B x + C p(x)=x^{3}+Ax^{2}+Bx+C We know, at first, that: S 1 = 1 2 , S 2 = 9 4 , S − 1 = 2 S_{1}=\frac{1}{2}, S_{2}=\frac{9}{4}, S_{-1}=2 By Newton's and Vieta's relations, we get: − A = 1 2 , − B C = 2 ⇒ B = − 2 C -A=\frac{1}{2}, \frac{-B}{C}=2 \Rightarrow B=-2C S 2 + A S 1 + B S 0 + C S − 1 = 0 S_{2}+AS_{1}+BS_{0}+CS_{-1}=0 9 4 + ( 1 2 ) ( − 1 2 ) + 3 ( − 2 C ) + 2 C = 0 \frac{9}{4}+\left(\frac{1}{2}\right)\left(\frac{-1}{2}\right)+3(-2C)+2C=0 ∴ C = 1 2 , B = − 1 , C = 1 2 \therefore C=\frac{1}{2}, B=-1, C=\frac{1}{2} p ( x ) = x 3 − 1 2 x 2 − x + 1 2 p(x)=x^{3}-\frac{1}{2}x^{2}-x+\frac{1}{2} p ( x ) = x 3 − 1 2 ( x + 1 ) 2 + 1 p(x)=x^{3}-\frac{1}{2}(x+1)^{2}+1 ⇒ p ( x ) = ( x + 1 ) ( x 2 − x + 1 ) − 1 2 ( x + 1 ) 2 \Rightarrow p(x)=(x+1)(x^{2}-x+1)-\frac{1}{2}(x+1)^{2} p ( x ) = ( x + 1 ) ( x 2 − 3 2 x + 1 2 ) p(x)=(x+1)\left(x^{2}-\frac{3}{2}x+\frac{1}{2}\right) p ( x ) = ( x + 1 ) ( x − 1 ) ( x − 1 2 ) p(x)=(x+1)(x-1)\left(x-\frac{1}{2}\right) ∴ 1 α = 1 , 1 β = − 1 1 γ = 1 2 ⇒ α = 1 , β = − 1 , γ = 2. \therefore \frac{1}{\alpha}=1, \frac{1}{\beta}=-1 \frac{1}{\gamma}=\frac{1}{2}\Rightarrow \alpha=1, \beta=-1, \gamma=2.

Dieuler Oliveira - 6 years, 10 months ago
Arif Ahmed
Oct 1, 2014

For the first one : − 2 < x < 1.91 -2<x<1.91 Therefore integral solutions for x are { − 1 , 0 , 1 } \{-1,0,1\}

Finding the individual values of α , β , γ \alpha, \beta ,\gamma is not required , playing around with the three given expressions yields values of α β γ \alpha\beta\gamma and α 3 + β 3 + γ 3 {\alpha }^{3} + {\beta}^{3} + {\gamma}^{3} .

α 3 + β 3 + γ 3 {\alpha }^{3} + {\beta}^{3} + {\gamma}^{3}

= ( α + β + γ ) ( α 2 + β 2 + γ 2 − α β − β γ − γ α ) + 3 α β γ = (\alpha + \beta + \gamma)({\alpha}^{2} + {\beta}^{2} + {\gamma}^{2} - \alpha\beta - \beta\gamma - \gamma\alpha) + 3\alpha\beta\gamma

= 2 ( 4 − 3 ( α β + β γ + γ α ) ) + 3 α β γ = \quad 2(4 - 3(\alpha\beta + \beta\gamma + \gamma\alpha)) + 3\alpha\beta\gamma

= 2 ( 4 − 3 α β γ 2 ) + 3 α β γ =\quad 2(4 - \frac {3\alpha\beta\gamma}{2}) + 3\alpha\beta\gamma

= 8 =\quad 8

Finding α β γ \alpha\beta\gamma : First expression can be written as : α β + β γ + γ α = α β γ 2 \alpha\beta+\beta\gamma+\gamma\alpha = \frac{\alpha\beta\gamma}{2}

Second expression : α 2 β 2 + β 2 γ 2 + γ 2 α 2 = 9 4 α 2 β 2 γ 2 { \alpha }^{ 2 }{ \beta }^{ 2 }+{ \beta }^{ 2 }{ \gamma }^{ 2 }+{ \gamma }^{ 2 }{ \alpha }^{ 2 }=\frac { 9 }{ 4 } { \alpha }^{ 2 }{ \beta }^{ 2 }{ \gamma }^{ 2 }

Squaring both sides of first expression : α 2 β 2 + β 2 γ 2 + γ 2 α 2 + 2 α β γ ( α + β + γ ) = α 2 β 2 γ 2 4 { \alpha }^{ 2 }{ \beta }^{ 2 }+{ \beta }^{ 2 }{ \gamma }^{ 2 }+{ \gamma }^{ 2 }{ \alpha }^{ 2 }+2\alpha \beta \gamma (\alpha +\beta +\gamma )=\frac { { \alpha }^{ 2 }{ \beta }^{ 2 }{ \gamma }^{ 2 } }{ 4 }

Using second equation we eventually get : α β γ = − 2 \alpha\beta\gamma = -2 Therefore , − 2 ≤ y ≤ 8 -2 \le y\le 8 and no. of integral solutions of y is 11 11

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