Solve x³+3xy²=-49 and x²-8xy+y²=8y-17x

Solve for real solutions for the following system of equations:

$x^3+3xy^2=-49\\x^2-8xy+y^2=8y-17x$

My solution:

From $x^2-8xy+y^2=8y-17x$, we do some algebraic manipulation to get

\[ (y-4x)^2-16x^2+x^2=8y-17x\\(y-4x)^2=8y-17x+15x^2-(1) \]

\[ x^2-8xy+y^2=8y-17x\\8xy=x^2+y^2-8y+17x-(2) \]

From $x^3+3xy^2=-49$, in trying to relate the $y$ term to $y-4x$, we get

\[ x^3+3xy^2=-49\\x^3+3x(y-4x+4x)^2=-49\\x^3+3x[(y-4x)^2+8x(y-4x)+16x^2]=-49\\x^3+3x(y-4x)^2+24x^2(y-4x)+48x^2=-49\\49(x^3+1)+3x[(y-4x)^2+8x(y-4x)]=0\\49(x^3+1)+3x(8y-17x+15x^2+8xy-32x^2)=0\text{   from (1)}\\49(x+1)(x^2-x+1)+3x[8y(x+1)-17x(x+1)]=0\\(x+1)[49(x^2-x+1)+3x(8y-17x)]=0\\(x+1)[49x^2-49x+49+24xy-51x^2]=0\\(x+1)[-2x^2-49x+49+3(8xy)]=0\\(x+1)[-2x^2-49x+49+3x^2+3y^2-24y+51x]=0\text{   from (2)}\\(x+1)(x^2+2x+1+3y^2-24y+48)=0\\(x+1)[(x+1)^2+3(y-4)^2]=0 \]

This is possible if and only if $x=-1$.

When $x=-1$, we get

\[ -1-3y^2=-49\\y^2=16\\y=\pm 4 \]

 

Find the sum \(a+b+c+d\).

Consider the following:

[MATH]f(x)=x^2-rax-sb[/MATH]

which has zeros \(c,d\).

[MATH]g(x)=x^2-rcx-sd[/MATH]

which has zeros \(a,b\).

Given \(f(x)\ne g(x)\), find the sum \(a+b+c+d\) in terms of \(r\) and \(s\) only.

Mark's solution:

Consider the following:

[MATH]f(x)=x^2-rax-sb[/MATH]

which has zeros \(c,d\).

[MATH]g(x)=x^2-rcx-sd[/MATH]

which has zeros \(a,b\).

Given \(f(x)\ne g(x)\), find the sum \(a+b+c+d\) in terms of \(r\) and \(s\) only.

We know:

[MATH]c+d=ra[/MATH]

[MATH]a+b=rc[/MATH]

Hence:

[MATH]a+b+c+d=r(a+c)[/MATH]

We also know:

[MATH]c^2-sb=a^2-sd[/MATH]

[MATH]sd-sb=a^2-c^2[/MATH]

[MATH]s(d-b)=(a+c)(a-c)[/MATH]

And we obtain by subtraction of the first 2 equations:

[MATH](d-b)-(a-c)=r(a-c)[/MATH]

Or:

[MATH]d-b=(r+1)(a-c)[/MATH]

Hence:

[MATH]s(r+1)(a-c)=(a+c)(a-c)[/MATH]

Since \(a\ne c\) (otherwise \(d=b\) and the two quadratics are identical) there results:

[MATH]a+c=s(r+1)[/MATH]

Hence:

[MATH]a+b+c+d=rs(r+1)[/MATH]

We know:

[MATH]c+d=ra[/MATH]

[MATH]a+b=rc[/MATH]

Hence:

[MATH]a+b+c+d=r(a+c)[/MATH]

We also know:

[MATH]c^2-sb=a^2-sd[/MATH]

[MATH]sd-sb=a^2-c^2[/MATH]

[MATH]s(d-b)=(a+c)(a-c)[/MATH]

And we obtain by subtraction of the first 2 equations:

[MATH](d-b)-(a-c)=r(a-c)[/MATH]

Or:

[MATH]d-b=(r+1)(a-c)[/MATH]

Hence:

[MATH]s(r+1)(a-c)=(a+c)(a-c)[/MATH]

Since \(a\ne c\) (otherwise \(d=b\) and the two quadratics are identical) there results:

[MATH]a+c=s(r+1)[/MATH]

Hence:

[MATH]a+b+c+d=rs(r+1)[/MATH]

We know:

[MATH]c+d=ra[/MATH]

[MATH]a+b=rc[/MATH]

Hence:

[MATH]a+b+c+d=r(a+c)[/MATH]

We also know:

[MATH]c^2-sb=a^2-sd[/MATH]

[MATH]sd-sb=a^2-c^2[/MATH]

[MATH]s(d-b)=(a+c)(a-c)[/MATH]

And we obtain by subtraction of the first 2 equations:

[MATH](d-b)-(a-c)=r(a-c)[/MATH]

Or:

[MATH]d-b=(r+1)(a-c)[/MATH]

Hence:

[MATH]s(r+1)(a-c)=(a+c)(a-c)[/MATH]

Since \(a\ne c\) (otherwise \(d=b\) and the two quadratics are identical) there results:

[MATH]a+c=s(r+1)[/MATH]

Hence:

[MATH]a+b+c+d=rs(r+1)[/MATH]

Minimize [MATH]\prod_{i=1}^{2017}x_i[/MATH].

If $x_1,x_2,...,x_{2017} \in\Bbb{R}^+$ and [MATH]\dfrac{1}{1+x_1}+\dfrac{1}{1+x_2}+...+\dfrac{1}{1+x_{2017}} = 1[/MATH], find the minimal possible value of [MATH]\prod_{i=1}^{2017}x_i[/MATH].

Solution:

By cyclic symmetry, we know the critical value is at the point:

[MATH]\left(x_1,\cdots,x_{2017}\right)=(2016,\cdots,2016)[/MATH]

And the objection function at that point is:

[MATH]f(2016,\cdots,2016)=2016^{2017}[/MATH]

Now, looking at another point on the constraint:

[MATH]\left(4032,4032,\cdots,4032,\frac{2016}{2017}\right)[/MATH]

We find the objective function at that point is:

[MATH]f\left(4032,4032,\cdots,4032,\frac{2016}{2017}\right)=\frac{2^{2016}2016^{2017}}{2017}>2016^{2017}[/MATH]

And so we conclude:

[MATH]f_{\min}=2016^{2017}[/MATH]

 

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