Loading [MathJax]/jax/element/mml/optable/BasicLatin.js

Sunday, October 11, 2020

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

Solve for real solutions for the following system of equations:

x3+3xy2=49x28xy+y2=8y17x

My solution:

From x28xy+y2=8y17x, we do some algebraic manipulation to get

(y4x)216x2+x2=8y17x(y4x)2=8y17x+15x2(1)

x28xy+y2=8y17x8xy=x2+y28y+17x(2)

From x3+3xy2=49, in trying to relate the y term to y4x, we get

x3+3xy2=49x3+3x(y4x+4x)2=49x3+3x[(y4x)2+8x(y4x)+16x2]=49x3+3x(y4x)2+24x2(y4x)+48x2=4949(x3+1)+3x[(y4x)2+8x(y4x)]=049(x3+1)+3x(8y17x+15x2+8xy32x2)=0 from (1)49(x+1)(x2x+1)+3x[8y(x+1)17x(x+1)]=0(x+1)[49(x2x+1)+3x(8y17x)]=0(x+1)[49x249x+49+24xy51x2]=0(x+1)[2x249x+49+3(8xy)]=0(x+1)[2x249x+49+3x2+3y224y+51x]=0 from (2)(x+1)(x2+2x+1+3y224y+48)=0(x+1)[(x+1)2+3(y4)2]=0

This is possible if and only if x=1.

When x=1, we get

13y2=49y2=16y=±4

 

Thursday, July 19, 2018

Find the sum a+b+c+d.

Consider the following:

f(x)=x2raxsb

which has zeros c,d.

g(x)=x2rcxsd

which has zeros a,b.

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

Mark's solution:
Consider the following:

f(x)=x2raxsb

which has zeros c,d.

g(x)=x2rcxsd

which has zeros a,b.

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

We know:

c+d=ra

a+b=rc

Hence:

a+b+c+d=r(a+c)

We also know:

c2sb=a2sd

sdsb=a2c2

s(db)=(a+c)(ac)

And we obtain by subtraction of the first 2 equations:

(db)(ac)=r(ac)

Or:

db=(r+1)(ac)

Hence:

s(r+1)(ac)=(a+c)(ac)

Since ac (otherwise d=b and the two quadratics are identical) there results:

a+c=s(r+1)

Hence:

a+b+c+d=rs(r+1)
We know:

c+d=ra

a+b=rc

Hence:

a+b+c+d=r(a+c)

We also know:

c2sb=a2sd

sdsb=a2c2

s(db)=(a+c)(ac)

And we obtain by subtraction of the first 2 equations:

(db)(ac)=r(ac)

Or:

db=(r+1)(ac)

Hence:

s(r+1)(ac)=(a+c)(ac)

Since ac (otherwise d=b and the two quadratics are identical) there results:

a+c=s(r+1)

Hence:

a+b+c+d=rs(r+1)
We know:

c+d=ra

a+b=rc

Hence:

a+b+c+d=r(a+c)

We also know:

c2sb=a2sd

sdsb=a2c2

s(db)=(a+c)(ac)

And we obtain by subtraction of the first 2 equations:

(db)(ac)=r(ac)

Or:

db=(r+1)(ac)

Hence:

s(r+1)(ac)=(a+c)(ac)

Since ac (otherwise d=b and the two quadratics are identical) there results:

a+c=s(r+1)

Hence:

a+b+c+d=rs(r+1)

Sunday, August 13, 2017

Minimize 2017i=1xi.



If x1,x2,...,x2017R+ and 11+x1+11+x2+...+11+x2017=1, find the minimal possible value of 2017i=1xi.


Solution:

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

(x1,,x2017)=(2016,,2016)

And the objection function at that point is:

f(2016,,2016)=20162017

Now, looking at another point on the constraint:

(4032,4032,,4032,20162017)

We find the objective function at that point is:

f(4032,4032,,4032,20162017)=22016201620172017>20162017

And so we conclude:

fmin
 

Monday, December 19, 2016

You can find ❤️ in ...... Math!