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Solving Hard Systems of Equations | Math Olympiad Techniques

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Solving Hard Systems of Equations | Math Olympiad Techniques
In this algebraic video of advanced mathematics as we tackle hard systems of equations using proven Math Olympiad techniques. This video will guide you through the strategies and methods needed to solve complex systems of equations, perfect for students preparing for math competitions or anyone looking to enhance their problem-solving skills. Join us and master the techniques that can help you excel in the Math Olympiad and beyond!
Topics covered:
How to solve system of equations?
System of equations
Algebra
Math Olympiad
Math Olympiad Training
Algebraic identities
Algebraic manipulations
Solving systems of equations
Math tutorial
Math Olympiad Preparation
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#matholympiad #systemofequations #education #mathenthusiast #mathtutorial #mathematics #mathskills #problemsolving #algebra
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4 сен 2024

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@woobjun2582 2 месяца назад

Known a property, (x +y)⁴ = x⁴ +y⁴ +4x³y +6x²y² +4xy³ that can be expressed in (x +y)⁴ = x⁴ +y⁴ +2xy(2x² +2y² +3xy) From the given eqns (x +y)⁴ =68 +2(128) =324 =(4)(81) =2²3⁴ that is x +y = 3√2 ...(eq1) Then, (x +y)² = x² +y² +2xy; x² +y² = (x +y)² -2xy; x² +y² = (3√2)² -2xy; x² +y² = 18 -2xy, and substituting into the 2nd given eqn, xy[2(18 -2xy) +3xy] =128; xy(36 -4xy +3xy) =128; xy(36 -xy) =128; 36xy -(xy)² =128; (xy)² -36xy +128 =0; (xy)² -4xy -32xy +128 =0; xy(xy -4) -32(xy -4) =0; (xy -4)(xy -32) =0, and thus, xy = 4 or xy = 32 ...(eq2) With eq1 and 2 x +y =3√2 & xy =4 ..(S1) x +y =3√2 & xy =32 ..(S2) Solving S1 x² -(3√2)x +4 =0, x = √2, 2√2 y = 2√2, √2 (x,y) =( √2, 2√2), (2√2, √2) Similarly solving S2 yields complex solutions.

@woobjun2582 2 месяца назад

Sorry I was not careful enough x +y = ±3√2 ...(eq1)

@mohammedsaysrashid3587 2 месяца назад

It was a wonderful introducing...thanks for sharing

@tejpalsingh366 2 месяца назад

Finally ,(X+ y)^ 4= 324 i.e x+y =+ - 3√2 X+ y = 2√2+2; 2+2√2; -2√2-2; -2-2√2; -3√2+0; 0- 3√2 Pair of 0 will be rejected as it makes l.h.s of 2nd eqn o. Rest r solns. Solns are (X;Y)= (2√2; 2); (2; 2√2); (-2; -2√2); ( -2√2; -2)

@user-ny6jf9is3t 2 месяца назад

Καταληγω στα συστηματα: χ+y=3ριζα2 χy=32, χ+y=3ριζα2 χy=4, χ+y=-3ριζα2 χy=32, χy=-3ριζα2 χy=4.etc

@johnstanley5692 Месяц назад

Alternative? let y=a*x => g1=(a^4+1)*x^4=68 => x^4=68/(a^4+1). g2=(2*a^3+3*a^2+2*a)*x^4 - 128 =0. Subs for x^4 and multiply by (a^4+1) to obtain ga=(-4) * (32*a^4 - 34*a^3 -51*a^2 - 34*a + 32)=0. Note from symmetry of original equations both 'a' and '1/a' will be solutions. a=2 is obvious. So 'ga' factors: ga= (a - 2)*( 2*a - 1) *(16*a^2 + 23*a +16). subs a=2 into g1 => 4 values of x -> { +/- sqrt(2), +/- sqrt(2)*i } (and y=2*x) (note a=1/2 merely swaps values of x and y, i.e. x -> { +/- 2*sqrt(2), +/- 2*sqrt(2)*i }(and y =x/2)

@johnstanley5692 Месяц назад

Another? g1= x^4+y^4-68; g2=x*y*(2*(x^2+y^2)+3*x); (x+y)^4 = (x^4+y^4)+2*x*y*(2*(x^2+y^2)+3*x*y) =68+2*128 = 324 =4*3^4. => (x+y)= +/- 3*sqrt(2). Now define g3=x+y-3*sqrt(2) (=0). g4=x+y+3*sqrt(2) (=0); now eliminate 'x' by dividing g1/g3 to obtain P1(y). Then divide g1/g4 to obtain P2(y). where P1(y)=0 &P2(y)=0. Here P1(y)=2*y^4 - 12*2^(1/2)*y^3 + 108*y^2 - 216*2^(1/2)*y + 256 P2(y) = 2*y^4 + 12*2^(1/2)*y^3 + 108*y^2 + 216*2^(1/2)*y + 256. P1(y)=0 => y={ [2^(1/2), 2*2^(1/2), (3*2^(1/2))/2 +/- (110^(1/2)*1i)/2 }, P2(y)=>y={-2^(1/2), -2*2^(1/2), - (3*2^(1/2))/2 + (110^(1/2)*1i)/2}