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For-what-value-of-k-x-y-z-2-k-x-2-y-2-z-2-can-be-resolved-into-linear-rational-factors-

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Question Number 20308 by Tinkutara last updated on 25/Aug/17
For what value of k, (x + y + z)^2 + k(x^2 + y^2 + z^2 ) can be resolved into linear rational factors?
Forwhatvalueofk,(x+y+z)2+k(x2+y2+z2)canberesolvedintolinearrationalfactors?
Answered by Tinkutara last updated on 28/Aug/17
(x+y+z)^2 +k(x^2 +y^2 +z^2 ) =x^2 +y^2 +z^2 +2xy+2yz+2zx+k(x^2 +y^2 +z^2 ) =z^2 [((x/z))^2 +((y/z))^2 +1+((2xy)/z^2 )+((2y)/z)+((2x)/z)+k((x/z))^2 +k((y/z))^2 +k] =z^2 [X^2 +Y^2 +1+2XY+2Y+2X+kX^2 +kY^2 +k] =z^2 [(k+1)X^2 +(k+1)Y^2 +2XY+2Y+2X+(k+1)] Δ=(k+1)(k+1)(k+1)+2(1)−(k+1)−(k+1)−(k+1)=0 =(k+1)^3 −3(k+1)+2=0 ⇒k=0,−3
(x+y+z)2+k(x2+y2+z2)=x2+y2+z2+2xy+2yz+2zx+k(x2+y2+z2)=z2[(xz)2+(yz)2+1+2xyz2+2yz+2xz+k(xz)2+k(yz)2+k]=z2[X2+Y2+1+2XY+2Y+2X+kX2+kY2+k]=z2[(k+1)X2+(k+1)Y2+2XY+2Y+2X+(k+1)]Δ=(k+1)(k+1)(k+1)+2(1)−(k+1)−(k+1)−(k+1)=0=(k+1)3−3(k+1)+2=0⇒k=0,−3
Commented by ajfour last updated on 28/Aug/17
Find the coefficients if k=−3 .
Findthecoefficientsifk=−3.

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