calculate value of expression.

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calculate value of expression.

Post Admin Tue Feb 01, 2011 9:53 pm

Admin: Admin; Posts : 103
Join date : 2011-02-01
Location : India

quadratic eqs....

Post mohit Sun Jul 17, 2011 2:08 pm

its answer is 25/6.

mohit: Guest

Re: calculate value of expression.

Post Admin Sun Jul 17, 2011 8:26 pm

$calculate value of expression. Gif.latex?$Here%20$\boxed{\mathbf{x+y+z=1}}$\\\\%20$\mathbf{(x+y+z)^2=x^2+y^2+z^2+2(xy+yz+zx)}$\\\\%20$\mathbf{2.(xy+yz+zx)=1-2\Leftrightarrow%20\boxed{\bold{(xy+yz+zx)=-\frac{1}{2}}}}$\\\\%20Similarly\\\\%20$\mathbf{x^3+y^3+z^3%20-%203xyz%20=%20(x+y+z)*\left(x^2+y^2+z^2-(%20xy+yz+zx)\right%20)}$\\\\%20$\mathbf{3-3xyz=1*\left(2+\frac{1}{2}\right)=\frac{5}{2}\Leftrightarrow%203xyz=\frac{1}{2}\Leftrightarrow%20\boxed{\bold{xyz=\frac{1}{6}}}}$\\\\%20Now%20Let%20We%20assume%20that%20$\mathbf{x,y,z}$%20be%20the%20Roots%20of%20Cubic%20equation%20$\mathbf{P(t)=0}$.\\\\%20$\mathbf{t^3-(x+y+z).t^2+(xy+yz+zx).t-xyz=0}$\\\\%20$\mathbf{t^3-t^2-\frac{3}{2}.t+\frac{1}{6}=0}$\\\\%20Now%20Multiply%20It%20by%20$\mathbf{t},$%20We%20Get\\\\%20$\mathbf{t^4-t^3-\frac{1}{2}.t^2-\frac{1}{6}.t=0}$\\\\%20Again%20Given%20equation%20has%20a%20roots%20$\mathbf{x,y,z}$%20and%20$\mathbf{t=0}$\\\\%20So%20If%20given%20equation%20has%20a%20roots%20$\mathbf{x,y,z},\;$%20.%20Then%20It%20will%20satisfy%20that%20equation$

$calculate value of expression. Gif.latex?\mathbf{\sum_{cyclic(x,y,z)}.x^4-\sum_{cyclic(x,y,z)}.x^3}-\frac{1}{2}.\sum_{\bold{cyclic(x,y,z)}}.\bold{x^2}-\frac{1}{6}.\sum_{\bold{cyclic(x,y,z)}}.\bold{x}=0$\\\\\\%20$\mathbf{\left(x^4+y^4+z^4\right)-\left(x^3+y^3+z^3\right)-\frac{1}{2}.\left(x^2+y^2+z^2\right)-\frac{1}{6}$

Admin: Admin; Posts : 103
Join date : 2011-02-01
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Re: calculate value of expression.

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