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If alpha,beta are the roots of x^(2)-3 ...

If `alpha,beta` are the roots of `x^(2)-3 x+5=0` and `gamma, delta` are the roots of `x^(2)+5 x-3=0`, then the equation whose roots are `alpha gamma+beta delta` and `alpha delta+beta gamma` is

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The correct Answer is:
`a^(2)l^(2)x^(2)-ablmx+(b^(2)-2ac) Ink +(m^(2)-2In)ac=0`
We have `alpha + beta=-b/a`
`alpha beta=c/aimpliesgamma + delta =-m/l` and `gamma delta=n/l`
Now sum of the roots
`=(alpha gamma+beta delta)+(alpha delta +beta gamma)=(alpha +beta)gamma +(alpha +beta) delta`
`=(alpha +beta)(gamma+delta)`
`=(-b/a)(-m/l)=(mbl)/(al)`
and product of the roots
`=(alpha gamma+beta delta)(alpha delta+beta gamma)`
`=(alpha^(2)+beta^(2))gamma delta +alpha beta (gamma^(2)+delta^(2))`
`{(alpha +beta)-2alpha beta} gamma beta+alpa beta {(gamma +delta)^(2)-2 gama delta}`
`={(-b/a)^(2)-(2c)/a}b/l+c/a{(m/l)^(2)-(2n)/l}`
`={(b^(2)-2ac)/(a^(2))}b/l+c/a{(m^(2)-2nl)/(l^(2))}=((b^(2)-2ac)ln+(m^(2)-2nl)ac)/(a^(2)l^(2))`
`:.` Required equation is
`x^(2)-((mb)/(al))x+((b^(2)-2ac)ln+(m^(2)-2nl)ac)/(a^(2)+l^(2))=0`
`impliesa^(2)l^(2)x^(2)-mbalc+(b^(2)-2ac)ln+(m^(2)-2nl)ac=0`
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