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If `alpha,beta`are the roots of lthe equation `2x 62-3x-6=0,`find the equation whose roots are `alpha^2+2a n dbeta^2+2.` |
Answer» Since `alpha, beta ` are roots of the equation `2x^(2) - 3x - 6 = 0, ` we have `alpha + beta = 3//2 and alpha beta = -3` `rArr alpha^(2) + beta^(2) = (alpha + beta)^(2) - 2alpha beta` = `(90/(4)+6 = (33)/(4)` Now , `(alpha^(2)+beta^(2))+(beta^(2)+2) = (alpha^(2)+ beta^(2)) + 4` `=(33)/(4) + 4 = (49)/(4)` and `(alpha ^(2) + 2 ) (beta^(2)+2)= alpha^(2) + beta^(2) + 2(alpha^(2)+ beta^(2))+4` `=(-3)^(2) + 2((33)/(4)) + 4` `= (59)/(2)` So , the equatioon whose roots are `alpha^(2) + 2 and beta^(2) + 2` is given by `x^(2) - x [(alpha^(2) + 2) + (beta^(2) + 2)] + (alpha^(2) + 2 ) (beta^(2) + 2 ) = 0` `rArr x^(2) - (49)/(4)x+(59)/(2) = 0` or `4x^(2) - 49x + 118 = 0` |
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