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If one root of equation (e -m)x+x1 0 be double of the other and if e be real, show that m |
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Answer» Sum of roots is -L/(L-m)=3r, product of roots is 1/(L-m)=2r^2 where r is one of the roots. So 1 / 2(L-m) = L^2 / 9(L-m)^2 9(L-m)=2L^2 2L^2 - 9L + 9m=0 For L to be real, the discriminant is non-negative. 81>=8X9m m<=9/8 The greatest value of m is therefore 9/8. |
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