Find the value of a for which roots of the equation (a−3)x2−2ax+5a=0 are positive.
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Solution
(a−3)x2−2ax+5a=0;FirstD≥0⇒(2a)2−4(5a)(a−3)≥0⇒a2−5a2+15a≥0⇒−4a2+15a≥0⇒a(4a−15)≤0[Multiplybothsidesby−1] ⇒aϵ[0,154]−(A) Now consider 2 cases, (a) or (b) (a) f(0)>0⇒0−0+5a>0⇒a>0(i) Coefficient of x2 less than 0 ⇒a−3<0⇒a<3(ii) From (i) and (ii), aϵ(0,3) (B)
(b) f(0)>0⇒0−0+5a>0 ⇒a>0(iii) Coefficient of x2 greater than 0 ⇒a−3>0⇒a>3(iii) From (iii) and (iv), aϵ(3,∞) (C) Hence, solution is A and B or C i.e., (0,3)∪(3,154)