find the all the values of 'a' for which root of the equation (a-3)x^2 - 2ax + 5a = 0 are positive ? plzz explain each and every step in detail don't skip any step fast answer

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Please find below the solution to the asked query:

a-3x2-2ax+5a=0If roots are real, thenD>0-2a2-4×5a×a-3>04a2-20aa-3>04a2-20a2+60a>0 -4a2+15a>0 Multiply both sides by -1, direction of inequality changes.4a2-15a<0a4a-15<0 Use method of interval-++++0----154+++a0,154 .....iAs both roots are positive, hence sum and product roots will be positive.Sum of roots>0--2aa-3>02aa-3>0aa-3>0-++++0-----3++++a-,03,....iiProduct of roots>05aa-3>0aa-3>0-++++0-----3++++a-,03,....iiiTaking intersectioncommon of i,ii,iii we geta3,154

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4a^2+20a-60>0                                         (Dividing by 2)
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