find the all the values of 'a' for which root of the equation (a-3)x^2 - 2ax + 5a = 0 - Maths - Logs Equations and Inequalities

Question

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

Shivam Mishra · Accepted Answer

Dear Student,
Please find below the solution to the asked query:

(a - 3) x^{2} - 2 ax + 5 a = 0 If roots are real, then D > 0 \Rightarrow {(- 2 a)}^{2} - 4 \times 5 a \times (a - 3) > 0 4 a^{2} - 20 a (a - 3) > 0 4 a^{2} - 20 a^{2} + 60 a > 0 - 4 a^{2} + 15 a > 0 Multiply both sides by - 1, direction of inequality changes . 4 a^{2} - 15 a < 0 a (4 a - 15) < 0 Use method of interval - \infty + + + + (0) - - - - (\frac{15}{4}) + + + \infty a \in (0, \frac{15}{4}) . . . . . (i) As both roots are positive, hence sum and product roots will be positive . Sum of roots > 0 - \frac{- (2 a)}{(a - 3)} > 0 \Rightarrow \frac{2 a}{a - 3} > 0 \Rightarrow \frac{a}{a - 3} > 0 - \infty + + + + (0) - - - - - (3) + + + + \Rightarrow a \in (- \infty, 0) \cup (3, \infty) . . . . (ii) Product of roots > 0 \frac{5 a}{a - 3} > 0 \Rightarrow \frac{a}{a - 3} > 0 - \infty + + + + (0) - - - - - (3) + + + + \Rightarrow a \in (- \infty, 0) \cup (3, \infty) . . . . (iii) Taking intersection (common) of (i), (ii), (iii) we get a \in (3, \frac{15}{4})

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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