Question

If alpha and beeta are the zeroes of (ax^2+bx+c) evaluate
1/(a*alpha+b) +1/(a*beeta+b)
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Answer 1

Step-by-step explanation:

hope this helps you this is ur answer

Answer 2

QUESTION

If α and  β are the zeroes of  ax²+bx+c  .  evaluate

$\frac{1}{(a*\alpha+b) } + \frac{1}{(a*\beta+b) }$  .

SOLUTION

Since α and  β  are zeroes of the given quadratic polynomial

therefore

α+β = -b/a

αβ = c/a

We have,

$\frac{1}{(a*\alpha+b) } +\frac{1}{(a*\beta+b)} \\\\( taking LCM )\\\\\frac{(a*\beta+b)+(a*\alpha+b)}{(a*\alpha+b)(a*\beta+b) }\\\\\frac{(a*\alpha+a*\beta +2b) }{(a^{2}\alpha\beta + ab\alpha+ab\beta+b^{2} ) }\\\\\frac{a(\alpha+\beta )+2b }{a^{2}\alpha\beta+ab(\alpha+\beta)+b^{2} } \\\\( putting \\\alpha+\beta = -b/a \\and \\ \alpha\beta = c/a)\\\\ \frac{a(\frac{-b}{a})+2b }{a^{2}(\frac{c}{a})+ab(\frac{-b}{a} )+b^{2}}\\\\\frac{-b+2b}{ac-b^{2}+b^{2} }\\\\\frac{b}{ac}$

ON EVALUATING WE WILL GET  (b/ac) .

so the answer will be

b/ac