Question

If a and b are zeros of the polynomial x2-mx+n then find value of a/b+b/a

Answer 1

Let the polynomial be denoted as P(x)

$P(x) = {x}^{2} - mx + n$

Now a , b are the zeroes of the polynomial . So we can say that :

$\blacksquare \: \: a + b = - ( \dfrac{ - m}{1} ) = m$

$\blacksquare \: \: a \times b = ( \dfrac{ n}{1} ) = n$

Now as per the question , we have to find the value of :

$\therefore \: \dfrac{a}{b} + \dfrac{b}{a}$

$= \dfrac{ {a}^{2} + {b}^{2} }{ab}$

$= \dfrac{ {(a + b)}^{2} - 2ab }{ab}$

$= \dfrac{ {m}^{2} - 2n}{n}$

$= \dfrac{ {m}^{2} }{n} - 2$

So final answer is :

$\boxed{ \bold{ \blue{ \dfrac{a}{b} + \dfrac{b}{a} = \dfrac{ {m}^{2} }{n} - 2}}}$

Answer 2

Answer:

m^2/n - 2

Step-by-step explanation:

Let the polynomial be denoted as P(x)

p(x) = x^2 - mx + n

Now,

a , b are the zeroes of the polynomial . So we can say that :

=> a+b = -{(-m)/1} = m .......... (i)

=> a×b = n/1 = n ...............(ii)

Now,

a/b + b/a

= (a^2 + b^2)/ab

= {(a + b)^2 - 2ab }/ab

= (m^2 - 2n)/n .......{ from (i) & (ii) }

= m^2/n - 2 Ans...

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