Math, asked by sowmiya35, 1 year ago

solve this question!!!

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Answers

Answered by FuturePoet
8

Solution :


Given ,

n and m are the zeros of the polynomial  F(x) = 3x^2 + 11x - 4


To Find

Value of m/n + n/m


Answer

⇒  3 (m)^2 + 11 (m) - 4 = 0

⇒ 3m^2 + 11m - 4 = 0

⇒ 3m^2 + 12m - m -4 = 0

⇒ 3m ( m + 4 ) - 1 ( m + 4) = 0

⇒ ( 3m - 1) ( m + 4) = 0

⇒  m = 1/3 or 4

So , we replace n in x place we get some value

m/n + n/m =  1 + 1

=> 2


Therefore , the value of m/n + n/m is 2


Anonymous: I think ur answer is incorrect.... how r u getting 1+1 ?
Answered by Anonymous
10

Polynomial given

Let f ( x ) be 3 x² + 11 x - 4


What is a zero ?

A zero is the value of x for which the function f ( x ) is simple equal to zero .

There are many ways to approach the answer .. Seeing your copy I guess you need the first method .


Method 1 :


3 x² + 11 x - 4 = 0

Comparing with a x² + b x + c = 0

a = 3

b = 11

c = -4

Sum of roots = - b / a

Product of roots = c / a

Given that m and n are the zeroes .

m + n = - b / a

= > m + n = - 11 / 3 .............................(1)

= > m n = - 4 / 3..............................(2)


Solving the problem :

m / n + n / m

L.C.M = m × n

= > ( m² + n² ) / ( m n )

Now note that :

m² + n² = m² + n² + 2 m n - 2 m n = ( m + n )² - 2 m n

Hence :

( m² + n² ) / m n

= > [ ( m + n )² - 2 m n ] / m n

Now put the values of ( 1 ) and ( 2 ) :

= > [ ( - 11 / 3 )² - 2 ( - 4 / 3 ) ] / ( - 4 / 3 )

= > [ 121 / 9 + 8 / 3 ] /  ( - 4 / 3 )

= > [ ( 121 + 8 × 3 ) / 9 ] / ( - 4 / 3 )

= > [ 121 + 24 ] / ( - 4 / 3 ) × 9

= > [ 145 ] / ( - 12 )

= > - 145 / 12



Method 2 :

3 x² + 11 x - 4 = 0

= > 3 x² + 12 x - x - 4 = 0

= > 3 x ( x + 4 ) - 1 ( x + 4 ) = 0

= > ( 3 x - 1 )( x + 4 ) = 0

Either

= > 3 x - 1 = 0

= > 3 x = 1

= > x = 1 / 3

Or :

= > x + 4 = 0

= > x = - 4

Let m be - 4 and n = 1 / 3

m / n + n / m

= > ( - 4 ) / ( 1 / 3 ) + ( 1 / 3 ) / ( - 4 )

= > - 12 - 1 / ( 12 )

= > ( - 144 - 1 ) / 12

= > - 145 / 12


ANSWER :

The value of m/n + n/m will be - 145 / 12


sowmiya35: thanks
Anonymous: welcome :-)
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