Math, asked by xyz89, 1 year ago

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Answered by Shanayarokz
1
Let the first Zeros be a
and second be 1/a


product \: of \: zero =  \frac{constant \: term}{coefficient \:  of \:  {x}^{2} }

a \times  \frac{1}{a}  =  \frac{6a}{ {a}^{2}  + 9}

1 =  \frac{6a}{ {a}^{2}  + 9}


a² + 9 = 6a

a² - 6a + 9 = 0

a² - 3a - 3a + 9 = 0

a ( a - 3 ) - 3 ( a - 3 ) = 0

( a - 3 ) ( a - 3 ) = 0



* ( a - 3 ) = 0
a = 3


* ( a - 3 ) = 0
a = 3


So , value of a is 3

Shanayarokz: it is given in the question that the Zeros are reciprocal of each other!
xyz89: but asa kyo
Shanayarokz: sorry ek mistake hogya ....a k place pr b hoga
Shanayarokz: i haven't seen that the question already contain a
xyz89: no problem
xyz89: ☺..
xyz89: mistake to kisi sa bhi ho sacti ha
Shanayarokz: :)
xyz89: mujsa bhi hue ha or hogi
xyz89: : )
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