Math, asked by rairaju121980, 15 hours ago

find the quadratic polynomial , the stun and product of whose zeroes are √3 and 1/√3 respectively

Answers

Answered by LoverBoy346
4

Step-by-step explanation:

We know that,

 {x}^{2}  - ( \alpha  +  \beta )x + ( \alpha  \times  \beta ) = 0

 {x}^{2}  -  \sqrt{3} x +  \frac{1}{ \sqrt{3} }  = 0

 \boxed{ \sqrt{3}  {x}^{2}  - 3x - 1 = 0}

\sqrt{3}  {x}^{2}  - 3x - 1 = 0 \:  \: is \: the \: required \: quadratic \: polynomial

Answered by icecreamqueen
1

Answer:

\sqrt 3x^{2} - 3x+1 = 0

Step-by-step explanation:

Given:

The sum of zeroes = \sqrt3

The product of zeroes = 1/\sqrt3

To find:

The quadratic equation.

Solution:

The general quadratic equation is given as:

x^{2} (Sum of zzeroes) + (Product of zeroes)=0

Substituting the value in the equation and then multiyplying the equation by \sqrt3

x^{2} -\sqrt3x + 1/\sqrt3 = 0

\sqrt3*x^{2} -\sqrt3* \sqrt3{x} + \sqrt3*1/\sqrt3 = 0

Multiyplying the terms we get,

\sqrt3x^{2} -3x+1=0

Hence the answer is \sqrt3x^{2} -3x+1=0

Hope it helps you☺️☺️☺️...

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