Question

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

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Answer 1

Step-by-step explanation:

We know that,

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

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

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

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

Answer 2

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☺️☺️☺️...