Question

find a quadratic polynomial with the given numbers as the sum and product of its zeroes respectively 2+√3 AND 1/2+√3

Answer 1

Answer:

x² - √3x - √3

Step-by-step explanation:

As per thw provided question, we have :

• Sum of zeroes = 2 + √3
• Product of zeroes = $\rm {\dfrac{1}{2+\sqrt{3}} }$

We have to find the quadratic polynomial. As we know that if α ans β are roots of the quadratic polynomial, then polynomial will ve

$\longmapsto \rm {x^2 - (\alpha + \beta)x + (\alpha.\beta)}$

Here, we have,

• Sum of zeroes = 2 + √3

$\longmapsto \bf {(\alpha + \beta) = 2 + \sqrt{3} }$

• Product of zeroes = $\rm {\dfrac{1}{2+\sqrt{3}} }$

$\longmapsto \rm {(\alpha \times \beta) =\dfrac{1}{2+\sqrt{3}} }$

Rationalising the denominator. In order to rationalise the denominator, we multiply the rationalising factor of the denominator with both the numerator and the denominator.

$\longmapsto \rm {(\alpha \times \beta) =\dfrac{1}{2+\sqrt{3}} \times \dfrac{2-\sqrt{3}}{2-\sqrt{3}} }$

Multiplying (2 - √3) which is the rationalising factor of the denominator with both the numerator and the denominator.

$\longmapsto \rm {(\alpha \times \beta) =\dfrac{1(2 - \sqrt{3} ) }{(2+\sqrt{3})(2 - \sqrt{3})} }$

Performing multiplication in the numerator and simplifying the denominator using the identity,

• $\rm (a - b)(a + b) = a^2 - b^2$

$\longmapsto \rm {(\alpha \times \beta) =\dfrac{2 - \sqrt{3} }{(2)^2 - (\sqrt{3})^2} }$

Putting the values of the squares of the numbers in the denominator.

$\longmapsto \rm {(\alpha \times \beta) =\dfrac{2 - \sqrt{3} }{4 - 3} }$

Performing subtraction in denominator.

$\longmapsto \bf {(\alpha \times \beta) =2 - \sqrt{3} }$

Now,substituting the values in the formula of the quadratic polynomial.

$\longmapsto \rm {x^2 - (2 + \sqrt{3})x + (2-\sqrt{3})}$

Removing the brackets. If there is minus sign before the bracket then the signs of the numbers in the brackets is changed.

$\longmapsto \rm {x^2 - 2 - \sqrt{3}x + 2-\sqrt{3} }$

Arranging the like terms.

$\longmapsto \rm {x^2 - \sqrt{3}x - \sqrt{3}+ 2-2 }$

Performing subtraction.

$\longmapsto \bf\underline {x^2 - \sqrt{3}x -\sqrt{3} }$

∴ The required quadratic polynomial is x² - √3x - √3.

Answer 2

Step-by-step explanation:

$\alpha = 2 + \sqrt{3} \: \: \: \: \: \beta = \frac{1}{2 + \sqrt{3} } = 2 - \sqrt{3}$

$qudratic \: polynomial \: having \: roots \\ \alpha \\ \beta$

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

$= {x}^{2} - (2 + \sqrt{3} + 2 - \sqrt{3}) x + (2 + \sqrt{3} )(2 - \sqrt{3} )$

$= {x}^{2} - 4x + 1$

so Required quadratic polynomial

$= {x}^{2} - 4x + 1$