Question

Find the polynomial, whose zeroes are 2 + √3 and 2 - √3

Answer 1

Deeply explained here,

General form of Quadratic polynomial with roots 'alpha' and 'beta' is

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

let

$\alpha = 2 + \sqrt{3}$

$\beta = 2 - \sqrt{3}$

therefore

$\alpha + \beta = 2 + \sqrt{3 } + 2 - \sqrt{3} = 4$

$\alpha \beta = (2 + \sqrt{3} )(2 - \sqrt{3} ) = 4 - 3$

Therefore

$\alpha \beta = 1$

substitute in that formula we get

${x}^{2} - (4)x + 1$

Therefore the required polynomial

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

Answer 2

Step-by-step explanation:

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sakshi154

sakshi154

04.12.2017

Math

Secondary School

answered

Write the polynomial whose zeroes are 2+√3 and 2-√3

2

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phillipinestest

The polynomial whose zeroes are \bold{2+\sqrt{3}} and \bold{2-\sqrt{3}} is \bold{x^{2}-4 x+1}

Solution:

Since there are 2 zeroes, the polynomial is of degree 2.

Let a and b represents the zeroes of the polynomial.

Let a=2+\sqrt{3} and b=2-\sqrt{3}

Since, the polynomial is of degree 2

Polynomial would be of the form \bold{x^{2}-(a+b) x+a b}

Therefore computing the values,

\bold{a+b=(2+\sqrt{3})+(2-\sqrt{3})=4}

\bold{a b=(2+\sqrt{3}) \times(2-\sqrt{3})=4-3=1}

Therefore, the polynomial is \bold{x^{2}-4 x+1}