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Write the polynomial whose zeros are $2 + \sqrt 3 $ and $2 - \sqrt 3 $
A). ${x^2} - 2x + 1$
B). ${x^2} - 4x + 1$
C). ${x^2} - 7x + 1$
D). ${x^2} - 4x - 1$

seo-qna
Last updated date: 21st Jul 2024
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Answer
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Hint: Zeros are nothing but the roots of the polynomial. Since there are two zeros the polynomial will be of degree two. Therefore we have to find a quadratic polynomial of degree two whose zeros (roots) are $2 + \sqrt 3 $ and $2 - \sqrt 3 $. By using the sum and product of the given two zeros we will find the required quadratic polynomial.

Complete step-by-step solution:
The standard form of a quadratic polynomial in $x$ is $a{x^2} + bx + c$. The value of $x$ which when substituted in the polynomial gives the value zero is known as the root or zero of the polynomial.
 let $\alpha = 2 + \sqrt 3 $ and $\beta = 2 - \sqrt 3 $ be the two zeros of the polynomial we need to find.
The sum of the zeros is , $\alpha + \beta = (2 + \sqrt 3 ) + (2 - \sqrt 3 )$ here $\sqrt 3 $ with opposite signs gets cancelled and we get $\alpha + \beta = 4$.
The product of the zeros is , $ab = (2 + \sqrt 3 ) \times (2 - \sqrt 3 ) = 4 - 2\sqrt 3 + 2\sqrt 3 - 3$ here $2\sqrt 3 $ with opposite signs gets cancelled and we get $\alpha \beta = 1$.
The formula to find the polynomial using sum and product of zeros is given by,
 ${x^2}$ - (sum of zeros )$x$ + product of zeros
$\Rightarrow {x^2} - (\alpha + \beta )x + \alpha \beta $
$ \Rightarrow {x^2} - 4x + 1 $
Therefore $2 + \sqrt 3 $ and $2 - \sqrt 3 $ are the zeros of ${x^2} - 4x + 1$. Hence option B is correct.

Note: We can also obtain the polynomial by forming the linear factors of the given zeros i.e. $(x - 2 + \sqrt 3 )(x - (2 - \sqrt 3 ))$ and then multiplying them together we get,
$ {x^2} - x(2 - \sqrt 3 ) - (2 + \sqrt 3 )x + (2 + \sqrt 3 )(2 - \sqrt 3 ) $
$ = {x^2} - 2x + \sqrt 3 x - 2x - \sqrt 3 x + 4 - 2\sqrt 3 + 2\sqrt 3 - 3 $
$ = {x^2} - 4x + 1 $
Which is the same as the polynomial we have obtained above, this process is little longer compared to the sum product formula which we have used. Another method is substituting the given zeros in all the options to check which option becomes zero by hit and trial method.