Find the quadric polynomial whose zeroes …

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Posted by Jeevith Arumugam 5 years, 8 months ago
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Yogita Ingle 5 years, 8 months ago
Since there are 2 zeroes, the polynomial is of degree 2.
Let a and b represents the zeroes of the polynomial.
Let {tex}a=2+\sqrt{3} and b=2-\sqrt{3}{/tex}
Since, the polynomial is of degree 2
Polynomial would be of the form {tex}{x^{2}-(a+b) x+a b} {/tex}
Therefore computing the values,
{tex}{a+b=(2+\sqrt{3})+(2-\sqrt{3})=4}{/tex}
{tex}{a b=(2+\sqrt{3}) \times(2-\sqrt{3})=4-3=1}{/tex}
Therefore, the polynomial is {tex}{x^{2}-4 x+1} {/tex}
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