Question

Form a quadratic polynomial whose one zero is (3+ $\sqrt{2}$) and sum of the zero is 6

Answer 1

Let a and b be the zeros
From the question, a = 3+√2, b = ?
And a + b = 6
3 + √2 + b= 6
b = 3 - √2
ab = (3+√2)(3-√2)
= 9-2
= 7
We get a+b and ab
Thus the required polynomial is
= k[x^2 - (a+b)x + ab]
= k[x^2 - 6x + 7]
Taking k = 1

x^2 - 6x + 7 is the polynomial

Answer 2

HEYA!
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✴QUADRATIC POLYNOMIAL✴
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Given that ,

Zero of the polynomial is 3 + √2

and sum of the zeroes = 6

$let \: \alpha \: \: and \: \beta \: \: be \: the \: roots$
$\alpha + \beta = 6 \\ \\ 3 + \sqrt{2} + \beta = 6 \\ \\ \beta = 6 - 3 - \sqrt{2} \\ \\ = 3 - \sqrt{2}$
Hence (3-√2 ) ( 3+√2 ) are the zeroes of the polynomial .


Sum of zeroes = 6

Product of zeroes = 9 - 2= 7


$putting \: \: in \: formula = > > \: \: x {}^{2} - sx + p \\ \\ where \: s = \: sum \: \: and \: \: p = product \: of \: zeroes$
Hence the polynomial is

$x {}^{2} - 6x + 7$
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