Question

-
23210
10) Find the aradhatic Polynomicus whose someone
of zeroes
= 4 and
Product
Of zero es=1​

Answer 1

x²-4x+1

Step-by-step explanation:

$Let \: the \: zeroes \: be \: x \: and \: y. \\ Given: \: x + y = 4 \\ xy = 1 \\To \: find: \: Quadratic \: polynomial \: with \: zeroes \: = x,y \\ Solution: y = 4 - x \\ x(4 - x) = 1 \\ 4x - {x}^{2} = 1 \\ - {x}^{2} + 4x - 1 = 0 \\ Multiply \: with \: - 1 \\ {x}^{2} - 4x + 1 = 0 \: (answer) \\ Proof: {x}^{2} - 4x + 1 = 0 \\ {x}^{2} - 4x + 4 - 3 = 0 \\ ( {x}^{2} - 2(2)(x) + {2}^{2} ) - {( \sqrt{3} )}^{2} = 0 \\ {(x - 2)}^{2} - {( \sqrt{3}) }^{2} = 0 \\ (x - 2 + \sqrt{3}) (x - 2 - \sqrt{3} ) = 0 \\ x = 2 - \sqrt{3} \: , \: x = 2 + \sqrt{3} \\ sum \: = 2 - \sqrt{3} + 2 + \sqrt{3} \\ = 4 \\ product \: = (2 - \sqrt{3} )( 2 + \sqrt{3} ) \\ = 1$