Question

Find the quadratic polynomial whose zeros are 5 +√2 and 5 - √2​

Answer 1

Answer:

$\huge{\red{\boxed{\boxed{\green{\sf{x^{2}-10x+23}}}}}}$

Step-by-step explanation:

$\huge{\red{\boxed{\boxed{\green{\underline{\red{\sf{SOLUTION-}}}}}}}}$

$two \: zeroes \: of \: a \: quadratic \: eqn \: are \\ \alpha = 5 + \sqrt{2} \: \\ \beta = \: 5 - \sqrt{2} \\ \\( x - \alpha )(x - \beta ) = 0 \\ = )(x -( 5 + \sqrt{2} ))(x - (5 - \sqrt{2} )) = 0 \\ = )(x - 5 - \sqrt{2} )(x - 5 + \sqrt{2} ) = 0\\ \therefore \: {a}^{2} - {b}^{2} = (a + b)(a - b)\\ = )(x - 5) ^{2} - (\sqrt{2})^{2} = 0 \\ = ) {x}^{2} + 25 - 10x - 2 = 0 \\ = ) {x}^{2} - 10x + 23 = 0$

$\huge{\red{\boxed{\boxed{\green{\sf{x^{2}-10x+23}}}}}}$

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