Question

Form the quadratic polynomial whose zeroes are 7+2 V2 and 7-2V2​

Answer 1

Answer:

this is the equation of the question

Answer 2

Answer:

$\huge{\boxed{\boxed{\sf{x^{2}-14x+41}}}}$

Step-by-step explanation:

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

$\alpha = 7 + 2 \sqrt{2} \\ \beta = 7 - 2 \sqrt{2} \\ according \: to \: the \: question \\( x - \alpha )(x - \beta )=0 \\ = )(x - (7 + 2 \sqrt{2} ))(x - (7 - 2 \sqrt{2} )) =0\\ = )(x - 7 - 2 \sqrt{2} )(x - 7 + 2 \sqrt{2} )=0 \\ \\we \: know \: {x}^{2} - {y}^{2} = (x + y)(x - y)=0 \\ \\so \: we \: conclude \: this \: in \: this \: answer \\ = )(x - 7)^{2} - (2 \sqrt{2} )^{2} =0 \\ = ) {x}^{2} + {7}^{2} - 14x - 4 \times 2 =0\\ = ) {x}^{2} + 49 - 14x - 8 =0\\ = ) {x}^{2} - 14x + 41=0$

$\huge{\boxed{\boxed{\sf{x^{2}-14x+41}}}}$

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