Math, asked by ifrafarukhi, 5 months ago

Find a quadratic polynomial with the given numbers as the sum and product of its

zeroes respectively 2 + √3 and 2 - √3​

Answers

Answered by lifekiller05
5

let \: the \:  \alpha  \: and \:  \beta  \: be \: zeros \: of  \\ polynomial

Now, \\ let \:  \alpha  =  \: 2 +  \sqrt{3} \:,\:  \beta=2- \sqrt{3}

we \: know \: that,\:  \\ sum \: of \: zero =  \alpha  +  \beta  \: and \\ product \: of \: zero =  \alpha  \times  \beta

so, \:    \: \alpha  +  \beta  =2 +  \sqrt{3}\:+2 -  \sqrt{3}   \:   \\   \:  \implies\alpha  +  \beta  = (2 + 2) = 4

  \alpha  \times  \beta  = (2 +  \sqrt{3})\times (2-\sqrt{3}) \\  \alpha  \times  \beta  = 4  + 2 \sqrt{3} - 2 \sqrt{3} - 3 \\\alpha  \times  \beta = (4 - 3) = ( \:  1)\:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:

we \: know \: that \: for \: any \: quadratic \\ euation \: the \: sum \: of zero \:  =  \frac{ - b}{a}  \\ and \: product \: of \:  \: zero =  \frac{c}{a}  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:  \:

so,sum \: of \: zero =  \frac{ - b}{a}  =  \frac{4}{1}  \\ sum \: of \: product =  \frac{c}{a} =  \frac{1}{1}

 \therefore \: a = 1 \:, \: b =  - 4 \:, \: c = 1

qadratic \: is \: of \: the \: form \:  \\ ax^{2} + bx + c = 0

1x^{2} - 4x+1 \: is \: the \: quadratic \:  \\ eqation

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