Math, asked by vipulchab1382, 1 year ago

Find a quadratic polynomial whose zeroes are 3+√5 and 3-√5

Answers

Answered by Mohitdon30
1

Answer:

Step-by-step explanation:

The quadratic polynomial with α and β as zeroes is x2 − (α + β)x + (αβ) Given zeroes of the polynomial are 3+√5 and 3-√5 Hence the quadratic polynomial is: x2 − (3+√5 + 3 −√5)x + (3+√5)(3 −√5) = x2 − 6x + (9 −5) = x2 − 6x + 4 Thus the quadratic polynomial is ( x2 − 6x + 4).

Answered by Anonymous
0

The quadratic polynomial whose zeroes are,

5 \sqrt{3} ,5 -  \sqrt{3}

 \alpha , \beta  \: is \: f(x) = k[ {x}^{2} - ( \alpha  +  \beta )x +  \alpha  \times  \beta  ]

where k is any non-zero real no.

THE QUADRATIC POLY POLYNOMIAL WHOSE ZEROES ARE

5 \sqrt{3} ,5 -  \sqrt{3}

 f(x) = k[ {x}^{2} - ( \alpha  +  \beta )x +  \alpha  \times  \beta  ]

 f(x) = k[ {x}^{2} - ( 5  \cancel{ +  \sqrt{3}}  + 5  \cancel{ -  \sqrt{3}} )x +    (5 +  \sqrt{3}   ) (5 -  \sqrt{3}  ) ]

 f(x) = k[ {x}^{2} -10x + ( {5)}^{2}  -  ({ \sqrt{3} )}^{2}  ]

 f(x) = k[ {x}^{2} -10x + (25  - 3)]

 f(x) = k[ {x}^{2} -10x + 22]

so, the QUADRATIC polynomial is

 f(x) = k[ {x}^{2} -10x + 22]

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