Math, asked by vatsshah012, 8 months ago

If alpha.and bet are Zeros of p(x)=x*2-x-2 then find polynomial whose zeros are 2alpha+3beta,3alpha+2beta​

Answers

Answered by Anonymous
1

HI !

NOTE :-

α² + β² can be written as (α + β)² - 2αβ

p(x) = 2x² - 5x + 7

a = 2 , b = - 5 , c = 7

α and β are the zeros of p(x)

we know that ,

sum of zeros = α + β

                     = -b/a

                     = 5/2

product of zeros = c/a

                          = 7/2

===============================================

2α + 3β and 3α + 2β are zeros of a polynomial.

sum of zeros = 2α + 3β+ 3α + 2β

                     = 5α + 5β

                     = 5 [ α + β]

                    = 5 × 5/2

                   = 25/2

product of zeros = (2α + 3β)(3α + 2β)

                         = 2α [ 3α + 2β] + 3β [3α + 2β]

                        = 6α² + 4αβ + 9αβ + 6β²

                        = 6α² + 13αβ +  6β²

                        = 6 [ α² + β² ] + 13αβ

                        = 6 [ (α + β)² - 2αβ ] + 13αβ

                        = 6 [ ( 5/2)² - 2 × 7/2 ] + 13× 7/2

                        = 6 [ 25/4 - 7 ] + 91/2

                        = 6 [ 25/4 - 28/4 ] + 91/2

                        = 6 [ -3/4 ] + 91/2

                       = -18/4 + 91/2

                       = -9/2 + 91/2

                       = 82/2

                       = 41

                                                                          -18/4 = -9/2 [ simplest form ]

a quadratic polynomial is given by :-

k { x² - (sum of zeros)x + (product of zeros) }

k {x² - 5/2x + 41}

k = 2

2 {x² - 5/2x + 41 ]

2x² - 5x + 82                   -----> is the required polynomial

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