Question

If α and β are the zeroes of the polynomial 2x²-5x+7, then find the quadratic polynomial whose zeroes are 3α+4β and 3β+4α ?

Answer 1

p(x) = x^2-bx+c

x^2- 25/2+41

taking 2 lcm

2x^2 -25+82/2 is the required polynomial.

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Answer 2

Answer:

Step-by-step explanation:

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

3α + 4β and 4α + 3β are zeros of a polynomial.

sum of zeros = 3α + 4β+ 4α + 3β

                      = 7α + 7β

                      = 7[ α + β]

                     = 7× 5/2

                    = 35/2

product of zeros=(3α+4β)(4α+ 3β)

= 3α [ 4α + 3β] + 4β [4α + 3β]

= 12α² + 9αβ + 16αβ + 12β²

= 12α² + 25αβ +  12β² 

= 12 [ α² + β² ] + 25αβ

= 12 [ (α + β)² - 2αβ ] + 25αβ

= 12 [ ( 5/2)² - 2 × 7/2 ] + 25× 7/2

= 12 [ 25/4 - 7 ] + 175/2

= 12 [ 25/4 - 28/4 ] + 175/2

= 12 [ -3/4 ] + 175/2

= -36/4 + 175/2

= -9+ 175/2

= 166/2

=83

a quadratic polynomial is given by :-

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

k {x² - 5/2x +83}

k = 2

2 {x² - 5/2x + 83 ]

2x² - 5x + 166 is the required polynomial...

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