if alpha and beta are the zeroes of quad. polynomial x2-x-2, find the polynomial whose zeroes are ( 2alpha+3beta )(3alpha+2beta)
Answers
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
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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
Answer:
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Step-by-step explanation:
x²-x-2
=>x²- 2x+x -2 =0
=>x(x-2) + 1(x-2) =0
=>(x+1) (x-2) =0
=>x = -1 , 2
Let alpha be -1 and beta be 2 then
( 2alpha+3beta )(3alpha+2beta)
(2(-1) + 3(2)) (3(-1) + 2 (2))
=> (-2+6) (-3+4)
=> 4(1) = 4