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Answers
▪2x² - 5x + 7
a = 2 , b = - 5 , c = 7
α and β are the zeros of given polynomial
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
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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@skb
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Given polynomial , 2x² - 5x + 7
By compared with ax^2 + bx + c , we get
a = 2
b = - 5
c = 7
α and β are the zeros of the polynomial
So ,
• sum of zeros
= α + β
= ( - b ) / a
= - ( - 5 ) / 2
= 5 / 2 -------------- ( i )
• product of zeros
= α × β
= c / a
= 7 / 2 ---------------- ( ii )
Given , zeros of the required polynomial are ( 2α + 3β ) and ( 3α + 2β )
Now ,
• sum of zeros
= ( 2α + 3β ) + ( 3α + 2β )
= 5α + 5β
= 5 ( α + β ) [ • Putting the value of ( 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 [ • Putting the values ]
= 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
= ( - 9 + 91 ) / 2
= 82 / 2
= 41
• One quadratic polynomial is given by :-
k { x² - (sum of zeros) x + (product of zeros) }
= k { x² - ( 25 / 2 ) x + 41 }
= 2 { x² - 25x / 2 + 41 } [ • k = 2 ]
= 2x² - 25x + 82
So , the required polynomial is 2x² - 25x + 82.
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