If α β are the zeroes of the polynomial 2x²-5x +7=0 then find a polynomial whose zeroes are 2α+3β,3α+2β
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α,β are the zeros of 2x²-5x+7=0. Then,
α+β=-(-5/2)=5/2 and α×β=7/2
Now, 2α+3β+3α+2β=5α+5β=5(α+β)=5×5/2=25/2 and
(2α+3β)(3α+2β)
=6α²+9αβ+4αβ+6β²
=6(α²+β²)+13αβ
=6{(α+β)²-2αβ}+13αβ
=6(α+β)²-12αβ+13αβ
=6(5/2)²+αβ
=6×25/4+7/2
=75/2+7/2
=(75+7)/2
=82/2
=41
The required equation :
x²-(sum of the roots)x+product of the roots=0
or, x²-(25/2)x+41=0
or, 2x²-25x+82=0
α+β=-(-5/2)=5/2 and α×β=7/2
Now, 2α+3β+3α+2β=5α+5β=5(α+β)=5×5/2=25/2 and
(2α+3β)(3α+2β)
=6α²+9αβ+4αβ+6β²
=6(α²+β²)+13αβ
=6{(α+β)²-2αβ}+13αβ
=6(α+β)²-12αβ+13αβ
=6(5/2)²+αβ
=6×25/4+7/2
=75/2+7/2
=(75+7)/2
=82/2
=41
The required equation :
x²-(sum of the roots)x+product of the roots=0
or, x²-(25/2)x+41=0
or, 2x²-25x+82=0
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