if alpha and beta are the zeros of the polynomial P(x) is = 2 x square - 8 x + a minus b such that (alpha + 1 )(beta + 1) is equal to zero then find the value of b
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Answered by
5
Hi there !!
p(x) = 2x² - 8x + a - b
α + β = -b/a = 8/2 = 4
αβ = a-b / 2
α and β are zeros of p(x)
Given :-
(α + 1) (β + 1 ) = 0
αβ + (α + β) + 1 = 0
a-b/2 + 4 + 1 = 0
a-b/2 + 5 = 0
a-b/2 + 10/2 = 0
[a-b+10]/2 = 0
a-b + 10 = 0
a- b = -10
Hence ,
the polynomial should be :-
p(x) = 2x²- 8x + 10
We cant find the value of "b"
one more relation is required .
I'm just finding the zeros of p(x)
2x² - 10x + 2x -10
= 2x (x -5) + 2(x -5)
=(2x +2) (x -5)
x = -2/2 = -1
x = 5
zeros are -1 and 5
p(x) = 2x² - 8x + a - b
α + β = -b/a = 8/2 = 4
αβ = a-b / 2
α and β are zeros of p(x)
Given :-
(α + 1) (β + 1 ) = 0
αβ + (α + β) + 1 = 0
a-b/2 + 4 + 1 = 0
a-b/2 + 5 = 0
a-b/2 + 10/2 = 0
[a-b+10]/2 = 0
a-b + 10 = 0
a- b = -10
Hence ,
the polynomial should be :-
p(x) = 2x²- 8x + 10
We cant find the value of "b"
one more relation is required .
I'm just finding the zeros of p(x)
2x² - 10x + 2x -10
= 2x (x -5) + 2(x -5)
=(2x +2) (x -5)
x = -2/2 = -1
x = 5
zeros are -1 and 5
Answered by
3
p(x) = 2x²-8x+a-b
α+β = c/a = a-b/2
αβ = -b/a = -(-8)/2 = 8/2 = 4
(α+1)(β+1)=0
(x+a)(x+b) = x²+(a+b)x+ab
(1)²+(α+β)1+αβ = 0
αβ+α+β+1 = 0
4+a-b/2+1=0
a-b/2+5=0
a-b/2+10/2=0
(a-b+10)/2=0
a-b+10=0×2
a-b+10=0
a-b=-10
values of a and b cannot be determined by this information.
I'm finding roots.
2x²-8x+(a-b)
= 2x²-8x-10 as a-b=-10
= 2x²+2x-10x-10
= 2x(x+1)-10(x+1)
= (2x-10)(x+1)
α = 5
β = -1
Hope it helps you ☺
Please mark as Brainliest ☺
α+β = c/a = a-b/2
αβ = -b/a = -(-8)/2 = 8/2 = 4
(α+1)(β+1)=0
(x+a)(x+b) = x²+(a+b)x+ab
(1)²+(α+β)1+αβ = 0
αβ+α+β+1 = 0
4+a-b/2+1=0
a-b/2+5=0
a-b/2+10/2=0
(a-b+10)/2=0
a-b+10=0×2
a-b+10=0
a-b=-10
values of a and b cannot be determined by this information.
I'm finding roots.
2x²-8x+(a-b)
= 2x²-8x-10 as a-b=-10
= 2x²+2x-10x-10
= 2x(x+1)-10(x+1)
= (2x-10)(x+1)
α = 5
β = -1
Hope it helps you ☺
Please mark as Brainliest ☺
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