Question

if alpha and beta are the zeroes of the polynomial 5x2-7x+2,then the sum of their reciprocals is:

give the answer step-wise.

Answer 1

we have 2 find (1/α + 1/β)

now 1/α + 1/β = (α + β)/ α β (taking LCM)

now by the given poly. we get

(α + β) = -b/a = 7/5

α β = c/a = 2/5

so, (α + β)/ α β = (7/5) / (2/5)


= 7/2

So, 1/α + 1/β = (α + β)/ α β = 7/2

Hence, 1/α + 1/β = 7/2

Answer 2

Answer:

the sum of their reciprocals is: 7/2

Step-by-step explanation:

The fashionable shape of a quadratic polynomial is given by ; ax² + bx + c = zero . The feasible values of variable ( or unknown ) for which the polynomial turns into 0 are referred to as its zeros. If α and ß are the zeros of any quadratic polynomial then that polynomial is given as ; x² - (α + ß)x + α•ß

α,β & γ are the zeroes of cubic polynomial P(x)=ax3+bx2+cx+d,(a=zero) then made of their zeroes . If α and β are roots of a Quadratic Equation ax2 + bx + c = zero then, α + β = -b/a.

Since, α and β are zeores of 5X²-7X+2

α+β = 7/5

αβ=2/5

now

1/α+1/β = α+β/αβ

=(7/5)/(2/5)

=7/2

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