Question

If alpha, beta are the zeros of the polynomial f(x) = x2 – 3x + 2, then find 1/ alpha + 1/beta. tumhare question me sabpe 2 answer hai ​

Answer 1

Step-by-step explanation:

$Answer:</p><p>\begin{gathered}\tt{\alpha + \beta = \frac{-b}{c} = 3}\\\end{gathered}α+β=c−b=3 \ \\begin{gathered}\tt{\alpha \beta = \frac{c}{a} = 2}\\\end{gathered}αβ=ac=2 \\begin{gathered}\tt{\frac{1}{\alpha} + \frac{1}{\beta} = \frac{\alpha + \beta}{\alpha \beta}}\\\end{gathered}α1+β1=αβα+β= \> \begin{gathered}\tt{\frac{3}{2}}\\\end{gathered}23</p><p>______________</p><p></p><p>$

Answer 2

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Given

α and β are the zeroes of the polynomial f(x)

f(x) = x² - 3x + 2

To find

1/α + 1/β

Solution

let us first find the values of :

Sum of the zeroes

=> α + β = -(coefficient of x) /coefficient of x²

=> α + β = -(-3)/1

=> α + β = 3

Product of the zeroes

=> α × β = constant term/ coefficient of x²

=> α × β = 2/1

=> α × β = 2

Now

1/α + 1/β = (α + β)/α × β {Taking LCM}

=> 3/2