Question

If α and βare the zeroes of a quadratic polynomial x2 + x – 2 then find the value

of 1/α - 1/β​

Answer 1

Solution

Given :-

Let,

• Quadratic Polynomial is , p(x) = x² + x - 2 .
• α and β are the zeroes of this Equation

Find :-

• Value of (1/α - 1/β )

Explantion

Using Formula

★ Sum of zeroes = -( coefficient of x)/(coefficient of x²)

★ Product of zeroes = ( constant part)/(coefficient of x ²)

So, Now

==> Sum of zeroes = -1/1

==> α + β = -1 _______________(1)

And,

==> Product of zeroes = -2/1

==> α . β = -2________________(2)

By, equ(1)

==> α = -1 - β _____________(3)

Keep in equ(2)

==> ( -1 - β ) . β = -2

multiply by (-ve) sign in both side

==> ( 1 + β ) . β = 2

==> β² + β - 2 = 0

==> β² + 2β - β - 2 = 0

==> β(β + 2) - 1( β + 2) = 0

==> ( β + 2)(β - 1) = 0

==> (β + 2) = 0 Or, ( β - 1) = 0

==> β = -2 Or, β = 1

keep value of β in equ(3)

When,

• β = -2

==> α = -1 - (-2)

==> α = -1 + 2

==> α = 1

When,

• β = 1

==> α = -1 - 1

==> α = -2

Now, Calculate value of (1/α - 1/β)

==> (1/α - 1/β) = ( β - α )/α β

But, we know

★ Difference of roots always be positive .

So,

==> (1/α - 1/β) = | ( β - α )| /α β

Now, keep required values

where

• α = 1 , β = -2

==> (1/α - 1/β) = | ( -2 - 1)|/(1×-2)

==> (1/α - 1/β) = | ( -3)|/(-2)

We know,

• Module give always positive value

==> (1/α - 1/β) = 3/(-2)

==> (1/α - 1/β) = -3/2

Hence

• Value will be = -3/2 .

Answer 2

Answer:

-3/2

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