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Question
If ๐ผ, ๐ฝ are the zeroes of the polynomial f(x) = x2 + x – 2, then `(∝/β-∝/β)`
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Solution
By using the relationship between the zeroes of the quadratic polynomial. We have
Sum of zeroes=`(-("Coefficient of x"))/("Coefficient of" x^2)` and Product of zeroes =`("Constant term")/("Coefficient of" x^2)`
∴ ๐ผ + ๐ฝ =`-1/1` and ๐ผ๐ฝ = `(-2)/1`
⇒ ๐ผ + ๐ฝ = −1 and ๐ผ๐ฝ = −2
Now, `(1/∝-1/β)^2=((β-∝)/(∝β ))`
=`((∝+β)^2-4∝β)/(∝β)^2` ` [โต(β-∝)^2=(∝+β)^2-4∝β]`
=`((-1)^2-4(-2))/((-2)^2)` `[โต∝+β=-1 and ∝β=-2]`
=`((-1)^2-4(-2))/4`
=`9/4`
โต` (1/∝-1/β)^2=9/4`
⇒`1/∝-1/β=+-3/2`
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