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If ЁЭЫ╝, ЁЭЫ╜ are the zeroes of the polynomial f(x) = x2 + x – 2, then `(∝/β-∝/β)`
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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)`
∴ ЁЭЫ╝ + ЁЭЫ╜ =`-1/1` and ЁЭЫ╝ЁЭЫ╜ = `(-2)/1`
⇒ ЁЭЫ╝ + ЁЭЫ╜ = −1 and ЁЭЫ╝ЁЭЫ╜ = −2
Now, `(1/∝-1/β)^2=((β-∝)/(∝β ))`
=`((∝+β)^2-4∝β)/(∝β)^2` ` [тИ╡(β-∝)^2=(∝+β)^2-4∝β]`
=`((-1)^2-4(-2))/((-2)^2)` `[тИ╡∝+β=-1 and ∝β=-2]`
=`((-1)^2-4(-2))/4`
=`9/4`
тИ╡` (1/∝-1/β)^2=9/4`
⇒`1/∝-1/β=+-3/2`
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