Plies solve my question .
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Hey !!
Solution ;- we have α+β =-2b/a
αβ = c/a
And , α +β + 2γ = - 2β /α
(α + γ )( β + γ) = c/α
=> 2γ = -2B/A + 2b/a , αβ + γ (α + β) + γ^2 =C/A
=> γ = -B/A + b/a , αβ + γ(α+β + γ) =C/A
Eliminating α, β and γ , we get
c/a + (-B/A + b/a ) [-2b/a - B/A + b/a ] =C/A
=> B^2/A^2 -C/A = b^2 /a^2 - c/a
=> a^2 (B^2 - AC ) = A^2 (B^2 - ac) prooved here
____________________________
Hope this helps !!
@Rajukumar111
Solution ;- we have α+β =-2b/a
αβ = c/a
And , α +β + 2γ = - 2β /α
(α + γ )( β + γ) = c/α
=> 2γ = -2B/A + 2b/a , αβ + γ (α + β) + γ^2 =C/A
=> γ = -B/A + b/a , αβ + γ(α+β + γ) =C/A
Eliminating α, β and γ , we get
c/a + (-B/A + b/a ) [-2b/a - B/A + b/a ] =C/A
=> B^2/A^2 -C/A = b^2 /a^2 - c/a
=> a^2 (B^2 - AC ) = A^2 (B^2 - ac) prooved here
____________________________
Hope this helps !!
@Rajukumar111
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