If `alpha,beta` are the roots of `ax^2+2bx+c=0` and `alpha+delta,beta+delta` be those of `Ax^2+2Bx+C=0` then prove that `(b^2-ac)/(B^2-AC)=(a/A)^2`

Options

A. `(b//B)^(2)`
B. `(a//A)^(2)`
C. `(a^(2)b^(2))//(A^(2)B^(2))`
D. `(ab)//(AB)`

Answer

B

Explanation

Since , `alpha and beta" are the roots of " ax^(2) + 2bx + c=0`
`:. " so" , alpha+beta = - (2b)/a and alphabeta= c/a`
Also `alpha + delta and beta + delta " are the roots of " `
`Ax^(2) + 2Bx+ C = 0`
` "so, sum of the roots" = alpha + beta + 2delta = -(2B)/A " and product of the roots" ( alpha + delta) ( beta + delta) = C/A`
` rArr -(2b)/a + 2delta = - (2B)/A`
`rArr delta = b/a - B/A" "`...(i) and ` (alpha +beta) (beta+ delta) = C/A`
` rArr alpha beta + (alpha + beta) delta + delta^(2) = C/A" "` ...(ii)
Putting value of `delta` from equation (i) in equation (ii),
` c/a-(2b)/a(b/a-B/A)+ (b/a- B/A)^(2) = C/A`
`rArr c/a - (2b^(2))/a^(2) + (2bB)/aA + (b/a)^(2)+(B/A)^(2)-(2bB)/(aA)=C/A`
`rArr c/a - (b/a)^(2) + (B/A)^(2)=C/A`
` B^(2)/A^(2) - C/A = b^(2)/a^(2) - c/a rArr (B^(2)-AC)/A^(2) = (b^(2) - ac)/a^(2)`
` rArr (b^(2)-ac)/(B^(2)-AC) = (a/A)^(2)`