If the roots of the equation `a x^2+b x+c=0` are of the form `(k+1)//ka n d(k+2)//(k+1),t h e n(a+b+c)^2` is equal to `2b^2-a c` b. `a 62` c. `b^2-4a c` d. `b^2-2a c`

Options

A. `2b^(2) - ac`
B. `a^(2)`
C. `b^(2) - 4ac`
D. `b^(2) - 2ac`

Answer

3

Explanation

We have `(k + 1)/(k) + (k + 2)/(k + 1) = - b/a" "(1)`
and `(k + 1)/(k) (k + 2)/(k + 1) = c/a`
`rArr (k + 2)/(k) = c/a` or `2/k = c/a - 1 = (c - a)/(a)` or `k = (2a)/(c - a)" "(2)`
Now, eliminate k. Putting the value of k in Eq. (1), we get
`(c + a)/(2a) + (2c)/(c + a) = -b/a`
or `(c + a)^(2) + 4ac = -2b (a + c)`
or `(a + c)^(2) + 2b (a + c) = -4ac`
Adding `b^(2)` to both sides, we have
`(a + b + c)^(2) = b^(2) - 4ac`