If `a >0` and discriminant of `a x^2+2b x+c` is negative, then `=a b a x+bb c b x+c a x+bb x+c0` is `+v e` b. `(a c-b)^2(a x^2+2b x+c)` c. `-v e` d. `0`

Options

A. `+ve`
B. `(ac-b)^(2))(ax^(2)+2bx+c)`
C. `-ve`
D. `0`

Answer

C

Explanation

here `a gt 0 " and " 4b^(2) -4ac lt 0 , i.e., ac -b^(2) gt 0`
`:. .ax^(2) +2bx +c gt 0 ,AA x in R `
Now
`Delta = |{:(a,,b,,ax+b),(b,,c,,bx+c),(0,,0,,-(ax^(2)+2bx+c)):}|`
`"[Operating " R_(3) to R_(3) -xR_(1)-R_(2)"]"`
`=-(ax^(2)+2bx+c)(ac-b^(2))`
` =- (+ve) (+ve) =-ve`