If `b=0,c<0,` then the roots of`x^2+bx+c=0` are numerically

Options

A. equal
B. equal and opposite in sign
C. distinct
D. no real roots

Answer

B

Explanation

Givent that `b=0` and `clt0` and quadratic equation
`x^(2)+bx+c=0` ……. (i)
Put `b=0` in Eq. (i) we get
`x^(2) +0+c=0`
`implies x^(2)=-c` `[("here" cgt0),( :'-cgt0)]`
`:. x=+-sqrt(-c)`
So, the roots of `x^(2)+bx+c=0` are numerically equal to opposite in sign.