If the roots of the equation `ax^2+bx+c=0(a!=0)` be equal then

Options

A. (a) `|p| ge |P|`
B. (b) `|p| le |P|`
C. (c) `|p| = |P|`
D. (d) All of these

Answer

2

Explanation

`|px^(2) + qx + r| le| Px^(2) + Qx + R| AA x in R" "(1)`
Form the graph we can see that this is possible only when both equations have same roots.
Thus, `alpha` and `beta` are roots of `Px^(2) + Qx + R = 0` and also of
`px^(2) + qx + r=0`
So, from (1),
`|p| |x-alpha| |x - beta| le |P| |x-alpha| |x-beta|`
`rArr |p| le |p|`
Also `|(4pr-q^(2))/(4p)| le |(4PR -Q^(2))/(4p)|`
`rArr |d| le |(p)/(P)| |D|`
`rArr |d| le |D| " "(because |p| le |P|)`