If the roots `alpha, beta` of the equatin `px ^(2) + qx +r=0` are real and of opposite sign (where p,q,r are real coefficient), then the roots of the equation `alpha (x - beta ) ^(2)+ beta ( x- alpha )^(2)=0` are :

Options

A. positive
B. negative
C. real and of opposite sign
D. Imaginary

Answer

C

Explanation

We have, `alpha+beta = - (q)/(p) and alpha beta = (r)/(p)`
Now, `alpha(x-beta)^(2)+beta(x-alpha)^(2) = 0" "...(i)`
`rArr" "x^(2)(alpha+beta) - 4 alpha beta x + alpha beta (alpha + beta) = 0`
`rArr" "pq x^(2) + 4 pr x + qr = 0" "[therefore alpha+beta=-(q)/(p)and alpha beta=(r)/(p)]`
Let D be the discriminant of this equation. Then, `D=16p^(2)r^(2) - 4pq^(2)r = 4pr(4pr - q^(2))`
Since roots of `px^(2) + qx + r = 0` are real and of opposite signs.
Disc `gt 0 and alpha beta lt 0`
`rArr" "q^(2) - 4pr gt 0 and (r)/(p) lt 0 rArr D gt 0`.
Thus, the roots of equation (i) real. Also,
Product of roots `=(r)/(p) lt 0 rArr` Roots are opposite signs