Solve for x : for equation having equal roots : p(q-r)x^2 + q(r-p)x + r(p-q) = 0
Answers
Answered by
1
let s and t are root of equation
s+t=-q (r-p)/p (q-r)
because s=t (equal root)
so, 2t=-q/p {(r-p)/(q-r)-----------(1)
st=r(p-q)/p (q-r)
=> t^2=r (p-q)/p (q-r)
q^2/4p^2 {(r-p)/(q-r)}^2=r (p-q)/p(q-r)
q^2 (r-p)^2=4pr (p-q)(q-r)
s+t=-q (r-p)/p (q-r)
because s=t (equal root)
so, 2t=-q/p {(r-p)/(q-r)-----------(1)
st=r(p-q)/p (q-r)
=> t^2=r (p-q)/p (q-r)
q^2/4p^2 {(r-p)/(q-r)}^2=r (p-q)/p(q-r)
q^2 (r-p)^2=4pr (p-q)(q-r)
Similar questions