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sunil kumar

· started a discussion

· 1 Months ago

are you top ranker (Question:110127.0)

please provide a tricky solution , this is not good to solve in exam. you know that already.

Question:
The roots of the equation (q - r)x2 + (r - p)x+ (p - q) = 0 are:
Options:
A) \(\cfrac{(r-p)}{(q-r)},\cfrac{1}{2}\)
B) \(\cfrac{(p-q)}{(q-r)},1\)
C) \(\cfrac{(q-r)}{(p-q)},1\)
D) \(\cfrac{(r-p)}{(p-q)},\cfrac{1}{2}\)
Solution:
Ans: (b) (q - r)x2 + (r - p)x + (p - q) = 0........... (i)

x =\(\cfrac{-(r-p)\pm\sqrt[]{(r-p)^2-4(q-r)(p-q)}}{2(q-r)}\)
\(\left [ since\ x = \cfrac{-b\pm\sqrt[]{b^2-4ac}}{2a} \right ]\)
\(\therefore\) x = (p - r) \(\pm\cfrac{\sqrt[]{r^2+p^2 -2pr-4(pq - rp -q^2+qr)}}{2(q-r)}\)

= \(\cfrac{(p-r)\pm\sqrt[]{r^2+p^2 -2pr-4pq+ 4pr+4q^2-4qr)}}{2(q-r)}\)

   

\(x = \cfrac{(p-r)\pm\sqrt[]{(p-2q+r)^2}}{2(q-r)}\)
\(\cfrac{(p-q)\pm{(p-2q+2r)}}{2(q-r)}\)                                   (ii)..............

By equation (ii) we get the roots of equation (i) roots are

x = \(\cfrac{p-r+p-2q+r}{2(q-r)}, \cfrac{p-r-p+2q-r}{2(q-r)}\)
x = \(\cfrac{p - q}{q-r}\), 1 

Knowledge Expert

· commented

· 1 Months ago

Dear student,
Your issue has been solved , please reattempt the question.
Keep learning,
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