Question

if the roots of the quadratic polynomial p(q-r)x2+q(r-p)x+r(p-q)=0 are equal. Then show that 1/p+1/r=2/q​

Answer 1

Questions ⤵

if the roots of the quadratic polynomial p(q-r)x2+q(r-p)x+r(p-q)=0 are equal. Then show that 1/p+1/r=2/q

sᴏʟᴜᴛɪᴏɴ ⤵

$p(q - r)x {}^{2} + q(r - p)x + r(p - q) = 0$

D = 0

so, the root are equal

D =

$b {}^{2} - 4ac$

$(q(r - p))x {}^{2} - 4(p(q - r))(r(p - q)) = 0 \\ q {}^{2} (r {}^{2} + p {}^{2} - 2pr) - 4((pq - pr)(pr - qr)) = 0 \\ q {}^{2} (r {}^{2} + p {}^{2} - 2pr) - 4(p {}^{2} q r - pq {}^{2} r - p {}^{2} r {}^{2} + pqr {}^{2} ) = 0 \\ q {}^{2} r {}^{2} + q {}^{2} p {}^{2} q {}^ -2 pq {}^{2} r - 4p {}^{2} qr + 4pq {}^{2} r + 4p {}^{2} r {}^{2} + 4pqr {}^{2} = 0 \\ q {}^{2} r {}^{2} + p {}^{2} q {}^{2} + 4p {}^{2} r {}^{2} - 4p {}^{2} q{}^{} r + 2pq {}^{2} r {}^{} + 4pqr {}^{2} = 0 \\ (pq + qr - 2pr) {}^{2} = 0 \: \: ((a + b + c) {}^{2} = a {}^{2} + b {}^{2} + c {}^{2} - 2ab + 2bc + 2ac) \\ pq + qr = 2pr \\ dividing \: by \: p = qr \\ 1 \div r + 1 \div p = 2 \div q$

hence, m = 2

Answer = 2

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