निम्न प्रत्येक द्विघात समीकरण में श का ऐसा मान ज्ञात कीजिए कि उसके दो बराबर मूल हों।
(I) 
(II) kx (x - 2) + 6 = 0
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हमें पता है कि द्विघात समीकरण में यदि दोनों मूल बराबर हो तो D= 0 होता है

(I)

(II) kx (x - 2) + 6 = 0
(I)
(II) kx (x - 2) + 6 = 0
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