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Why are the roots of (2x^2-(3q+1)x+q=0) always real and distinct?समीकरण (2x^2-(3q+1)x+q=0) के मूल हमेशा वास्तविक और भिन्न क्यों हैं?
Answer and explanationउत्तर और व्याख्या
Correct answer:सही उत्तर: Because (D=9q^2-2q+1>0)क्योंकि (D=9q^2-2q+1>0)
Here (D=(3q+1)^2-8q=9q^2-2q+1). Its own discriminant ((-2)^2-4(9)(1)<0), so it is always positive.यहाँ (D=(3q+1)^2-8q=9q^2-2q+1) है। इसका अपना विविक्तकर ((-2)^2-4(9)(1)<0) और मान सदैव धनात्मक है।
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