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If the roots of the quadratic equation \(3x^2+mx+12=0\) are real and equal, what is the value of \(m^2\)?यदि समीकरण \(3x^2+mx+12=0\) के मूल वास्तविक और समान हैं, तो \(m^2\) का मान क्या होगा?
Answer and explanationउत्तर और व्याख्या
Correct answer:सही उत्तर: \(144\)\(144\)
For a quadratic equation \(ax^2+bx+c=0\) to have real and equal roots, its discriminant must be zero: \(D=b^2-4ac=0\). Here, \(a=3\), \(b=m\), and \(c=12\), so \(m^2-4(3)(12)=0\), giving \(m^2=144\). Exam tip: For equal roots, immediately set the discriminant equal to zero.द्विघात समीकरण \(ax^2+bx+c=0\) के समान वास्तविक मूलों के लिए विविक्तकर \(D=b^2-4ac=0\) होता है। यहाँ \(a=3\), \(b=m\) और \(c=12\) हैं, इसलिए \(m^2-4(3)(12)=0\), अर्थात \(m^2=144\)। परीक्षा-युक्ति: समान मूल दिखें तो सबसे पहले \(D=0\) लगाएँ।
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