यदि \(2x^2+mx+8=0\) की जड़ें (1:2) के अनुपात में हैं, तो (m) के संभव मान क्या हैं?
If the roots of \(2x^2+mx+8=0\) are in the ratio (1:2), what are the possible values of (m)?
Explanation opens after your attempt
Correct Answer
A. \(6\sqrt{2}\) और \(-6\sqrt{2}\)\(6\sqrt{2}\) and \(-6\sqrt{2}\)
Concept
Let the roots be (r) and (2r), then \(2r^2=4\) gives \(r=\pm\sqrt{2}\). Since \(3r=-\frac{m}{2}\), we get \(m=\pm6\sqrt{2}\).
Why this answer is correct
The correct answer is A. \(6\sqrt{2}\) और \(-6\sqrt{2}\) / \(6\sqrt{2}\) and \(-6\sqrt{2}\). Let the roots be (r) and (2r), then \(2r^2=4\) gives \(r=\pm\sqrt{2}\). Since \(3r=-\frac{m}{2}\), we get \(m=\pm6\sqrt{2}\).
Exam Tip
जड़ें (r) और (2r) मानें, तब \(2r^2=4\) से \(r=\pm\sqrt{2}\) मिलता है। योग \(3r=-\frac{m}{2}\), इसलिए \(m=\pm6\sqrt{2}\)।
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