यदि \(3x^2+mx+10=0\) की जड़ें (2:5) के अनुपात में हैं, तो (m) के संभव मान क्या हैं?
If the roots of \(3x^2+mx+10=0\) are in the ratio (2:5), what are the possible values of (m)?
Explanation opens after your attempt
Correct Answer
A. \(7\sqrt{3}\) और \(-7\sqrt{3}\)\(7\sqrt{3}\) and \(-7\sqrt{3}\)
Concept
Let the roots be (2r) and (5r). From \(10r^2=\frac{10}{3}\), \(r=\pm\frac{1}{\sqrt{3}}\), so \(7r=-\frac{m}{3}\) gives \(m=\pm7\sqrt{3}\).
Why this answer is correct
The correct answer is A. \(7\sqrt{3}\) और \(-7\sqrt{3}\) / \(7\sqrt{3}\) and \(-7\sqrt{3}\). Let the roots be (2r) and (5r). From \(10r^2=\frac{10}{3}\), \(r=\pm\frac{1}{\sqrt{3}}\), so \(7r=-\frac{m}{3}\) gives \(m=\pm7\sqrt{3}\).
Exam Tip
जड़ें (2r) और (5r) मानें। \(10r^2=\frac{10}{3}\) से \(r=\pm\frac{1}{\sqrt{3}}\), इसलिए \(7r=-\frac{m}{3}\) से \(m=\pm7\sqrt{3}\)।
Login to save your score, XP, coins and progress.