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Find the values of m so that the expression x 2 -5mx+4m 2 +1 is always positive for real values of x. Please answer this problem as soon as possible.


2 years ago

Susmita
425 Points
							Whenever u get a problem like this try to write in whole square form.x2-5mx+4m2+1$=x^2 -2.x.\frac{5m}{2}+(\frac{5m}{2})^2-(\frac{5m}{2})^2 +4m^2 +1$$=(x -\frac{5m}{2})^2-(\frac{5m}{2})^2 +4m^2 +1$$=(x -\frac{5m}{2})^2+1-\frac{9m^2}{4}$So for it to be always positive$1-\frac{9m^2}{4}=0$$Or,m=\pm \frac{2}{3}$Please approve if you are help.thank you.

2 years ago
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### Course Features

• 101 Video Lectures
• Revision Notes
• Test paper with Video Solution
• Mind Map
• Study Planner
• NCERT Solutions
• Discussion Forum
• Previous Year Exam Questions