Hard Mathematics Quadratic Equations Class 10 Level 36

यदि \(x^2+px+25=0\) का एक मूल दूसरे मूल का दुगुना है और दोनों ऋणात्मक हैं, तो (p) क्या होगा?

If one root of \(x^2+px+25=0\) is double the other and both are negative, what is (p)?

Explanation opens after your attempt
Correct Answer

A. \( \frac{15\sqrt{2}}{2}\)

Step 1

Concept

Let the roots be (-r) and (-2r), then \(2r^2=25\) and \(p=3r=\frac{15\sqrt{2}}{2}\). In exams, do not forget to rationalize the denominator.

Step 2

Why this answer is correct

The correct answer is A. \( \frac{15\sqrt{2}}{2}\). Let the roots be (-r) and (-2r), then \(2r^2=25\) and \(p=3r=\frac{15\sqrt{2}}{2}\). In exams, do not forget to rationalize the denominator.

Step 3

Exam Tip

मूलों को (-r) और (-2r) मानें, तो \(2r^2=25\) और \(p=3r=\frac{15\sqrt{2}}{2}\) है। परीक्षा में हर को परिमेय बनाना न भूलें।

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Mathematics Answer, Explanation and Revision Hints

यदि \(x^2+px+25=0\) का एक मूल दूसरे मूल का दुगुना है और दोनों ऋणात्मक हैं, तो (p) क्या होगा? / If one root of \(x^2+px+25=0\) is double the other and both are negative, what is (p)?

Correct Answer: A. \( \frac{15\sqrt{2}}{2}\). Explanation: मूलों को (-r) और (-2r) मानें, तो \(2r^2=25\) और \(p=3r=\frac{15\sqrt{2}}{2}\) है। परीक्षा में हर को परिमेय बनाना न भूलें। / Let the roots be (-r) and (-2r), then \(2r^2=25\) and \(p=3r=\frac{15\sqrt{2}}{2}\). In exams, do not forget to rationalize the denominator.

Which concept should I revise for this Mathematics MCQ?

Let the roots be (-r) and (-2r), then \(2r^2=25\) and \(p=3r=\frac{15\sqrt{2}}{2}\). In exams, do not forget to rationalize the denominator.

What exam hint can help solve this Mathematics question?

मूलों को (-r) और (-2r) मानें, तो \(2r^2=25\) और \(p=3r=\frac{15\sqrt{2}}{2}\) है। परीक्षा में हर को परिमेय बनाना न भूलें।

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