\(10x^2-41x+40=0\) और \(15x^2-47x+30=0\) में कौनसा मूल समान है?

Which root is common to \(10x^2-41x+40=0\) and \(15x^2-47x+30=0\)?

Explanation opens after your attempt
Correct Answer

A. \(x=\frac{5}{2}\)

Step 1

Concept

The first equation has roots \(\frac{5}{2},\frac{8}{5}\), and the second has roots \(\frac{5}{2},\frac{4}{5}\). In exams, solve both equations separately for the common root.

Step 2

Why this answer is correct

The correct answer is A. \(x=\frac{5}{2}\). The first equation has roots \(\frac{5}{2},\frac{8}{5}\), and the second has roots \(\frac{5}{2},\frac{4}{5}\). In exams, solve both equations separately for the common root.

Step 3

Exam Tip

पहले समीकरण के मूल \(\frac{5}{2},\frac{8}{5}\) और दूसरे के मूल \(\frac{5}{2},\frac{4}{5}\) हैं। परीक्षा में समान मूल के लिए दोनों समीकरण अलग हल करें।

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Mathematics Answer, Explanation and Revision Hints

\(10x^2-41x+40=0\) और \(15x^2-47x+30=0\) में कौनसा मूल समान है? / Which root is common to \(10x^2-41x+40=0\) and \(15x^2-47x+30=0\)?

Correct Answer: A. \(x=\frac{5}{2}\). Explanation: पहले समीकरण के मूल \(\frac{5}{2},\frac{8}{5}\) और दूसरे के मूल \(\frac{5}{2},\frac{4}{5}\) हैं। परीक्षा में समान मूल के लिए दोनों समीकरण अलग हल करें। / The first equation has roots \(\frac{5}{2},\frac{8}{5}\), and the second has roots \(\frac{5}{2},\frac{4}{5}\). In exams, solve both equations separately for the common root.

Which concept should I revise for this Mathematics MCQ?

The first equation has roots \(\frac{5}{2},\frac{8}{5}\), and the second has roots \(\frac{5}{2},\frac{4}{5}\). In exams, solve both equations separately for the common root.

What exam hint can help solve this Mathematics question?

पहले समीकरण के मूल \(\frac{5}{2},\frac{8}{5}\) और दूसरे के मूल \(\frac{5}{2},\frac{4}{5}\) हैं। परीक्षा में समान मूल के लिए दोनों समीकरण अलग हल करें।