Hard Mathematics Quadratic Equations Class 10 Level 30

यदि (\(n^2-16\)x-2-3x+7=0) द्विघात समीकरण है, तो (n) पर सही शर्त क्या है?

If (\(n^2-16\)x-2-3x+7=0) is a quadratic equation, what is the correct condition on (n)?

Explanation opens after your attempt
Correct Answer

C. \(n\neq \pm4\)

Step 1

Concept

For the equation to be quadratic, \(n^2-16\neq0\) is needed. Hence both \(n\neq4\) and \(n\neq-4\) are necessary.

Step 2

Why this answer is correct

The correct answer is C. \(n\neq \pm4\). For the equation to be quadratic, \(n^2-16\neq0\) is needed. Hence both \(n\neq4\) and \(n\neq-4\) are necessary.

Step 3

Exam Tip

द्विघात होने के लिए \(n^2-16\neq0\) होना चाहिए। इसलिए \(n\neq4\) और \(n\neq-4\) दोनों जरूरी हैं।

Question me issue ya doubt hai?

Answer, explanation, typing mistake ya suggestion directly hamari team ko bhejein. 📱Helpline (Call / WhatsApp): +91 7272824365

Related Mathematics Questions

FAQs

Mathematics Answer, Explanation and Revision Hints

यदि (\(n^2-16\)x-2-3x+7=0) द्विघात समीकरण है, तो (n) पर सही शर्त क्या है? / If (\(n^2-16\)x-2-3x+7=0) is a quadratic equation, what is the correct condition on (n)?

Correct Answer: C. \(n\neq \pm4\). Explanation: द्विघात होने के लिए \(n^2-16\neq0\) होना चाहिए। इसलिए \(n\neq4\) और \(n\neq-4\) दोनों जरूरी हैं। / For the equation to be quadratic, \(n^2-16\neq0\) is needed. Hence both \(n\neq4\) and \(n\neq-4\) are necessary.

Which concept should I revise for this Mathematics MCQ?

For the equation to be quadratic, \(n^2-16\neq0\) is needed. Hence both \(n\neq4\) and \(n\neq-4\) are necessary.

What exam hint can help solve this Mathematics question?

द्विघात होने के लिए \(n^2-16\neq0\) होना चाहिए। इसलिए \(n\neq4\) और \(n\neq-4\) दोनों जरूरी हैं।

Student Class Required

Select your class first

Quiz questions, daily challenge and practice pages will open according to your selected class. Class 11/12 ke liye stream bhi select karein.