Hard Mathematics Quadratic Equations Class 10 Level 35

\(x^4-10x^2+9=0\) के वास्तविक हल कौनसे हैं?

What are the real solutions of \(x^4-10x^2+9=0\)?

Explanation opens after your attempt
Correct Answer

A. \(x=\pm1,\pm3\)

Step 1

Concept

From \(y^2-10y+9=0\), (y=1,9), so \(x^2=1,9\) and \(x=\pm1,\pm3\). In exams, do not forget to return to (x).

Step 2

Why this answer is correct

The correct answer is A. \(x=\pm1,\pm3\). From \(y^2-10y+9=0\), (y=1,9), so \(x^2=1,9\) and \(x=\pm1,\pm3\). In exams, do not forget to return to (x).

Step 3

Exam Tip

\(y^2-10y+9=0\) से (y=1,9), इसलिए \(x^2=1,9\) और \(x=\pm1,\pm3\) हैं। परीक्षा में वापस (x) के मान निकालना न भूलें।

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FAQs

Mathematics Answer, Explanation and Revision Hints

\(x^4-10x^2+9=0\) के वास्तविक हल कौनसे हैं? / What are the real solutions of \(x^4-10x^2+9=0\)?

Correct Answer: A. \(x=\pm1,\pm3\). Explanation: \(y^2-10y+9=0\) से (y=1,9), इसलिए \(x^2=1,9\) और \(x=\pm1,\pm3\) हैं। परीक्षा में वापस (x) के मान निकालना न भूलें। / From \(y^2-10y+9=0\), (y=1,9), so \(x^2=1,9\) and \(x=\pm1,\pm3\). In exams, do not forget to return to (x).

Which concept should I revise for this Mathematics MCQ?

From \(y^2-10y+9=0\), (y=1,9), so \(x^2=1,9\) and \(x=\pm1,\pm3\). In exams, do not forget to return to (x).

What exam hint can help solve this Mathematics question?

\(y^2-10y+9=0\) से (y=1,9), इसलिए \(x^2=1,9\) और \(x=\pm1,\pm3\) हैं। परीक्षा में वापस (x) के मान निकालना न भूलें।

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