समीकरण \(\sqrt{x}+x^2=0\) को सामान्य रूप में द्विघात क्यों नहीं माना जाता?
Why is \(\sqrt{x}+x^2=0\) not considered a quadratic equation in the usual form?
Explanation opens after your attempt
D. क्योंकि इसमें \(\sqrt{x}\) पद हैBecause it has a \(\sqrt{x}\) term
Concept
The term \(\sqrt{x}\) shows a fractional power of the variable, so it is not in usual quadratic form. A quadratic equation has only \(x^2\), (x), and constant terms.
Why this answer is correct
The correct answer is D. क्योंकि इसमें \(\sqrt{x}\) पद है / Because it has a \(\sqrt{x}\) term. The term \(\sqrt{x}\) shows a fractional power of the variable, so it is not in usual quadratic form. A quadratic equation has only \(x^2\), (x), and constant terms.
Exam Tip
\(\sqrt{x}\) चर की भिन्न घात दिखाता है इसलिए यह सामान्य द्विघात रूप में नहीं है। द्विघात में केवल \(x^2\), (x) और स्थिर पद होते हैं।
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