यदि (\(t^2-64\)x-2+(t-8)x+5=0) द्विघात समीकरण है, तो (t) पर सही शर्त क्या है?
If (\(t^2-64\)x-2+(t-8)x+5=0) is a quadratic equation, what is the correct condition on (t)?
Explanation opens after your attempt
C. \(t\neq \pm8\)
Concept
For the equation to be quadratic, the coefficient of \(x^2\) must not be (0). Here \(t^2-64\neq0\), so \(t\neq\pm8\).
Why this answer is correct
The correct answer is C. \(t\neq \pm8\). For the equation to be quadratic, the coefficient of \(x^2\) must not be (0). Here \(t^2-64\neq0\), so \(t\neq\pm8\).
Exam Tip
द्विघात होने के लिए \(x^2\) का गुणांक (0) नहीं होना चाहिए। यहाँ \(t^2-64\neq0\), इसलिए \(t\neq\pm8\)।
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