Expert Mathematics Quadratic Equations Class 10 Level 30

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
Correct Answer

C. \(t\neq \pm8\)

Step 1

Concept

For the equation to be quadratic, the coefficient of \(x^2\) must not be (0). Here \(t^2-64\neq0\), so \(t\neq\pm8\).

Step 2

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 \(t\neq\pm8\).

Step 3

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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Mathematics Answer, Explanation and Revision Hints

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)?

Correct Answer: C. \(t\neq \pm8\). Explanation: द्विघात होने के लिए \(x^2\) का गुणांक (0) नहीं होना चाहिए। यहाँ \(t^2-64\neq0\), इसलिए \(t\neq\pm8\)। / For the equation to be quadratic, the coefficient of \(x^2\) must not be (0). Here \(t^2-64\neq0\), so \(t\neq\pm8\).

Which concept should I revise for this Mathematics MCQ?

For the equation to be quadratic, the coefficient of \(x^2\) must not be (0). Here \(t^2-64\neq0\), so \(t\neq\pm8\).

What exam hint can help solve this Mathematics question?

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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