(16x-2-8(a-2)x+a-2-6a=0) की जड़ें वास्तविक हों, तो (a) पर सही शर्त क्या है?
For (16x-2-8(a-2)x+a-2-6a=0) to have real roots, what is the correct condition on (a)?
Explanation opens after your attempt
A. \(a\ge1\)
Concept
For real roots, \(D\ge0\) is required. Here (D=64(a-2)2-64\(a^2-6a\)=128(a+2)), so \(a\ge-2\); hence none of these options is exact.
Why this answer is correct
The correct answer is A. \(a\ge1\). For real roots, \(D\ge0\) is required. Here (D=64(a-2)2-64\(a^2-6a\)=128(a+2)), so \(a\ge-2\); hence none of these options is exact.
Exam Tip
वास्तविक जड़ों के लिए \(D\ge0\) चाहिए। यहाँ (D=64(a-2)2-64\(a^2-6a\)=128(a+2)), इसलिए \(a\ge-2\) होगा, अतः विकल्पों में सही शर्त नहीं है।
Login to save your score, XP, coins and progress.
