Algebra practice question for Railways, SSC. Review the correct answer and explanation, then continue with related practice below.
If 16 × 4^(x+2) − 16 × 2^(x+1) + 1 = 0, then x = ?
Correct Answerसही उत्तर
A. −4A. −4
Explanationव्याख्या
Since 4 = 2², we have 4^(x+2) = 2^(2x+4).
Therefore, 16 × 2^(2x+4) − 16 × 2^(x+1) + 1 = 0.
Since 16 = 2⁴, this becomes 2^(2x+8) − 2^(x+5) + 1 = 0.
Let y = 2^(x+4).
Then y² − 2y + 1 = 0 ⇒ (y − 1)² = 0 ⇒ y = 1.
Thus, 2^(x+4) = 1 = 2⁰, so x + 4 = 0 ⇒ x = −4.
चूँकि 4 = 2² है, इसलिए 4^(x+2) = 2^(2x+4)। अतः समीकरण 16 × 2^(2x+4) − 16 × 2^(x+1) + 1 = 0 होगा। चूँकि 16 = 2⁴ है, इसलिए 2^(2x+8) − 2^(x+5) + 1 = 0। मान लेते हैं y = 2^(x+4)। तब y² − 2y + 1 = 0 ⇒ (y − 1)² = 0 ⇒ y = 1। अतः 2^(x+4) = 1 = 2⁰, इसलिए x + 4 = 0 ⇒ x = −4।
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