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प्रश्न
Solve the following quadratic equation:
`3^((x + 2)) + 3^(-x) = 10`
योग
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उत्तर
`3^((x + 2)) + 3^(-x) = 10`
`3^(x) · 9 + 1/3^(x) = 10`
Let 3x be equal to y.
∴ `9y + 1/y = 10`
⇒ 9y2 + 1 = 10y
⇒ 9y2 – 10y + 1 = 0
⇒ (y – 1) (9y – 1) = 0
⇒ y – 1 = 0 or 9y – 1 = 0
⇒ y = 1 or y = `1/9`
⇒ 3xx = 1 or 3x = `1/9`
⇒ 3x = 30 or 3x = 3–2
⇒ x = 0 or x = –2
Hence, 0 and –2 are the roots of the given equation.
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