Question

solve quadratic equations (9^x+2) - (6×3^x+1) + 1 = 0​

Answer 1

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Answer 2

Answer:

Given equation 9x + 2 - 6.3x + 1 + 1 = 0

⇒ 9x.92 – 6.3x.31 + 1 = 0

⇒ 81.(32)x - 18.3x + 1 = 0

⇒ 81.32x – 18.3x + 1 = 0

Putting 3x = y, then it becomes 81y2 – 18y + 1 = 0

⇒ 81y2 – 9y – 9y + 1 = 0

⇒ 9y(9y – 1) – 1(9y – 1) = 0

⇒ (9y- 1)(9y – 1) = 0

⇒ 9y = 1 ⇒ y = 1/9

But 3x = 1/9 = 1/32 = 3-2

∴ x = - 2

Hence, the required root is – 2.