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प्रश्न
Solve for x : `9^(x + 2) -6.3^(x + 1) + 1 = 0`.
उत्तर
Given equation `9^(x + 2) -6.3^(x + 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)/(3^2) = 3^-2`
∴ x = -2
Hence, the required root is -2.
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संबंधित प्रश्न
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Determine the nature of the roots of the following quadratic equation :
(x - 1)(2x - 7) = 0
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Find the discriminant of the following equations and hence find the nature of roots: 16x2 - 40x + 25 = 0
(x2 + 1)2 – x2 = 0 has ______.
State whether the following quadratic equation have two distinct real roots. Justify your answer.
2x2 + x – 1 = 0
Complete the following activity to determine the nature of the roots of the quadratic equation x2 + 2x – 9 = 0 :
Solution :
Compare x2 + 2x – 9 = 0 with ax2 + bx + c = 0
a = 1, b = 2, c = `square`
∴ b2 – 4ac = (2)2 – 4 × `square` × `square`
Δ = 4 + `square` = 40
∴ b2 – 4ac > 0
∴ The roots of the equation are real and unequal.
Statement A (Assertion): If 5 + `sqrt(7)` is a root of a quadratic equation with rational co-efficients, then its other root is 5 – `sqrt(7)`.
Statement R (Reason): Surd roots of a quadratic equation with rational co-efficients occur in conjugate pairs.
