Advertisements
Advertisements
प्रश्न
Solve for x : `9^(x + 2) -6.3^(x + 1) + 1 = 0`.
Advertisements
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.
संबंधित प्रश्न
Determine the nature of the roots of the following quadratic equation:
(x - 2a)(x - 2b) = 4ab
Determine the nature of the roots of the following quadratic equation:
(b + c)x2 - (a + b + c)x + a = 0
Find the value of k for which the roots are real and equal in the following equation:
3x2 − 5x + 2k = 0
Find the values of k for which the roots are real and equal in each of the following equation:
4x2 - 2(k + 1)x + (k + 4) = 0
Determine the nature of the roots of the following quadratic equation :
x2 -5x+ 7= 0
For what values of k, the roots of the equation x2 + 4x +k = 0 are real?
Choose the correct answer from the given four options :
If the equation 3x² – kx + 2k =0 roots, then the the value(s) of k is (are)
If α, β are roots of the equation x2 + 5x + 5 = 0, then equation whose roots are α + 1 and β + 1 is:
If the coefficient of x2 and the constant term of a quadratic equation have opposite signs, then the quadratic equation has real roots.
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.
