[中四]方程 急問 10分

(1) Find the roots of 4^x - 6 (2^x) + 8 = 0 .

4^x - 6 (2^x) + 8 = 0

(2^x)^2 - 6(2^x) + 8 = 0

(2^x - 4)( 2^x - 2 ) = 0

2^x = 4 or 2^x = 2

Therefore x = 2 or x = 1

(2) Solve log(x+3) + log(x-3) = 3log2 + 2log3

log(x+3) + log(x-3) = 3log2 + 2log3

log (x+3)(x-3) = log (2^3) + 1og (3^2)

log (x^2 - 9 ) = log (8)(9)

x^2 - 9 = 72

x^2 = 81

x = -9 (rejected) or x = 9

x= -9 is rejected as log a is not defined for a < 0.

(3) Let the length of the square be x cm.

(x + 3 )(x + 5) = 840

x^2 + 3x + 5x + 15 = 840

x^2 + 8x - 825 = 0

(x - 25)(x + 33) = 0

x = 25 or x = -33 (rejected)

Therefore the length of the square is 25cm.

(4) Find the number of intersections of y = 4x - 1 and y = 2^x - 5x + 8 .

y = 4x - 1 --------- (1)

y = 2^x - 5x + 8 ---------- (2)

Put (1) into (2).

4x - 1 = 2^x - 5x + 8

2x^2 - 9x + 9 = 0

Delta = (-9)^2 - 4(2)(9) = 9 > 0

Therefore the equations have 2 intersections.

(5) Find the value of k if x^2 + y =25 and 6x - y = k have only one solution.

y = 25 - x^2 ------------ (1)

y = 6x - k ------------ (2)

Put (1) into (2).

25 - x^2 = 6x - k

x^2 + 6x + (-k - 25) = 0

Delta = 0 (given)

(6)^2 - 4(1)(-k - 25) = 0

36 + 4k +100 = 0

4k = -136

Therefore k = -34

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