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16? using 16, you have (3x2 + 4)(3x2 + 4) = 9x2 + 24x + 16
3x2(x2 - 3)
3x2 - 2y2 = 9x2 + 4y2 - 12xy (subtract 3x2 and add 2y2 to both sides) 0 = 9x2 -3x2 + 4y2 + 2y2 - 12xy 0 = 6x2 + 6y2 - 12xy (divide by 6 to both sides) 0 = x2 + y2 - 12xy or x2 - 2xy + y2 = 0 (x - y)(x - y) = 0 x - y = 0 x = y
The answe is 27x^4 + 6x^3 - 61x^2 - 68x - 42
Given, 3x2 - 4x = -2 Then, 9x2 - 12x = -6; 9x2 - 12x + 4 = -2 {completing the square} ; 3x - 2 = ±i√2 {sq rt of both sides} ; and 3x = 2 ± i√2. Therefore, x = ⅓(2 ± i√2).
16? using 16, you have (3x2 + 4)(3x2 + 4) = 9x2 + 24x + 16
(3x2 + 2x + 1)
-6x + 9x2 - 24 = 3(3x2 - 2x - 8) = 3(3x2 - 6x + 4x - 8) = 3(3x(x - 2) + 4(x - 2)) = 3(3x + 4)(x - 2)
While it is possible to factor 3x2 from both of these and get 3x2(4 - 1), it's a lot easier to subtract 3x2 from 12x2 and get 9x2
3x2(x2 - 3)
3x2 - 2y2 = 9x2 + 4y2 - 12xy (subtract 3x2 and add 2y2 to both sides) 0 = 9x2 -3x2 + 4y2 + 2y2 - 12xy 0 = 6x2 + 6y2 - 12xy (divide by 6 to both sides) 0 = x2 + y2 - 12xy or x2 - 2xy + y2 = 0 (x - y)(x - y) = 0 x - y = 0 x = y
The answe is 27x^4 + 6x^3 - 61x^2 - 68x - 42
Given, 3x2 - 4x = -2 Then, 9x2 - 12x = -6; 9x2 - 12x + 4 = -2 {completing the square} ; 3x - 2 = ±i√2 {sq rt of both sides} ; and 3x = 2 ± i√2. Therefore, x = ⅓(2 ± i√2).
Answer this question…A. x4 + 2x3 + 9x2 + 4 B. x4 + 4x3 + 9x2 + 4 C. x4 + 2x3 + 9x2 + 4x + 4 D. x4 + 2x3 + 9x2 - 4x + 4
= 53 9x3 + 9x2 + 5x1+ 3x1 27 + 18 + 5 + 3
1
9x2 + 12x + 4 = 9x2 + 6x + 6x + 4 = 3x(3x + 2) + 2(3x + 2) = (3x + 2)(3x + 2) = (3x + 2)2