Posted by Unknown on Monday, August 20, 2012 at 8:07pm.
1. 4y^4 + 9 = 13y^2.
4y^4 - 13y^2 + 9 = 0
Uwse the AC method of factoring.
A*C = 4*9 = 36 = (-1)*(-36) = (-4)*(-9).
Use the pair of factors whose sum = -13.
4y^4 + (-4y^2-9y^2) +9 = 0
Arrange the 4 terms to form 2 factorable pairs:
(4y^4-4y^2) - (9y^2-9) = 0
4y^2(y^2-1) - 9(y^2-1) = 0
(y^2-1)(4y^2-9) = 0
(y+1)(y-1)(2y+3)(2y-3) = 0.
y+1 = 0, Y = -1.
y-1 = 0, Y = 1.
2y+3 = 0, Y = -3/2.
2y-3 = 0, Y = 3/2.
Solution set: Y = -1,1,-3/2, and 3/2.
2.
3. (x-5)^2 + 2(x-5)-35 = 0.
x^2-10x+25 +2x-10-35 = 0
x^2-8x-20 = 0
(x+2)(x-10) = 0
x+2 = 0, X = -2.
x-10 = 0, X = 10.
Solution set: X = -2, and 10.
4. Same procedure as #3.
5. 2(x^2-5) - 13(x^2-5) +20 = 0.
(x^2-5)(2-13) + 20 = 0
-11x^2 + 55 + 20 = 0
-11x^2 + 75 = 0
2x^2 -10 -13x^2 + 65 + 20 = 0
-11x^2 + 75 = 0
-11x^2 = -75
x^2 = 6.818
X = 2.61, and-2.61.
8. x^-2 + 4x^-1 = 12.
1/x^2 + 4/x = 12
1 + 4x = 12x^2
12x^2 - 4x - 1 = 0.
A*C = 12*(-1) = -12 = 1(-12) = 2(-6).
Select the pair of factors whose sum=-4:
12x^2 + (2x-6x) -1 = 0
Arrange the 4 terms into 2 factorable
pairs:
(12x^2-6x) + (2x-1) = 0
6x(2x-1) + 1(2x-1) = 0
(2x-1)(6x+1) = 0.
2x-1 = 0, X = 1/2.
6x+1 = 0, X = -1/6.
Solution set: X = -1/6, and 1/2.
9. x^4 - 8x^2 + 7 = 0.
(x^2-1)(x^2-7) = 0
(x+1)(x-1)(x^2-7) = 0.
x+1 = 0, X = -1.
x-1 = 0, X = 1.
x^2-7 = 0, x^2 = 7, X = 2.65, and -2.65.
Solution set: X=-1,1,2.65, and -2.65.
10. Same procedure as #9.