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Quadratic Equations ;
1. 4y^4+9=13y^2
2. x3x 1/2+2=0
3. (x5)^2+2(x5)35=0
4. (x2)^23(x2)+2=0
5. 2(x^25)  13(x^25)+20=0
6. 2x 2/3 + 5 1/3=12
7.6x 2/3 5x 1/36=0
8.x^2 + 4x^1=12
9.x^48x^2+7=0
10.x^42x^235=0

Algebra. 
Henry,
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^29y^2) +9 = 0
Arrange the 4 terms to form 2 factorable pairs:
(4y^44y^2)  (9y^29) = 0
4y^2(y^21)  9(y^21) = 0
(y^21)(4y^29) = 0
(y+1)(y1)(2y+3)(2y3) = 0.
y+1 = 0, Y = 1.
y1 = 0, Y = 1.
2y+3 = 0, Y = 3/2.
2y3 = 0, Y = 3/2.
Solution set: Y = 1,1,3/2, and 3/2.
2.
3. (x5)^2 + 2(x5)35 = 0.
x^210x+25 +2x1035 = 0
x^28x20 = 0
(x+2)(x10) = 0
x+2 = 0, X = 2.
x10 = 0, X = 10.
Solution set: X = 2, and 10.
4. Same procedure as #3.
5. 2(x^25)  13(x^25) +20 = 0.
(x^25)(213) + 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, and2.61.
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 + (2x6x) 1 = 0
Arrange the 4 terms into 2 factorable
pairs:
(12x^26x) + (2x1) = 0
6x(2x1) + 1(2x1) = 0
(2x1)(6x+1) = 0.
2x1 = 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^21)(x^27) = 0
(x+1)(x1)(x^27) = 0.
x+1 = 0, X = 1.
x1 = 0, X = 1.
x^27 = 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.