Question 165943: Here are 3 equations that I need you to check my answers on:
#1) Solve this equation in qudratic form:
y^4 + 5y^2 - 36 = 0
y^2 + 5y - 36 = 0
(y + 9)(y-4)
y +9 = 0 y-4=9
y = -9 y = 4
y^2 = -9 y^2 = 4
sqr. rt. of y^2 = +/- sqr. rt. of 9 sqr. rt. of y^2 = +/- sqr. rt. of 4
y = =/- 9i y = +/- 2
solutions are 9i, -9i, 2, -2
#2) Solve a radical equation this is reducible to a quadratic equation
sqr. rt. of x-4 + x = 6
sqr. rt. of x-4 = 6-x
(sqr. rt. of x-4)^2 = (6-x)^2
x-4 = (6-x)^2
x-4 = 36 - 6x - 6x = X^2
x-4 = x^2 - 12x + 36
0 = x^2 - 13x + 40
0 = (x-5) (x - 8)
x = 5 x = 8
5 is a solution, 8 is not.
#3) Solve a fractional equation that is reducible to a quadratic equation:
2/y+1 + 1/y-1 = 1
(y+1) (y-1) (2/y+1 + 1/y-1) = 1(y+1)(y-1)
2(y^2-1) (y+1)(y-1) + (y+1)(y-1)
Here I am stuck on this. Please help. Thanks.
Answer by Fombitz(32388)   (Show Source):
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