SOLUTION: 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)

Algebra ->  Quadratic Equations and Parabolas -> SOLUTION: 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)       Log On


   

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) About Me  (Show Source):
You can put this solution on YOUR website!
#1) Almost, sqrt%289%29=3
y%5E2+=+-9
y=0+%2B-+sqrt%28-9%29
y=0+%2B-+3%2Asqrt%28-1%29
y=0+%2B-+3i
.
.
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#2) Correct
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.
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#3)You're on the right track.
Distribute on the left hand side.
%28y%2B1%29%28y-1%29+%282%2F%28y%2B1%29+%2B+1%2F%28y-1%29%29+=+1%28%28y%2B1%29%28y-1%29%29

+2%2A%28y-1%29+%2B+%28y%2B1%29+=+y%5E2-1
+2y-2%2By%2B1+=+y%5E2-1
+3y-1+=+y%5E2-1
+3y=y%5E2
You can factor out a y but before you do that remember that y=0 is a solution.
y=3
Check the answer.
2%2F%28y%2B1%29%2B1%2F%28y-1%29=1
2%2F%280%2B1%29%2B1%2F%280-1%29=1
2-1=1
True.
2%2F%28y%2B1%29%2B1%2F%28y-1%29=1
2%2F%283%2B1%29%2B1%2F%283-1%29=1
2%2F4%2B1%2F2=1
True.
y=0, y=3.