Can someone write all process to find a, here are the task:
25=\frac{24^{2}}{a^{2}}+\frac{a^{2}}{2^{2}}
+130511 \(25=\frac{24^{2}}{a^{2}}+\frac{a^{2}}{2^{2}}\)
25 = 576/a^2 + a^2 / 4 getting a commom denominator on the right, we have
25 = [ 576 * 4 + a^4 ] / [ 4a^2] cross-multiply
25 * 4a^2 = 576 *4 + a^4
100a^2 = 2304 + a^4 rearrange
a^4 - 100a^2 - 2304 = 0 this can be factored as
(a^2 - 36) (a^2 - 64) = 0 setting both factors to 0, we have that
a^2 - 36 = 0 which means that a = ± 6 and
a^2 - 64 = 0 which means that the other two soliutions are a = ± 8
So a = { -8, -6, 6, 8 }
a^4 - 100a^2 - 2304 = 0 this can be factored as
(a^2 - 36) (a^2 - 64) = 0
But how it can be factored? if (-36)*(-64)=2304 but you write -2304, and i need to get just one solution.
Thanks anyway for your help.
Solve for a:
25 = a^2/4+576/a^2
25 = a^2/4+576/a^2 is equivalent to a^2/4+576/a^2 = 25:
a^2/4+576/a^2 = 25
Bring a^2/4+576/a^2 together using the common denominator 4 a^2:
(a^4+2304)/(4 a^2) = 25
Multiply both sides by 4 a^2:
a^4+2304 = 100 a^2
Subtract 100 a^2 from both sides:
a^4-100 a^2+2304 = 0
Substitute x = a^2:
x^2-100 x+2304 = 0
The left hand side factors into a product with two terms:
(x-64) (x-36) = 0
Split into two equations:
x-64 = 0 or x-36 = 0
Add 64 to both sides:
x = 64 or x-36 = 0
Substitute back for x = a^2:
a^2 = 64 or x-36 = 0
Take the square root of both sides:
a = 8 or a = -8 or x-36 = 0
Add 36 to both sides:
a = 8 or a = -8 or x = 36
Substitute back for x = a^2:
a = 8 or a = -8 or a^2 = 36
Take the square root of both sides:
Answer: |a = 8 or a = -8 or a = 6 or a = -6
