I don't know where this leads but you can try substituting t + 1 = u to see if this helps simplify.
Hey everyone!
So there is this equation that's giving me a big headache. The task goes like this:
Solve the following equations:
(t+1)^4 + (t+5)^4 = 82
Book result: t1= -4, t2= -2, t3= -3+5i, t4= -3-5i
I tried to do it in many ways but I can't solve it.
Here is a similar task, if you think it could help you:
(k-7)^4 - 13(k-7)^2 + 36 = 0
z = (k-7)
z^2 - 13z + 36 = 0 [from this we get a=1, b=-13, c=36 and we are going to use the quadratic formula]
z = [ -b +/- √( b^2 - 4ac ) ] / 2a
z = [ 13 +/- √( 169 - 144 ) ] / 2
z = [ 13 +/- 5 ] / 2
z1 = 9
z2 = 4
(k-7)^2 = 9 |√
|k-7|= 3
k1-7 = 3
k1 = 10
k2-7 = -3
k2= 4
(k-7)^2 = 4 |√
|k-7|= 2
k3 = 9
k4 = 5
Results: k1=10, k2=4, k3=9, k4=5.
When you expand this you get a quartic with integer coefficients of the form:
So the rational root theorm tell you that the rational roots (if any) of this are amoung
Check these to find the rational roots, and what you find is that there is a typo: -2 is not a root.
CB
OK, just to let you know, I managed to solve it!
Here it is:
(t+1)^4 + [t + 5]^4 = 82
t+1 = p
p^4 + [p + 4]^4 = 82
p^4 + [p ^2+ 16 + 8p]^2 = 82
p^4 + p^4 + 64p^2 + 256 + 16^3 + 32 p^2 + 256p - 82 = 0
2p^4 + 16p^3 + 96 p^2 + 256 p + 174 = 0
p = -1 => p + 1 = 0 {The left side MUST b equal to the right side of the equation}
2p^3 (p+1) + 14p^2 (p+1) + 82 p (p+1) + 174 (p+1) = 0
(p+1)[2p^3 + 14p^2+ 82 p +174] = 0
p = -3 => p + 3 = 0 {The left side MUST b equal to the right side of the equation}
(p+1)[2p^2 (p+3) + 8p (p+3) + 58 (p+3) ] = 0
(p+1)(p+3) [2p^2 + 8p + 58] = 0
t1 + 1 = p1
t1 + 1 = -1
t1 = -2
t2 + 1 = p2
t2 + 1 = -3
t2 = -4
2p^2 + 8p + 58 = 0 |:2
p^2 + 4p + 29 = 0 => a = 1, b = 4, c = 29
p3,4 = [ -b +/- √( b^2 - 4ac ) ] / 2a
p3,4 = [ -4 +/- √( 16 - 116 ) ] / 2
p3,4 = 2(-2 +/- 5i) / 2
p3 = -2 + 5i
p4 = -2 - 5i