# Math Help - Bi-square equation

1. ## Bi-square equation

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.

(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.

2. I don't know where this leads but you can try substituting t + 1 = u to see if this helps simplify.

3. Originally Posted by IoNForce
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)^2 = 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.
When you expand this you get a quartic with integer coefficients of the form:

$t^4+...-56=0$

So the rational root theorm tell you that the rational roots (if any) of this are amoung $\pm7, \pm 14, \pm 28, \pm 56. \pm 2, \pm 4, \pm 8$

Check these to find the rational roots, and what you find is that there is a typo: -2 is not a root.

CB

4. 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

5. Originally Posted by IoNForce
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
Makes it easier for us if you post the right question from the start, or give a new post pointing out you have changed the question.

CB

6. Yeah, sorry I had a typo and I corrected it as soon as I noticed...
I'll keep it in mind if it happens again.