Solve x^{6}-7x^{3}-8=0 algebraically. Find real solutions only, if they exist.

Kathleen Rausch

Kathleen Rausch

Answered question

2021-12-26

Solve x67x38=0 algebraically. Find real solutions only, if they exist.

Answer & Explanation

Elois Puryear

Elois Puryear

Beginner2021-12-27Added 30 answers

Put u=x3 then rewrite the equation
u27u8=0
u28u+u8=0
u(u8)+1(u8)=0
(u+1)(u8)=0
u+1=0,u8=0
u=1 or 8u=8
Put u=x3 then solve x
u=1x3=1
(x+1)(x2x+1)=0
x+1=0 or x2x+1=0
x=1 or x=+1±142
x=1±32
x=1±3i2
[ax2+bx+c=0 x=b±b24ac2a]
and u=8
x3=8x3=(2)3x2(2)3=0
(x2)(x2+2x+4)=0
a3b3=(ab)(a2+ab+b2)
x2=0 or x2+2x+2=0
x=2 or x=2±4162
x=2±122
x=2±12i2
x=2±23i2
x=1±3i
Hence real roots are -1 and 2.
Orlando Paz

Orlando Paz

Beginner2021-12-28Added 42 answers

x67x{3}8=0
x3=t
t27t8=0
D=49+32=81
t1=7+92=8
t2=792=1
x3=8
x3=1
x1=2
x2=1

karton

karton

Expert2022-01-04Added 613 answers

x67x38=0Let P3=a, then a27a8=07+49+32a=?a=8a=1x3=8 and x3=1х=2 х=1
Answer: 2, -1

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