Question

Solve the equation D⁴+4D³+8D²+8D+1=0

Answer 1

Step-by-step explanation:

Quartic equations can be hard to solve. In your particular case, you need first to know the quartic identity:

( + )⁴ = ⁴ + 4³ + 6²² + 4³ + ⁴

which is true for any value and .

From here setting = and =-1 yields:

( − 1)⁴ = ⁴ − 4³ + 6² − 4 + 1

Thus starting from your equation:

⁴ − 4³ + 8² − 8 + 4 = 0

⁴ − 4³ + 6² − 4 + 1 + 2² − 4 + 3 = 0

( − 1)⁴ + 2² − 4 + 3 = 0 ←←← Since ⁴−4³+6²−4+1=(−1)⁴

( − 1)⁴ + 2² − 4 + 2 + 1 = 0

( − 1)⁴ + 2(² − 2 + 1) + 1 = 0

( − 1)⁴ + 2( − 1)² + 1 = 0 ←←← Since ²−2+²=(−)²

Making a variable change =(−1)², the equation becomes:

² + 2 + 1 = 0

( + 1)² = 0 ←←← Since ²+2+²=(+)²

Using zero product property, you conclude that:

+ 1 = 0

= -1

Replacing by =(−1)² you obtain:

( − 1)² = -1

which clearly shows there is no real solution to your initial equation (since no squared real value can be negative).

Solving it in the complex plane ℂ would follow:

( − 1)² = -1

( − 1)² − ² = 0 ←←← where is the imaginary constant such that ²=-1

( − 1 − )( − 1 + ) = 0 ←←← Since ²−²=(−)(+)

And finally zero product property yields the complex solutions:

{ ₀ = 1 − (multiplicity 2)

{ ₁ = 1 + (multiplicity 2)