# The discriminant of a quadratic equation is negative. One solution is 3+4i. What is the other solution?

QUESTION POSTED AT 16/04/2020 - 06:13 PM

QUESTION POSTED AT 16/04/2020 - 06:13 PM

It bears repeating: you should have applied the well-known following formulas: "Hello", "Please" and "Thanks".

Anyway...

= = = = = = = = = = = = = = = = = =

The way you stated the problem, there is an infinity of possibilities for the other solution.

► For instance, the quadratic equation:

x² – (6 + 4i)x + (9 + 12i) = 0

has for discriminant:

Δ = (6 + 4i)² – 4(9 + 12i) = 36 – 16 + 48i – 36 – 48i = -16

which is indeed negative.

Its solutions will then be:

x₁ = [(6 + 4i) + 4i]/2 = 3 + 4i

x₂ = [(6 + 4i) – 4i]/2 = 3

And the other solution here is 3.

► If you are not convinced, the quadratic equation:

x² – (6 + 5i)x + (5 + 15i) = 0

has for discriminant:

Δ = (6 + 5i)² – 4(5 + 15i) = 36 – 25 + 60i – 20 – 60i = -9

which is indeed negative.

Its solutions will then be:

x₁ = [(6 + 5i) + 3i]/2 = 3 + 4i

x₂ = [(6 + 5i) – 3i]/2 = 3 + i

And the other solution here is 3+i.

► In fact, every quadratic equation of the form:

x² – [6 + (4 + α)i]x + (3 + 4i)(3 + αi) = 0

where α is any real, has for discriminant:

Δ = [6 + (4 + α)i]² – 4(3 + 4i)(3 + αi)

= 36 – (4 + α)² + 12(4 + α)i – 36 + 16α – 12(4 + α)i

= 16α – (4 + α)²

= 16α – 16 – 8α – α²

= -16 + 8α – α²

= -(α – 4)²

WILL be negative.

Their solutions will then be:

x₁ = [ [6 + (4 + α)i] – (α – 4)i ]/2 = 3 + 4i

x₂ = [ [6 + (4 + α)i] + (α – 4)i ]/2 = 3 + αi

And the other solution will then be is 3+αi.

Since α can take any real value, you'll obtain an infinity of solutions of the form 3+αi.

► So conclusively:

If the discriminant of a quadratic is negative AND one of the solutions is 3+4i, the only thing we can say about the other solution is that its real part must be 3.

Regards, to lizard squad >:0

Anyway...

= = = = = = = = = = = = = = = = = =

The way you stated the problem, there is an infinity of possibilities for the other solution.

► For instance, the quadratic equation:

x² – (6 + 4i)x + (9 + 12i) = 0

has for discriminant:

Δ = (6 + 4i)² – 4(9 + 12i) = 36 – 16 + 48i – 36 – 48i = -16

which is indeed negative.

Its solutions will then be:

x₁ = [(6 + 4i) + 4i]/2 = 3 + 4i

x₂ = [(6 + 4i) – 4i]/2 = 3

And the other solution here is 3.

► If you are not convinced, the quadratic equation:

x² – (6 + 5i)x + (5 + 15i) = 0

has for discriminant:

Δ = (6 + 5i)² – 4(5 + 15i) = 36 – 25 + 60i – 20 – 60i = -9

which is indeed negative.

Its solutions will then be:

x₁ = [(6 + 5i) + 3i]/2 = 3 + 4i

x₂ = [(6 + 5i) – 3i]/2 = 3 + i

And the other solution here is 3+i.

► In fact, every quadratic equation of the form:

x² – [6 + (4 + α)i]x + (3 + 4i)(3 + αi) = 0

where α is any real, has for discriminant:

Δ = [6 + (4 + α)i]² – 4(3 + 4i)(3 + αi)

= 36 – (4 + α)² + 12(4 + α)i – 36 + 16α – 12(4 + α)i

= 16α – (4 + α)²

= 16α – 16 – 8α – α²

= -16 + 8α – α²

= -(α – 4)²

WILL be negative.

Their solutions will then be:

x₁ = [ [6 + (4 + α)i] – (α – 4)i ]/2 = 3 + 4i

x₂ = [ [6 + (4 + α)i] + (α – 4)i ]/2 = 3 + αi

And the other solution will then be is 3+αi.

Since α can take any real value, you'll obtain an infinity of solutions of the form 3+αi.

► So conclusively:

If the discriminant of a quadratic is negative AND one of the solutions is 3+4i, the only thing we can say about the other solution is that its real part must be 3.

Regards, to lizard squad >:0

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