Dividing by variable when solving algebraic equation

[FallentoReason]

It may have a repeated solution, however. Some quadratic equations have no real solutions, but they all have at least one complex solution.

Could I have an example?

Any equation where the constant is positive will do.

E.g. x^2 + 4

Let's test where it intersects the x axis:

x^2 + 4 = 0

=> x^2 = -4

therefore, x = 2i

[GrandizerII]

It may have a repeated solution, however. Some quadratic equations have no real solutions, but they all have at least one complex solution.

Could I have an example?

x^2 +2x + 1 = 0 => (x+1)^2 = 0 => x = -1 (repeated solution due to the left side of the original equation being a perfect square).

To be clear, repeated solution does not apply to the specific form of quadratic equation that Vic referred to: (i.e., x^2 = constant). Aside from when the constant is 0, there will always be two solutions to such an equation.

[robvalue]

Could I have an example?

Any equation where the constant is positive will do.

E.g. x^2 + 4

Let's test where it intersects the x axis:

x^2 + 4 = 0

=> x^2 = -4

therefore, x = 2i

Also x = -2i

[FallentoReason]

Any equation where the constant is positive will do.

E.g. x^2 + 4

Let's test where it intersects the x axis:

x^2 + 4 = 0

=> x^2 = -4

therefore, x = 2i

Also x = -2i

Oops, thanks for the correction!

[robvalue]

No worries, so easily done

For the general equation

ax^2 + bx + c = 0

where a isn't zero, and a, b and c are real numbers:

you can calculate the value of

d = b^2 - 4ac

When d > 0 you have two real solutions
When d = 0 you have one real solution
When d < 0 you have two complex solutions

The solution(s) are given by

x = [ -b + sqrt(d) ] / 2a

and

x = [ -b - sqrt(d) ] / 2a

sqrt is square root. As you can see, when d = 0, these two solutions are the same. When d < 0 we're taking the square root of a negative number, and as such the result will be complex.

[If you allow a, b and c to be complex the solution formulae still work.]

[FallentoReason]

Ah yeah, good old quadratic equation. Clearly I haven't touched it in a while

[Alex K]

Any equation where the constant is positive will do.

E.g. x^2 + 4

Let's test where it intersects the x axis:

x^2 + 4 = 0

=> x^2 = -4

therefore, x = 2i

Also x = -2i

And here I go with an imaginary heart attack.

[robvalue]

It's like classic Shakespeare

Teacher: What is the square root of 4?

Pupil: Easy! 2!

Teacher: And...

Pupil: ...

Pupil: ...

Pupil: Oh, and minus 2!

Teacher: Too late. *Pulls out gun*

[robvalue]

Ah yeah, good old quadratic equation. Clearly I haven't touched it in a while

One of those things that's forever etched into my brain. I can't remember what I did five fucking minutes ago, but I can recite all that like a parrot.

I dare you to go look up the general solution to a cubic equation

[FallentoReason]

Also x = -2i

And here I go with an imaginary heart attack.

Would it kill you if I said I have an engineering degree...? *gulp!*

In my defence, I study architecture at the moment. Maybe maths just wasn't meant to be