Negative square roots and real numbers

[spardawg]

Decide whether the complex number is a solution of the equation:

1.) x^2-2x+2=0 1+i <-------(complex number)

-All I know is that you have to plug in 1+i to x. but i dont know how to add the 2xi or something to the other numbers to get an actual solution. Please help! Thank you!

 

[JeffM]

Decide whether the complex number is a solution of the equation:

1.) x^2-2x+2=0 1+i <-------(complex number)

-All I know is that you have to plug in 1+i to x. but i dont know how to add the 2xi or something to the other numbers to get an actual solution. Please help! Thank you!

Complex numbers have the form of a + bi, where a and b are real and i is the square root of minus 1.

Now do you know the quadratic formula?

Do you know what the discriminant is?

 

[Mrspi]

Decide whether the complex number is a solution of the equation:

1.) x^2-2x+2=0 1+i <-------(complex number)

-All I know is that you have to plug in 1+i to x. but i dont know how to add the 2xi or something to the other numbers to get an actual solution. Please help! Thank you!

Ok...substitute (1 + i) for x:

x[sup:1hu0srdn]2[/sup:1hu0srdn] - 2x + 2 = 0

(1 + i)[sup:1hu0srdn]2[/sup:1hu0srdn] - 2(1 + i) + 2 = 0

Now, you have to do some algebraic manipulation. (1 + i)[sup:1hu0srdn]2[/sup:1hu0srdn] means (1 + i)(1 + i), or 1 + 2i + i[sup:1hu0srdn]2[/sup:1hu0srdn]

Remember that i[sup:1hu0srdn]2[/sup:1hu0srdn] = -1.....

So, 1 + 2i + i[sup:1hu0srdn]2[/sup:1hu0srdn] is 1 + 2i - 1, or just 2i.....

2i - 2(1 + i) + 2 = 0

Can you finish simplifying??

 

[spardawg]

Thank you so much, but at the end where you left off you distribute -2(1+i), and how do you combine 2i and 2i? is it 4i?

 

[JeffM]

Thank you so much, but at the end where you left off you distribute -2(1+i), and how do you combine 2i and 2i? is it 4i?

Yes 2i + 2i = 4i.

But as mrspi wisely points out below, you really needed to ask about 2i - 2i = 0. And Had I been wise, I would have let mrspi handle the whole thing.

 

[Mrspi]

Thank you so much, but at the end where you left off you distribute -2(1+i), and how do you combine 2i and 2i? is it 4i?

We had

2i - 2(1 + i) + 2 = 0

Distribute...multiply each term inside the parentheses by -2....and you'll have this:

2i - 2 - 2i + 2 = 0

Adding terms containing "i" is just like combining ANY like terms. 2i + 2i = 4i, OR AS WE HAVE IN THIS CASE, 2i - 2i = 0.

2i - 2 - 2i + 2 = 0 becomes
0 = 0

Which is (obviously, I hope!) true. This tells us that (1 + i) is indeed a root or solution of the original equation.