Explain this answer please....?

Question

Say if I was to factor x^2+2x-1 and I used the quadratic equation and I end up with 1 plus or minus i square root of 3 over 2. So if that's simplified its 1/2 Plus or minus 1/2 i square root of three. This answer is complete arbitrary but why is like i(square root of 3) over 2 = to 1/2 i square root 3. It is it because there's always a 1 infront of square root 3?

Answer 1

Yes, half of x = x divided by 2. That's what half MEANS. So anytime, (1/2)x = x/2. It's also the rule for multiplying fractions: (1/2)x = (1/2)(x/1) = (1x)/(1*2) = x/2.

Answer 2

I don't see how you get
1 plus or minus i square root of 3 over 2

When I use the quadratic formula on
x^2+2x-1= 0

I have A = 1, B = 2 , and C = -1.

Substituting into
(-B +/- sqrt( b*b - 4 * A * C))/ (2*A)
I get
(-2 +/- sqrt( 4 - 4*1*(-1))) / (2*1)
= (-2 +/- sqrt(8))/2

The discriminant is positive, so there are no imaginary terms.

Answer 3

First off, that equation you gave has no imaginary roots.

By the "completing the square" method:
x^2 + 2x - 1 = 0
x^2 + 2x + 4 = 5
(x + 2)^2 = 5
x + 2 = +/- sqrt(5)
x = +/- sqrt(5) - 2

However, for your specific question, the answer is that i*sqrt(3)/2 = (1/2)i*sqrt(3) because they are completely identical.

Say i*sqrt(3) = z.

z/2 is the same thing as (1/2)z = (0.5)z

Dividing something by 2 is the same thing as multiplying something by one half.

Try it out yourself on some arbitrary numbers to convince yourself that this is true.

Answer 4

Dividing by 2 is the same as multiplying by it's inverse.. or 1/2..

some people consider this as factoring out 1/2 from the equation,

so when you divide: sqrt(3)/ 2.. it is the same as multiplying sqrt(3)* (1/2)... then it is normal to re-arrange so the fraction is in front and you have (1/2)sqrt(3)

isn't it GREAT that you have so many ways to write the SAME thing in mathematics?!?! hehe

Answer 5

I've a feeling that you may have got your answer wrong...

Using quadratic equation for the expression x^2 + 2x - 1..

a= 1, b= 2, c= -1

x= [-b +- (sqrt(b^2 - 4ac))]/2a

Substitute the values...

x= [-2 +- (sqrt(2^2 - 4(1)(-1)))] / 2(1)
= [-2 +- (sqrt(4 + 4))] / 2
= [-2+- sqrt(8)] / 2

x will have to values...

[-2 + sqrt(8)] / 2 and
[-2 - sqrt(8)] / 2

From these values you cannot assume it to be -2/2 +- sqrt(8).. which if you so-called simplify -1 +- sqrt(8).. the final answer will be totally different..

I hope you are somehow clearer... remember though.. The denominator 2 belong to both -2 and sqrt(8)...

If you really want to simplify [-2 + sqrt(8)] / 2...
it will be -1 + (sqrt(8)/2)...


Cheers.. (",)