Help

sorry, I am washed out today this is what I have so far
x^2 + 61 =12x
x^2 - 12x + 61 =12x - 12x
x^2 - 12x + 61 = 0

okay tell me is this correct so far?
 
Sissy Devane said:
sorry, I am washed out today this is what I have so far
x^2 + 61 =12x
x^2 - 12x + 61 =12x - 12x
x^2 - 12x + 61 = 0

okay tell me is this correct so far?
Click to expand...

Now apply quadratic formula - and solve for 'x'. Watch out for those "imaginary" numbers (negative numbers under square root)
 
now I am going to use the quadratic formula
a=1, b=-12, c=61

-(-12)+ - sqrt(-12)^2 - 4(1)(61) divided by 2(1)

then I have
12 + - sqrt 144- 244
12+ - sqrt of-100
12 + - 10
answers are 22 and 2 ?
Is this how you complete this problem?
 
Sissy Devane said:
then I have
12 + - sqrt 144- 244
Click to expand...


Hi Sissy:

You forgot to write down the division by 2.

x = [12 ± sqrt(-100)]/2
sqrt of-100
- 10

'
Click to expand...


You did not take the square root of -1.

sqrt(-1) = i

x = [12 ± 10i]/2

Can you simplify this expression?

Cheers,

~ Mark

 
I have a problem with coming trying to make the problem go further
Could I divide the 2 into the 12 and 10?
 
Sissy Devane said:
Could I divide the 2 into the 12 and 10?
Click to expand...


Yes, both 12 and 10 are divisible by 2 because they are both even numbers.

When we have a fraction with addition or subtraction in the numerator, then each term gets divided by the denominator.

Here's how it looks symbollically.

\(\displaystyle \frac{A + B}{C} = \frac{A}{C} + \frac{B}{C}\)

\(\displaystyle \frac{A - B}{C} = \frac{A}{C} - \frac{B}{C}\)


 
okay therefore that means this problem can go another step? so the finally answer would look like this
x=6+5i
right?
 
Sissy Devane said:
so the finally answer would look like this
x=6+5i
Click to expand...


That's one of the two solutions. (Complex solutions always come in conjugate pairs.)

Look at the Quadratic Formula, again, and make sure that you understand the abbreviation "±".

 


Yes, we can use the abbreviation "plus or minus", anytime it's applicable! The Quadratic Formula shows that it's applicable in this case because the formula itself uses the abbreviation to indicate that one solution for x comes from an addition and that a second solution of x comes from a subtraction.

However, I would not type it as "6 + - 5i" because somebody might interpret that to mean adding 6 to the opposite of 5i (i.e., only one expression for x), instead of the abbreviation for two separate expressions for x.

We can type this pair of Complex conjugate solutions in any of the following ways (really, any clear way is valid).

x = 6 +/- 5i

x = 6 +or- 5i

x = 6 ± 5i

x = 6 + 5i or x = 6 - 5i

I get an impression that you could use some more algebra practice. You could try work through the verification of your results, to make sure that both expressions for x lead to a true statement when substituted into the original quadratic equation.

In other words, we need expressions for x such that x^2 - 12x + 61 = 0.

Is that what happens with the following after multiplying everything out (i.e., "expanding the righthand sides") and combining like-terms (i.e., "simplifying")?

(6 + 5i)^2 - 12(6 + 5i) + 61 = 0

and

(6 - 5i)^2 - 12(6 - 5i) + 61 = 0

In order to expand (6 ± 5i)^2, you need to have learned the FOIL algorithm (or something similar).

In order to expand -12(6 ± 5i), you need to understand the Distributive Property.

In order to simplify i^2, you need to understand the definition of i: i = sqrt(-1).

In order to simplify the results (to verify that zero is obtained), you need to understand the concept of like-terms.

Cheers 8-)

 
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