SOLUTION: Find a Quadratic equation with roots of (4+i) and (4-i)?

Algebra ->  Rational-functions -> SOLUTION: Find a Quadratic equation with roots of (4+i) and (4-i)?      Log On


   

Question 216530: Find a Quadratic equation with roots of (4+i) and (4-i)?
Found 2 solutions by stanbon, drj:
Answer by stanbon(75887) About Me  (Show Source):
You can put this solution on YOUR website!
Find a Quadratic equation with roots of (4+i) and (4-i)?
----------------
Rewrite:
y = (x-(x+i))(x-(x-i))
------------------
y = ((x-4)-i)((x-4)+i)
--
Notice this has the form (a-b)(a+b)
---
y = [(x-4)^2 - (i)^2]
---
y = [x^2-8x+16+1]
---
y = x^2-8x+17
=================
Cheers,
Stan H.

Answer by drj(1380) About Me  (Show Source):
You can put this solution on YOUR website!
Find a Quadratic equation with roots of (4+i) and (4-i)?

Step 1. We note that i=sqrt%28-1%29 and i%5E2=-1

Step 2. To get a quadratic equation we use the given roots and then multiply %28x-%284-i%29%29%28x-%284%2Bi%29%29

Step 3. Multiply and simplify

%28x-%284-i%29%29%28x-%284%2Bi%29%29=%28%28x-4%29%2Bi%29%28%28x-4%29-i%29%29

Step 4. We recognize the difference of squares given as %28A-B%29%28A%2BB%29=A%5E2-B%5E2.

Step 5. Then let A=x-4 and B=i

%28x-%284-i%29%29%28x-%284%2Bi%29%29=%28%28x-4%29%5E2-i%5E2%29

%28x-%284-i%29%29%28x-%284%2Bi%29%29=x%5E2-8x%2B16-%28-1%29%29

%28x-%284-i%29%29%28x-%284%2Bi%29%29=x%5E2-8x%2B17%29

We can verify the above quadratic equation by using the quadratic formula given as

x+=+%28-b+%2B-+sqrt%28+b%5E2-4%2Aa%2Ac+%29%29%2F%282%2Aa%29+

where a=1, b=-8, and c=17

Solved by pluggable solver: SOLVE quadratic equation with variable
Quadratic equation ax%5E2%2Bbx%2Bc=0 (in our case 1x%5E2%2B-8x%2B17+=+0) has the following solutons:

x%5B12%5D+=+%28b%2B-sqrt%28+b%5E2-4ac+%29%29%2F2%5Ca

For these solutions to exist, the discriminant b%5E2-4ac should not be a negative number.

First, we need to compute the discriminant b%5E2-4ac: b%5E2-4ac=%28-8%29%5E2-4%2A1%2A17=-4.

The discriminant -4 is less than zero. That means that there are no solutions among real numbers.

If you are a student of advanced school algebra and are aware about imaginary numbers, read on.


In the field of imaginary numbers, the square root of -4 is + or - sqrt%28+4%29+=+2.

The solution is x%5B12%5D+=+%28--8%2B-+i%2Asqrt%28+-4+%29%29%2F2%5C1+=++%28--8%2B-+i%2A2%29%2F2%5C1+

Here's your graph:
graph%28+500%2C+500%2C+-10%2C+10%2C+-20%2C+20%2C+1%2Ax%5E2%2B-8%2Ax%2B17+%29



Ignoring the above graph, we have the same complex roots as given.

Step 6. ANSWER: %28x-%284-i%29%29%28x-%284%2Bi%29%29=x%5E2-8x%2B17%29

I hope the above steps were helpful.

For FREE Step-By-Step videos in Introduction to Algebra, please visit
http://www.FreedomUniversity.TV/courses/IntroAlgebra and for Trigonometry visit
http://www.FreedomUniversity.TV/courses/Trigonometry.

Good luck in your studies!

Respectfully,
Dr J