SOLUTION: Please help. I do not understand 2 problems and I have been working on them for hours now?? Thanks so much! 1. Solve w^4-12w^2-2=0 2. Write the quadratic equation in the var

Algebra ->  Rational-functions -> SOLUTION: Please help. I do not understand 2 problems and I have been working on them for hours now?? Thanks so much! 1. Solve w^4-12w^2-2=0 2. Write the quadratic equation in the var      Log On


   

Question 442413: Please help. I do not understand 2 problems and I have been working on them for hours now?? Thanks so much!
1. Solve w^4-12w^2-2=0

2. Write the quadratic equation in the variable x having the given numbers as solutions. Type the equation in standard form ax^2+bx+c=0.
-(sqrt)2, 7 (sqrt)2
Found 3 solutions by ewatrrr, rwm, htmentor:
Answer by ewatrrr(24785) About Me  (Show Source):
You can put this solution on YOUR website!
 

Hi

w^4-12w^2-2=0   |Let w^2 = x

  x^2 - 12x - 2 = 0

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

x+=+%2812+%2B-+sqrt%28+152+%29%29%2F%282%29+

x+=+%2812+%2B-+sqrt%28+4%2A38+%29%29%2F%282%29+

x+=++6+%2B-+sqrt%2838%29+     | w = ± sqrt%28x%29 

  w = ± sqrt%286+%2B-+sqrt%2838%29%29

  w = ± 3.4878 and w = ± .1644i  Two real and two imaginary roots



%28x%2Bsqrt%282%29%29%28x-7sqrt%282%29%29=+0 = x%5E2+-+6sqrt%282%29x+-+14+=+0









ewatrrr@aol.com

private/pay tutoring Inquiries only, please.

Answer by rwm(914) About Me  (Show Source):
You can put this solution on YOUR website!
The first has two real and two complex solutions.
it can be written as
(w^2-6)^2-38 = 0
w = -sqrt(6+sqrt(38))
w = +sqrt(6+sqrt(38))
w = -i sqrt(sqrt(38)-6)
w = +i sqrt(sqrt(38)-6)
second
(x+sqrt(2))*(x-7sqrt(2))=0
x^2-6sqrt(2)x-14=0

Answer by htmentor(1343) About Me  (Show Source):
You can put this solution on YOUR website!
1. Let x = w^2
Then we can write the equation as
x^2 - 12x - 2 = 0
Since the factorization is not obvious, solve using the quadratic formula:
x+=+%2812+%2B-+sqrt%2812%5E2+%2B+8%29%29%2F2
x+=+%2812+%2B-+sqrt%28152%29%29%2F2
152 can be factored as 4*38, so we can simplify the radical:
x+=+%2812+%2B-+2sqrt%2838%29%29%2F2
This simplifies to x+=+6+%2B-+sqrt%2838%29
But x+=+w%5E2, so w+=+sqrt%28x%29
w+=+sqrt%286+%2B-+sqrt%2838%29%29,-sqrt%286+%2B-+sqrt%2838%29%29
If we are restricting ourselves to real numbers, we can't take the square root
of a negative number so the two solutions are:
w+=+sqrt%286+%2B+sqrt%2838%29%29,-sqrt%286+%2B+sqrt%2838%29%29 [approximately +-3.5]
This is a bit messy, but as a check of our result we can graph the function and look for the zero crossings.
We see that the function crosses the x-axis around x = -3.5 and x = 3.5.
graph%28400%2C400%2C-5%2C5%2C-40%2C40%2Cx%5E4-12x%5E2-2%29
2. If the two roots are -sqrt%282%29,7%2Asqrt%282%29, then the equation can be factored as:
%28x%2Bsqrt%282%29%29%28x-7sqrt%282%29%29+=+0
Multiply using FOIL and collect terms:
x%5E2+-6sqrt%282%29x+-+14+=+0