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
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) (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 | w = ±
w = ±
w = ± 3.4878 and w = ± .1644i Two real and two imaginary roots
=
Answer by rwm(914) (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) (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:
152 can be factored as 4*38, so we can simplify the radical:
This simplifies to
But , so ,
If we are restricting ourselves to real numbers, we can't take the square root
of a negative number so the two solutions are: , [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.
2. If the two roots are ,, then the equation can be factored as:
Multiply using FOIL and collect terms: