Non linear equation

[tklopfstein]

Solve for w: w-15w²-16=0
-W²(w²+15+16w²)=0
-w²(15² + 15) = 0
-w²(5w+5)(3w+3)=0

w= -1 w= -1

answer should be + - 4 or + - i

Where did I go wrong?

 

[Gene]

I'm unsure what you did to that poor equation in your first step. I'm guessing that it was
w⁴ - 15w² - 16 =0
when you started
(On Webtv (which I use) it came out w? - 15w² - 16)
That factors into
(w^2-16)(w^2+1) = 0
Can you take it from there?

 

[stapel]

w-15w²-16=0
-W²(w²+15+16w²)=0...
Where did I go wrong?

You went wrong in your first step. Try multiplying that back out. Do you get what you started with?

. . . . .-w^2(w^2 + 15 + 16w^2) = -w^4 - 15w^2 - 16w^4 = 15w^4 - 15w^2 = 15w^2(w^2 - 1)

In your "factoring", you changed the equation. And a different equation will have different solutions.

You also have a problem here:

-w²(15² + 15) = 0
-w²(5w+5)(3w+3)=0

Again, check your "factoring" by multiplying back:

. . . . .(5w + 5)(3w + 3) = (5w)(3w) + (5)(3w) + (5w)(3) + (5)(3) = 15w^2 + 30w + 15

...not 15w^2 + 15. Perhaps it might be beneficial to review factoring...?

Eliz.

 

[tklopfstein]

thank you that makes sense. When the instructor was going over some same problems she had factored out exponents to make every thing an exponent of 2, so I thought that's what I shuold do. Thanks so much for the help,all of you really appreciate it.

 

[tklopfstein]

w4 - 15w² - 16 =0

(w²-16) (w²+1)=0
(w+4) (w-4) (w+1) (w-1)=0
W= 4 or -4 w=-1

w= 4 or -4 w= -1


How does the -1 become + - i

 

[stapel]

...(w²+1)=0
(w+1) (w-1)=0...
How does the -1 become + - i

How does the SUM of squares factor?

Eliz.

 

[tklopfstein]

sum of squares factors by making the right side of the equation a square root but in this case, i is not in square root form right?

 

[stapel]

Sums of square do not factor in the real numbers.

"Factoring" and "solving" are two different processes.

The number "i" is the square root of -1; that is, sqrt(-1) = i.

Eliz.