First you want to subtract 36
so it looks like this ![\sqrt[4] {(4x+164)^3}=64](https://tex.z-dn.net/?f=%5Csqrt%5B4%5D%20%7B%284x%2B164%29%5E3%7D%3D64)
Then you want to cancel out the square root 4 by raising that to the 4th power (you must do this to both sides)
which is equal to 
Then you take the cube root to both sides [tex]\sqrt[3]{(4x+164)^3}=\sqrt[3]{16777216}[tex]
Then you end up with the equation 4x+164=256
Then subtract 164 to both sides
4x=92
then divide 92 by 4
Then you get x=23
I believe the answer is D
Answer:
- 0 real zeros
- 2 complex zeros
Step-by-step explanation:
The "fundamental theorem of algebra" says a polynomial of degree n will have n zeros. If the polynomial has real coefficients, the complex zeros will appear in conjugate pairs.
The graph of this quadratic (degree = 2) does not cross the x-axis, so there are no real values of x that make y=0. That means the two zeros are both complex.
To solve that we first need to write it as an equation where x is the number:
85 + x^2 = (x-17)^2
85 + x^2 = x^2 - 34x + 289
Now organize the equation by gathering like terms in the same side lf the equation:
You can shift 85 to the other side by subtracting 85 from both sides of the equation:
85 - 85 + x^2 =x^2 34x +289 - 85
x^2 =x^2 + 34x +204
And shift x^2 to the other side by subtracting x^2 from both sides:
x^2 - x^2 =x^2 -x^2 + 34x +204
0 = 34x + 204
Shift 34x to the other side by subtracting 34x from sides
-34x = 34x - 34x + 204
-34x = 204
And now isolate x by dividing both sides of the equation by -34
-34/-34x = 204/-34
x = -6
Thats your awnser , the number is -6
I hope you understood my brief explanation...
