Treat x^4 as the square of p: x^4 = p^2.
Then x^4 - 5x^2 - 36 = 0 becomes p^2 - 5p - 36 = 0.
This factors nicely, to (p-9)(p+4) = 0. Then p = 9 and p = -4.
Equating 9 and x^2, we find that x=3 or x=-3.
Equating -4 and x^2, we see that there's no real solution.
Show that both x=3 and x=-3 are real roots of x^4 - 5x^2 - 36 = 0.
<h2>
Answer:</h2>
A. 1/3 liters
B. 3/1 liters
<h2>Show-MY-Work</h2>
The graph shows it's 1 block up, as in 1 liter of water, and 3 blocks to the right, as in 3 liters of flour.
<u><em>I hope this helps you.</em></u>
<h2>
Extra:</h2>
I'd like to get brainliest, as I need it to rank up.
Sincerely,
Kai " The Brainliest" Pro
Just letting you know, you don’t have a picture attached to your problem so unfortunately no one can help you.
Thought you'd want to know: If you're talking about parabolas, it's parabolas, not probables. ;)
The standard equation of a a quadratic is y = ax^2 + bx + c. We need to find the values of the coefficients a, b and c.
Taking the first point: When x=3, y=0, so write 0 = a(3)^2 + b(3) + c, or
0 = 9a + 3b + 1c
Do the same for points (-2,3) and (-1,4).
You will have obtained three linear equations in a, b and c:
3= a(-2)^2 + b(-2) + c, or 3 = 4a - 2b + 1c, also
4 = a(-1)^2 + b(-1) + 1c, or 1a - 1b + 1c.
I used matrix operations to solve this system. The results are:
a= -2/5, b= 1/5, c= 21/5
and so the function f(x) is f(x) = (-2/5)x^2 + (1/5)x + 21/5.
Do this need to be simplified or factored?
