How do I answer this
f(x) = x - 3 (x + 4)^(1/2)
and am asked how to solve for x when y = -5
I have no idea how to do this and know that I could just solve it by guess and checking but I want to know how to do this algebraically when plugging in the -5 and trying to solve for x like so
-5 = x - 3 (x + 4)^(1/2)
and I don't know how to do that
also why is 22/7 pi???? is it just by luck??? kind of random but just have to know
To solve the equation f(x) = x - 3 (x + 4)^(1/2) when y = -5, you need to substitute -5 for f(x) and solve for x algebraically. Here's how you can proceed:
1. Start with the equation: f(x) = x - 3 (x + 4)^(1/2)
2. Substitute -5 for f(x): -5 = x - 3 (x + 4)^(1/2)
3. To simplify the equation, you can first isolate the radical term. Add 3 (x + 4)^(1/2) to both sides of the equation:
-5 + 3 (x + 4)^(1/2) = x
4. Now, the equation can be rearranged to isolate the radical term:
3 (x + 4)^(1/2) = x + 5
5. Square both sides of the equation to eliminate the radical:
[3 (x + 4)^(1/2)]^2 = (x + 5)^2
Simplifying both sides gives: 9(x + 4) = (x + 5)^2
6. Expand and simplify the equation further:
9x + 36 = x^2 + 10x + 25
7. Rearrange the equation as a quadratic equation by moving all terms to one side:
x^2 + (10x - 9x) + (25 - 36) = 0
Simplifying gives: x^2 + x - 11 = 0
At this point, you can solve the quadratic equation using various methods such as factoring, completing the square, or using the quadratic formula. Once you find the solutions for x, you can determine if they satisfy the original equation f(x) = -5.
Regarding your question about the value of pi being represented as 22/7, it is not a matter of luck or randomness. The value 22/7 is an approximation for the mathematical constant pi (π) which represents the ratio of a circle's circumference to its diameter. The true value of pi is an irrational number, meaning it cannot be expressed exactly as a finite decimal or a fraction. The approximation 22/7 is commonly used in calculations because it is a close approximation to pi and is easy to remember, even though it is not as precise as the actual value.