# Solving y=8x-2x^2 and y=4x-x^2 for x - Thank You!

• shoopa
In summary, to solve for x in terms of y, you need to use the quadratic formula and rearrange the equation so that x is in the form of (x + b)^2. This can be done by dividing both sides of the equation by a constant and adding a constant to both sides. Then, by taking the square root of both sides, you can get x alone. Remember to add \pm when taking the square root and that a and b can be negative. However, it's important to note that the function y(x) may only be invertible on a certain interval, not on all real numbers.
shoopa
please help, how can i make y=8x-2x^2 go to x= something y

same with y=4x-x^2 goes to x=

thank you for help!

Always by completing the square for quadratic functions.

The method is pretty straight forward. Just divide both sides so that x^2 has a coefficient of 1. Then add on a constant C to both sides so that the right hand side is a square (remember square quadratics have the form X^2 + 2*b + b^2 = (x+b)^2). Then you just need to square root both sides and then it becomes clear how to get the x alone. You may want to add a $$\pm$$ when you take the square root if you are doing that in this class.

so for y = ax^2 +bx just divide both sides by a, add (b/(2a))^2. Then the right hand side can be written out as (x + b/(2a))^2. Then just square root and add stuff around. Remember nothing says a or b can't be negative. a just can't be 0.

You can express "x" in terms of "y" only on a certain interval.That is to say that the function "y(x)" is invertible only on an interval,and not on R.

Daniel.

Yes, but I think that that won't be a good function, since it doesn't map -to one on any interval.

EDIT:
Ah, I see now that you mean interval of x. Sorry.

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I think that I can correctly derive (without testing) that

$$x = \sqrt { \frac {y} {2} } + \frac {8} {y}$$

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I don't understand. 8*1-2*1^2=6 but 6*1-1^2=5
So the two functions differ at 1 right?
The first case you should get $$x=2\pm 1/2\,\sqrt {16-2\,y}$$
and the second cased
$$x=3/4\pm 1/4\,\sqrt {9-4\,y}$$
Two different answers for two different equations.

y=8x-2x^2--> 8x-2x^2-y=0--> 2x^2-8x+y=0
use quadratic formula!

never mind! that was wrong

no! it's not! it's right! (sorry for not being sure )

'Twas not.It was correct.How else could this problem b solved?

Daniel.

u should be strongwilled baby garfield, the step is right, the use of the quadratic formula gives x in terms of y and that satisfies ur problem dexter

$$y = 8x - 2x^2$$

$$\Rightarrow 2x^2 - 8x + y = 0$$

Apply to quadratic formula: $$x = \frac{ -b \pm \sqrt{b^2 - 4ac}}{2a}$$

a = 2, b = -8 and c = y

$$x = \frac{ 8 \pm \sqrt{(-8^2) - (4 \times 2 \times y)}}{2 \times 2}$$

$$\Rightarrow x = \frac{ 8 \pm \sqrt{64 - 8y}}{4}$$

$$\Rightarrow x = 2 \pm \sqrt{4 - \frac{1}{2}y}$$

The Bob (2004 ©)

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$$y = 4x - x^2$$

$$\Rightarrow x^2 - 4x + y = 0$$

Apply to quadratic formula: $$x = \frac{ -b \pm \sqrt{b^2 - 4ac}}{2a}$$

a = 1, b = -4 and c = y

$$x = \frac{ 4 \pm \sqrt{(-4^2) - (4 \times 1 \times y)}}{2 \times 1}$$

$$\Rightarrow x = \frac{ 8 \pm \sqrt{16 - 4y}}{2}$$

$$\Rightarrow x = 4 \pm \sqrt{4 - y}$$

The Bob (2004 ©)

## What are the equations being solved?

The equations being solved are y=8x-2x^2 and y=4x-x^2.

## What is the value of x that solves these equations?

The value of x that solves these equations can be found by setting the two equations equal to each other and then solving for x.

## What is the process for solving these equations?

The process for solving these equations involves setting the two equations equal to each other, simplifying the resulting equation, and then using the quadratic formula or factoring to find the value(s) of x.

## Are there any other ways to solve these equations?

Yes, there are other ways to solve these equations such as graphing both equations and finding the point(s) of intersection, or using substitution to solve for one variable and then plugging that value into the other equation to solve for the remaining variable.

## What is the importance of solving these equations?

Solving these equations allows us to find the values of x that make both equations true, which can be useful in various mathematical and scientific applications. It also helps us understand the relationship between the two equations and how they intersect on a graph.

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