- #1

shoopa

- 5

- 0

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

thank you for help!

You are using an out of date browser. It may not display this or other websites correctly.

You should upgrade or use an alternative browser.

You should upgrade or use an alternative browser.

- Thread starter shoopa
- Start date

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.

- #1

shoopa

- 5

- 0

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

thank you for help!

Mathematics news on Phys.org

- #2

Moo Of Doom

- 366

- 1

Always by completing the square for quadratic functions.

- #3

snoble

- 127

- 0

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.

- #4

- 13,358

- 3,454

Daniel.

- #5

Berislav

- 239

- 0

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.

EDIT:

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

Last edited:

- #6

eNathan

- 352

- 2

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

[tex]x = \sqrt { \frac {y} {2} } + \frac {8} {y}[/tex]

[tex]x = \sqrt { \frac {y} {2} } + \frac {8} {y}[/tex]

Last edited:

- #7

snoble

- 127

- 0

So the two functions differ at 1 right?

The first case you should get [tex]x=2\pm 1/2\,\sqrt {16-2\,y}[/tex]

and the second cased

[tex]x=3/4\pm 1/4\,\sqrt {9-4\,y}[/tex]

Two different answers for two different equations.

- #8

baby_garfield

- 7

- 0

y=8x-2x^2--> 8x-2x^2-y=0--> 2x^2-8x+y=0

use quadratic formula!

use quadratic formula!

- #9

baby_garfield

- 7

- 0

never mind! that was wrong

- #10

baby_garfield

- 7

- 0

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

- #11

- 13,358

- 3,454

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

Daniel.

Daniel.

- #12

abia ubong

- 70

- 0

- #13

The Bob

- 1,126

- 0

[tex]y = 8x - 2x^2[/tex]

[tex]\Rightarrow 2x^2 - 8x + y = 0[/tex]

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

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

[tex]x = \frac{ 8 \pm \sqrt{(-8^2) - (4 \times 2 \times y)}}{2 \times 2}[/tex]

[tex]\Rightarrow x = \frac{ 8 \pm \sqrt{64 - 8y}}{4}[/tex]

[tex]\Rightarrow x = 2 \pm \sqrt{4 - \frac{1}{2}y}[/tex]

The Bob (2004 ©)

[tex]\Rightarrow 2x^2 - 8x + y = 0[/tex]

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

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

[tex]x = \frac{ 8 \pm \sqrt{(-8^2) - (4 \times 2 \times y)}}{2 \times 2}[/tex]

[tex]\Rightarrow x = \frac{ 8 \pm \sqrt{64 - 8y}}{4}[/tex]

[tex]\Rightarrow x = 2 \pm \sqrt{4 - \frac{1}{2}y}[/tex]

The Bob (2004 ©)

Last edited:

- #14

The Bob

- 1,126

- 0

[tex]\Rightarrow x^2 - 4x + y = 0[/tex]

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

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

[tex]x = \frac{ 4 \pm \sqrt{(-4^2) - (4 \times 1 \times y)}}{2 \times 1}[/tex]

[tex]\Rightarrow x = \frac{ 8 \pm \sqrt{16 - 4y}}{2}[/tex]

[tex]\Rightarrow x = 4 \pm \sqrt{4 - y}[/tex]

The Bob (2004 ©)

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

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.

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.

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.

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.

- Replies
- 2

- Views
- 1K

- Replies
- 3

- Views
- 2K

- Replies
- 5

- Views
- 1K

- Replies
- 10

- Views
- 1K

- Replies
- 2

- Views
- 2K

- Replies
- 3

- Views
- 974

- Replies
- 3

- Views
- 1K

- Replies
- 1

- Views
- 3K

- Replies
- 2

- Views
- 2K

- Replies
- 4

- Views
- 2K

Share: