Solving 4th Degree Equation: √x+y=7, √y+x=11

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In summary, the conversation is about solving an equation with two square roots and two variables. The participants discuss different approaches and solutions, including using the quadratic formula, isolating one variable and substituting it into the other equation, and creating two quadratic equations. The final solution is x=9 and y=4.
  • #1
obing
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√x+y=7
√y+x=11
solving this equation using x=-b√b2-4ac/2a for finding the root is impossible

the equation turns out to be of 4th degree

or two quadratic eq. can anybody help

this isn't a homework , just wanted to know ...;)
 
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  • #2
If you're trying to solve for x in these equations, isolate y in either the first or second and then substitute it into the other equation.

[itex]\sqrt{x}+y=7[/itex] (1)

[itex]\sqrt{y}+x=11[/itex] (2)

Rearranging (2) to make y the subject:

[itex]\sqrt{y}=11-x[/itex]

[itex]y=(11-x)^2[/itex]

Now substitute into (1):

[itex]\sqrt{x}+(11-x)^2=7[/itex]

To simplify this it's important to try get rid of the square root. This would imply squaring, but remember if we just squared both sides as it is, something like [itex](\sqrt{a}+b)^2=a+2b\sqrt{a}+b^2[/itex] so we don't really get rid of the root, we just create more problems.

So instead, isolate the [itex]\sqrt{x}[/itex] and then square both sides.

Yes you will get a quartic (4th degree) equation and it is generally very complicated to solve, unless you're in luck and this question was devised to have nice, rational roots in which case you could then use the rational root theorem to find them.
But I doubt this since it sounds like you created the problem yourself.
 
  • #3
http://www.wolframalpha.com/input/?i=sqrt%28x%29%3D7-%2811-x%29^2

According to this, one real solution, 9, and one complex solution, which I won't attempt to copy...
 
  • #4
Certainly, x=9 is a solution.

If you continue to solve and finally end with the quartic and then solve that for x, you're given 4 real solutions. Some of these however could be extraneous solutions.

But look at what wolfram gives as solutions to the original system of equations:
http://www.wolframalpha.com/input/?i=sqrt(x)+y=7,+sqrt(y)+x=11"

Also, graphmatica draws [itex]\sqrt{y}+x=11[/itex] as the parabola [itex]y=(11-x)^2[/itex] which is fine except that for x>11 it does not satisfy the first equation. And this really stumped me at first.

Well, I'm stumped. Is the only solution x=9 or are there more complex solutions?
 
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  • #5
thanks guys :)

ive found a way out


if we take x= l2

and y=n2

then √l2 +y= 7 (√x+y=7)


√n2+x =11 (√y+x=11)

which becomes l2+y-7=0 (getting rid of root)



n2+x-11=0

2 quadratic equn. :)

ans sloving it gives x=9 y=4
 

Related to Solving 4th Degree Equation: √x+y=7, √y+x=11

1. How do you solve a 4th degree equation?

To solve a 4th degree equation, you must first rearrange the equation to have all terms on one side and a zero on the other. Then, you can use various methods such as factoring, the quadratic formula, or the method of substitution to find the roots of the equation.

2. What is the significance of the square root in this equation?

The square root in this equation indicates that the variable must be a positive number, as negative numbers do not have a real square root. This helps to narrow down the possible solutions to the equation.

3. Can you solve this equation algebraically?

Yes, this equation can be solved algebraically by manipulating the equation and isolating the variables on one side. However, depending on the complexity of the equation, it may be easier to solve using numerical methods.

4. Is there more than one solution to this equation?

Yes, there can be more than one solution to this equation. In fact, a 4th degree equation can have up to 4 distinct solutions.

5. What are some possible methods to solve this equation?

Some possible methods to solve this equation include factoring, the quadratic formula, the method of substitution, or using graphing or calculator tools. Depending on the specific equation, some methods may be easier or more efficient than others.

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