Hope this helps!Solving Simultaneous Equations with Square Roots

In summary, to solve these equations without solving a 4th degree polynomial, it is helpful to use new variables to remove the square roots and create a new equation. This new equation can then be split into two quadratic equations and solved for two unknown values, c and d. From there, the solutions for x and y can be found by substituting c and d back into the original equations.
  • #1
benf.stokes
71
0
Hi

How do you solve this equations analyticaly without solving a 4th degree polynomial
sqrt(x)+y=11; x+sqrt(y)=7

Thanks
 
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  • #2
I'll give a general outline on how to solve this:

First of all, use new variables. a = sqrt(x) and b = sqrt(y). That will remove those ugly square roots for now.

Then, add the 2 equations and form a new equation a + a² + b + b² = 18. In this equation, 18 can be split up into 2 unknown values c and d to form 2 quadratic equations. Of these 2 values, you only need to know 1 to calculate the other, that is d = 18 - c. The roots of these 2 quadratic equations can be written in terms of these 2 values. Keeping in mind that the solutions can only be positive, that should lead you to the right answer
 
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  • #3
I may be misunderstanding but you seem to be saying that if you have [itex]a+ a^2+ b+ b^2= 18= c+ d[/itex] then you must have [itex]a+ a^2= c[/itex] and [itex]b+ b^2= d[/itex]. That is not true.
 
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  • #4
HallsofIvy said:
I may be misunderstanding but you seem to be saying that is you have [itex]a+ a^2+ b+ b^2= 18= c+ d[/itex] then you must have [itex]a+ a^2= c[/itex] and [itex]b+ b^2= d[/itex]. That is not true.

Do u want to explain why this is not true? If I define c as being a+a², and d as being b+b², then a + a² + b + b² = c + d = 18

Edit: nvm, you probably did misunderstand me^^
 
  • #5
Kyouran said:
I'll give a general outline on how to solve this:

First of all, use new variables. a = sqrt(x) and b = sqrt(y). That will remove those ugly square roots for now.

Then, add the 2 equations and form a new equation a + a² + b + b² = 18. In this equation, 18 can be split up into 2 unknown values c and d to form 2 quadratic equations. Of these 2 values, you only need to know 1 to calculate the other, that is d = 18 - c. The roots of these 2 quadratic equations can be written in terms of these 2 values. Keeping in mind that the solutions can only be positive, that should lead you to the right answer

can you plees put full steps for this equation plees
 
  • #6
Hello I'm 14 years old so i don't know what a 4th degree polynomial is... however i agree with the guy named Kyouran. This is how i did it:

sqrt(x) + y = 11 ...(1)
sqrt(y) + x = 7 ...(2)

Let sqrt(x) = a, and sqrt(y) = b.

The eqs will then be:
a + b^2 = 11 ...(3)
b + a^2 = 7 ...(4)

Add them together and you get:

a + a^2 + b + b^2 = 18

The way i solved it is just by doing guess and check. This only takes common sense:

I first tried a = 2 and b = 3 which when substituted into the equation equals 18:

2 + 2^2 + 3 + 3^2 = 18

However when subbing them into eqs 3 and 4, they weren't correct:

So i then switched the values around saying a = 3, and b = 2 which when subbed into eqs 3 and 4 were correct!

And so, as said before...

Since I let sqrt(x) = a, and the sqrt(y) = b:

sqrt (x) = 3
and so therefore x = 9

sqrt (y) = 2
and so therefore y = 4

therefore x = 9, y = 4 ;)

Now you will find, that when subbing these x and y values into your original eqs (1) and (2), they are sure to be correct!
 
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Related to Hope this helps!Solving Simultaneous Equations with Square Roots

1. What are hard simultaneous equations?

Hard simultaneous equations refer to a system of equations that cannot be easily solved using basic algebraic methods. They often involve multiple variables and complex coefficients, making them more challenging to solve.

2. How do you solve hard simultaneous equations?

There are several methods for solving hard simultaneous equations, such as substitution, elimination, and the graphical method. These methods involve manipulating the equations to eliminate variables or create new equations that are easier to solve.

3. What makes simultaneous equations difficult?

Simultaneous equations can be difficult for several reasons. They may involve multiple variables, coefficients that are not easily simplified, or equations that are not in a standard form. Additionally, some systems of equations may have no solution or infinite solutions, making them more challenging to solve.

4. Are there any tips for solving hard simultaneous equations?

One helpful tip for solving hard simultaneous equations is to start by simplifying the equations as much as possible. This can involve combining like terms, factoring, or rearranging the equations. It is also important to carefully track your steps and check your solutions to prevent making mistakes.

5. Can technology be used to solve hard simultaneous equations?

Yes, technology such as graphing calculators and computer software can be useful for solving hard simultaneous equations. They can quickly and accurately solve systems of equations that may be difficult or time-consuming to solve by hand. However, it is still important to understand the underlying concepts and methods for solving simultaneous equations.

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