- #1

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How do you solve this equations analyticaly without solving a 4th degree polynomial

sqrt(x)+y=11; x+sqrt(y)=7

Thanks

- Thread starter benf.stokes
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- #1

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How do you solve this equations analyticaly without solving a 4th degree polynomial

sqrt(x)+y=11; x+sqrt(y)=7

Thanks

- #2

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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

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

HallsofIvy

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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

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Do u want to explain why this is not true? If II 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.

Edit: nvm, you probably did misunderstand me^^

- #5

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can you plees put full steps for this equation pleesI'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

- #6

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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!

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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