Question

If

Answer 1

Answer:

  x + y = 2

Step-by-step explanation:

You want to solve an equation in x and y that is the sum of two square roots equal to a different sum of two square roots.

Solution

This necessarily gets messy. You must square the equation, rearrange it to isolate the radicals, and repeat that process two more times. A number of terms cancel, so the result is a 4th degree equation in x and y that can be factored to the square of the product of two linear trinomials.

Here goes:

  $2x^2 +x +y +9 +2\sqrt{(5 + x + x^2 - y)(4 + x^2 + 2 y)}\\\\=2 x^2 +3x + 3 y+5 + 2\sqrt{(3 + 2 x + x^2)(2 + x + x^2 + 3 y)}\qquad\text{square both sides}\\\\\\x+y-2=\sqrt{(5 + x + x^2 - y)(4 + x^2 + 2 y)}-\sqrt{(3 + 2 x + x^2)(2 + x + x^2 + 3 y)}\\\text{ separate radicals from non-radical terms}$

Repeating the process of squaring both sides and separating the radical terms, we get ...

  $22 + 15 x + 15 x^2 + 4 x^3 + 2 x^4 + 19 y + 6 x y + 4 x^2 y - 3 y^2\\\\=2\sqrt{(3 + 2 x + x^2)(5 + x + x^2 - y)(4 + x^2 + 2 y)(2 + x + x^2 + 3 y)}$

Ater one final squaring of both sides, we get an 8th-degree equation that can be factored with some difficulty to ...

  $(2 + x - x^2 - 7 y + 2 x y + 3 y^2)^2 = 0\\\\(x -3y +1)^2(x+y-2)^2=0$

Zero product rule

Solutions to this are values of x and y that make the factors zero. The second factor is zero when ...

  (x +y -2) = 0   ⇒   x +y = 2 . . . . . . . . the value the problem is looking for

The first factor is zero when ...

  x -3y +1 = 0   ⇒   x = 3y -1

Using this expression to substitute for x in the original radical equation, we get ...

  $2\sqrt{5 - 4 y + 9 y^2}=2\sqrt{2 + 9 y^2}\\\\-4y+3=0\qquad\text{divide by 2, square, subtract 2+9y^2 }\\\\y=\dfrac{3}{4}\\\\x=3\cdot\dfrac{3}{4}-1=\dfrac{5}{4}\\\\x+y=\dfrac{5}{4}+\dfrac{3}{4}=2$

The solution to this equation is x +y = 2.

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Additional comment

As we did with x=3y-1, we can substitute x=2-y into the original radical equation. Doing so gives a tautology, good for all values of x and y.

Attached is a graph of the radical equation. It is the line x+y=2.

Answer 2

The value of 'x + y' from solving the square roots is 2

What is the value of 'x+ y' ?

$\sqrt{x^{2} +2y +4} +\sqrt{x^{2} +x-y+5} =\sqrt{x^{2} +x+3y+2 } +\sqrt{x^{2} +2x+3} \\\\\text{Squaring on both sides},\\\\x^{2} +2y +4 + x^{2} +x-y+5= x^{2} +x+3y+2 +x^{2} +2x+3\\\\2x^{2} +x +y+9 =2x^{2} +3x+3y+5\\\\3x-x +3y-y= 9-5\\\\2x +2y = 4\\\\2 (x + y) = 4\\\\x +y = 2$

Thus , the value of x + y from solving the square roots is 2

What is a square root?

• The opposite of squaring an integer is finding its square root.

• The result of multiplying a number by itself yields its square value, while the square root of a number can be found by looking for a number that, when squared, yields the original value.

• The factor of a number that, when multiplied by itself, yields the original number is known as the square root of the number.

• Special exponents include squares and square roots.

• Every number has two square roots, one of a positive value and one of a negative value, because the square of any number is always a positive number.

To learn more about square roots, refer:

brainly.com/question/428672

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