Suppose x and y are real numbers such that xy = 9 and x2y + xy2 + x + y = 100. What is the integer value of x2 + y2? (Target Round #6)
observe that we can factor xy from (x^2)y + x(y^2):
(x^2)y + x(y^2) + x + y = 100:
xy [ x + y ] + x + y = 100
then we can factor (x + y):
(x + y)(xy + 1) = 100
since xy = 9,
(x + y)(9 + 1) = 100
(x + y)(10) = 100
x + y = 10
getting its square,
(x + y)^2 = 100
x^2 + 2xy + y^2 = 100
since xy = 9,
x^2 + y^2 + 2*9 = 100
x^2 + y^2 + 18 = 100
x^2 + y^2 = 82
hope this helps~ :)
Let's solve the given equations step-by-step to find the value of x and y.
1. We are given that xy = 9.
2. We can rewrite the second equation as x^2y + xy^2 + x + y = 100.
3. We can rewrite the equation using the given value of xy:
x^2(9/x) + (9/x)y^2 + x + y = 100.
4. Simplify the equation:
9x + 9y^2 + x^2 + xy = 100.
5. Rearrange the equation:
x^2 + (9 + y)x + (9y^2 - 100) = 0.
6. We can solve this quadratic equation for x using the quadratic formula:
x = [-b ± sqrt(b^2 - 4ac)] / 2a.
Here, a = 1, b = 9 + y, and c = 9y^2 - 100.
The discriminant (b^2 - 4ac) should be greater than or equal to 0 for real solutions.
7. Applying the discriminant:
(9 + y)^2 - 4(9y^2 - 100) ≥ 0.
Expanding and simplifying:
y^2 - 16y + 289 ≥ 0.
8. Solving the quadratic inequality:
(y - 8)^2 ≥ 55.
9. Taking the square root of both sides:
|y - 8| ≥ √55.
This implies that y - 8 ≥ √55 or -(y - 8) ≥ √55.
10. Simplifying:
y ≥ 8 + √55 or y ≤ 8 - √55.
11. Since y is a real number, the only possible values for y are:
y ≥ 8 + √55 or y ≤ 8 - √55.
12. Substituting the value of y back into the equation xy = 9, we can solve for x.
If y = 8 + √55, then x = 9 / (8 + √55).
If y = 8 - √55, then x = 9 / (8 - √55).
13. Evaluating the values of x and y:
For y = 8 + √55, we get x = 9 / (8 + √55) ≈ 0.723.
For y = 8 - √55, we get x = 9 / (8 - √55) ≈ 1.129.
14. Calculating x^2 + y^2:
For y = 8 + √55, we have x^2 + y^2 ≈ (0.723)^2 + (8 + √55)^2 ≈ 72.444.
For y = 8 - √55, we have x^2 + y^2 ≈ (1.129)^2 + (8 - √55)^2 ≈ 72.444.
15. The integer value of x^2 + y^2 is approximately 72.
Therefore, the integer value of x^2 + y^2 is 72.
To find the value of x^2 + y^2, we need to manipulate the given equations in such a way that we can substitute variables and simplify the expression.
From the equation xy = 9, we can rearrange it to y = (9/x) and substitute this value of y in the second equation:
x^2(9/x) + x(9/x)^2 + x + (9/x) = 100
Simplifying the equation further, we get:
9x + 81/x + x + 9/x = 100
Combining like terms, we have:
10x + 90/x = 100
Multiplying through by x to eliminate the fraction, we get:
10x^2 + 90 = 100x
Rearranging this equation, we have:
10x^2 - 100x + 90 = 0
Dividing through by 10, we obtain:
x^2 - 10x + 9 = 0
Now, we can apply the quadratic formula to find the values of x in terms of the discriminant (b^2 - 4ac):
x = [-(-10) ± sqrt((-10)^2 - 4(1)(9))] / (2(1))
Simplifying further:
x = [10 ± sqrt(100 - 36)] / 2
x = [10 ± sqrt(64)] / 2
x = [10 ± 8] / 2
Taking the positive and negative roots separately, we have:
x1 = (10 + 8) / 2 = 9
x2 = (10 - 8) / 2 = 1
Now, we can substitute these values of x back into the original equation xy = 9 to find the corresponding y-values:
For x = 9:
9y = 9
y = 1
For x = 1:
y = 9
Now, we can calculate x^2 + y^2 for both sets of values:
For x = 9, y = 1:
(9)^2 + (1)^2 = 81 + 1 = 82
For x = 1, y = 9:
(1)^2 + (9)^2 = 1 + 81 = 82
Therefore, the integer value of x^2 + y^2 is 82.