One integer is 5 less than another. The sum of their squares is 157. Find the integers.
To solve this problem, we can use algebraic equations to represent the given information. Let's call the first integer "x" and the second integer "y".
From the information given, we can write the first equation:
x = y - 5
We also know that the sum of their squares is 157. Mathematically, we can represent this as:
x^2 + y^2 = 157
Now we have a system of two equations. We can solve this system by substituting the value of x from the first equation into the second equation.
Substituting x = y - 5, we have:
(y - 5)^2 + y^2 = 157
Expanding and simplifying the equation, we get:
y^2 - 10y + 25 + y^2 = 157
2y^2 - 10y + 25 = 157
Rearranging the equation, we have:
2y^2 - 10y + 25 - 157 = 0
2y^2 - 10y - 132 = 0
Now we can solve the quadratic equation by factoring, using the quadratic formula, or by completing the square. Since this equation can be factored easily, we will factor it:
2(y^2 - 5y - 66) = 0
Factorizing further, we get:
2(y - 11)(y + 6) = 0
Now, we can solve for y by setting each factor equal to zero:
y - 11 = 0 or y + 6 = 0
Solving each equation, we find:
y = 11 or y = -6
Now that we have obtained the possible values for y, we can substitute these values back into the first equation to find the corresponding values of x.
If y = 11, then from the first equation:
x = y - 5 = 11 - 5 = 6
So, one set of integers is x = 6 and y = 11.
If y = -6, then from the first equation:
x = y - 5 = -6 - 5 = -11
So, another set of integers is x = -11 and y = -6.
Therefore, the two pairs of integers that satisfy the given conditions are (6, 11) and (-11, -6).
To summarize,
The integers are (6, 11) and (-11, -6).