Two consecutive integers are squared. The sum of these squares of these squares is 545. What are the integers?
Fast way:
Half of 545 = 272 must fall between two perfect sequres, namely 256 and 289, which means that the integers are 16 and 17.
Standard way:
Let x be one of the integers, then x+1 is the other. So that:
x^2+(x+1)^2=545
2x^2+2x-544=0
x^2+x-272=0
(x+17)(x-16)=0
So x=16 or x=-17.
The consecutive integers are therefore:
16, 17, or
-17, -16
To solve this problem, let's represent the two consecutive integers using variables. Let's call the first integer "x," and the second integer will be "x + 1" since they are consecutive.
According to the problem, we need to square these two numbers and find their sum, which is equal to 545.
So we can set up the equation:
x^2 + (x + 1)^2 = 545
Now let's solve for x:
x^2 + (x + 1)^2 = 545
x^2 + (x^2 + 2x + 1) = 545
2x^2 + 2x + 1 = 545
2x^2 + 2x - 544 = 0
Now we can solve this quadratic equation. We can either use factoring, completing the square, or the quadratic formula. In this case, let's use factoring:
2x^2 + 2x - 544 = 0
Factoring out a common factor of 2:
2(x^2 + x - 272) = 0
Now, let's factor the quadratic expression inside the parentheses:
2(x - 16)(x + 17) = 0
Setting each factor to zero:
x - 16 = 0 OR x + 17 = 0
Solving each equation:
x = 16 OR x = -17
So, the two consecutive integers are 16 and 17, or -17 and -16.
Let's represent the two consecutive integers as x and x+1.
The square of the first integer, x, is x^2.
The square of the second integer, x+1, is (x+1)^2.
According to the problem, the sum of these squares is 545:
x^2 + (x+1)^2 = 545
Expanding the equation:
x^2 + (x^2 + 2x + 1) = 545
Combining like terms:
2x^2 + 2x + 1 = 545
Subtracting 545 from both sides:
2x^2 + 2x + 1 - 545 = 0
Simplifying:
2x^2 + 2x - 544 = 0
Dividing the entire equation by 2 to simplify it further:
x^2 + x - 272 = 0
Now we can solve this quadratic equation using factoring, completing the square, or the quadratic formula. In this case, we can factor the equation:
(x + 17)(x - 16) = 0
Setting each factor equal to zero:
x + 17 = 0
or
x - 16 = 0
Solving for x:
x = -17
or
x = 16
So the two consecutive integers can be either -17 and -16, or 16 and 17.