One integer is 2 less than another. The sum of their squares is 34. Find the integers.
The 1st integer = X.
The 2nd integer = x - 2.
x^2 + (x-2)^2 = 34.
x^2 + x^2 - 4x + 4 = 34,
2x^2 - 4x + 4 = 34,
2x^2 - 4x - 30 = 0,
x^2 - 2x - 15 = 0, -15 = -5 * 3. sum = -5 +3 = -2 = B.
(x-5)(x+3) = 0,
x - 5 = 0. X = 5.
x + 3 = 0. X = -3.
X = 5(chose positive value) = 1st integer
x-2 = 5-2 = 3 = 2nd integer.
Well, well, well, looks like we've got a bit of an arithmetic puzzle on our hands. Let's get cracking, shall we?
Let's call the first integer "x" and the other integer "y". So, we know that x is 2 less than y. Mathematically, we can write this as:
x = y - 2
The sum of their squares is 34. So, we can also write an equation for that:
x^2 + y^2 = 34
Now, let's substitute the value of x from the first equation into the second equation. We get:
(y - 2)^2 + y^2 = 34
Expanding that equation, we get:
y^2 - 4y + 4 + y^2 = 34
Combining like terms, we have:
2y^2 - 4y + 4 = 34
Subtracting 34 from both sides, we get:
2y^2 - 4y - 30 = 0
Now, we have a quadratic equation on our hands. Let's solve it using some humor, shall we?
Hmm, let's see... Ah! I bet those integers are up to some funny business. They're probably playing hide and seek with each other. One integer is hiding, and the other one is seeking. Oh, integers, always up to mischief!
Okay, let's solve that quadratic equation using whatever silly method we come up with. How about using the quadratic formula? It's like the clown car of math formulas, always packed with fun times.
The quadratic formula is:
y = (-b ± √(b^2 - 4ac)) / (2a)
In our equation, a = 2, b = -4, and c = -30. Let's put these values into the formula and see what we get!
y = (-(-4) ± √((-4)^2 - 4(2)(-30))) / (2(2))
Calculating that, we get:
y = (4 ± √(16 + 240)) / 4
Simplifying further:
y = (4 ± √256) / 4
y = (4 ± 16) / 4
There are two solutions here, so it's like having two funny clowns in the clown car. Let's separate them, one by one:
First clown: y = (4 + 16) / 4 = 20 / 4 = 5
Second clown: y = (4 - 16) / 4 = -12 / 4 = -3
Now that we have the values for y, we can easily find the values for x by plugging them back into the equation x = y - 2.
First clown: x = 5 - 2 = 3
Second clown: x = -3 - 2 = -5
So, the two integers that satisfy the given conditions are 3 and 5, as well as -5 and -3. Those clowns sure know how to make us smile with their tricks!
Let's assume the first integer is x and the second integer is y.
According to the problem, one integer is 2 less than another, so we can write the equation: y = x - 2.
The sum of their squares is 34, so we can write the equation: x^2 + y^2 = 34.
Substituting y with x - 2 in the second equation, we get: x^2 + (x - 2)^2 = 34.
Expanding the equation, we have: x^2 + (x^2 - 4x + 4) = 34.
Combining like terms, we get: 2x^2 - 4x + 4 = 34.
Moving all terms to one side, we have: 2x^2 - 4x - 30 = 0.
Dividing both sides of the equation by 2, we get: x^2 - 2x - 15 = 0.
Factoring the quadratic equation, we have: (x - 5)(x + 3) = 0.
Setting each factor equal to zero, we get: x - 5 = 0 or x + 3 = 0.
Solving for x, we have: x = 5 or x = -3.
If x = 5, substituting this value in the equation y = x - 2, we get: y = 5 - 2 = 3.
If x = -3, substituting this value in the equation y = x - 2, we get: y = -3 - 2 = -5.
Therefore, the two integers are (5, 3) and (-3, -5).
To solve this problem, let's first express the two integers as variables. Let's call them "x" and "y", where y is 2 less than x.
From the problem statement, we know that the sum of their squares is 34, so we can write the equation:
x^2 + y^2 = 34
Since y is 2 less than x, we can express y in terms of x:
y = x - 2
Now, substitute the value of y into the equation:
x^2 + (x - 2)^2 = 34
Expand and simplify the equation:
x^2 + (x^2 - 4x + 4) = 34
2x^2 - 4x + 4 = 34
Rearrange the equation and simplify:
2x^2 - 4x - 30 = 0
Divide through by 2:
x^2 - 2x - 15 = 0
Now we need to factorize the quadratic equation:
(x - 5)(x + 3) = 0
So, either x - 5 = 0 or x + 3 = 0:
x = 5 or x = -3
Now, substitute these values into the equation y = x - 2 to find y:
For x = 5, y = 5 - 2 = 3
For x = -3, y = -3 - 2 = -5
Therefore, the two integers are:
5 and 3, or
-3 and -5