The sum of two squares of TWO consecutive even integers is 340. Find the integers.
first we represent unknowns using variables,,
let x = first number
let x+2 = second number
note that since they are consecutive but both even numbers, the difference between them is 2, that's why the second number is x+2
now we set-up the equation,, since the sum of the squares (meaning both number are raised to 2) is equal to 340,
(x)^2 + (x+2)^2 = 340
x^2 + x^2 + 4x + 4 = 340
combine similar terms and simplify:
2x^2 + 4x + 4 - 340 = 0
2x^2 + 4x - 336 = 0
we can factor out 2 and cancel it:
2(x^2 + 2x - 168) = 0
x^2 + 2x - 168 = 0
since this quadratic equation is factorable,
(x + 14)(x - 12) = 0
x = -14 and x = 12
thus there are two pairs of answers:
(i) x = -14 and x+2 = -12
(ii) x = 12 and x+2 = 14
hope this helps~ :)
So the average of the two squares is 340/2=170.
Sqrt(170)=13 (approx.)
What would you propose for the two consecutive even integers?
Let's assume the first even integer is represented by "x", and the next consecutive even integer is "x + 2".
The sum of their squares can be written as:
x^2 + (x + 2)^2 = 340
Expanding the equation, we get:
x^2 + (x^2 + 4x + 4) = 340
Combining like terms:
2x^2 + 4x + 4 = 340
Subtracting 340 from both sides:
2x^2 + 4x - 336 = 0
Dividing both sides by 2:
x^2 + 2x - 168 = 0
Now, let's factorize the quadratic equation:
(x - 12)(x + 14) = 0
Setting each factor equal to zero and solving for x:
x - 12 = 0 or x + 14 = 0
If x - 12 = 0, then x = 12.
If x + 14 = 0, then x = -14.
Since we are looking for even integers, x = 12 is the solution.
Therefore, the two consecutive even integers are 12 and 14.
To solve this problem, we need to find two consecutive even integers whose squares can be added to equal 340. Let's proceed step by step.
Step 1: Represent the first even integer as 'x'.
Step 2: Since the integers are consecutive, the next even integer can be represented as 'x + 2'.
Step 3: Write the equation that represents the sum of their squares: x^2 + (x + 2)^2 = 340.
Step 4: Simplify and expand the equation: x^2 + (x^2 + 4x + 4) = 340. Combining like terms, we get 2x^2 + 4x + 4 = 340.
Step 5: Subtract 340 from both sides of the equation: 2x^2 + 4x + 4 - 340 = 0. Simplify further: 2x^2 + 4x - 336 = 0.
Step 6: Divide the equation by 2 to simplify it: x^2 + 2x - 168 = 0.
Step 7: Factor the quadratic equation: (x - 12)(x + 14) = 0.
Step 8: Solve for x by setting each factor equal to zero: x - 12 = 0 or x + 14 = 0.
Step 9: Solve for x in each equation: x = 12 or x = -14.
Both solutions are possible, which means there are two sets of consecutive even integers that satisfy the given conditions.
Using x = 12, the integers are 12 and 14.
Using x = -14, the integers are -14 and -12.
So, the two pairs of consecutive even integers whose squares sum up to 340 are (12, 14) and (-14, -12).