three consecutive even integers are such that the square of the thrid is 76 more than the square of the second. find the three integers.

Question

anonymous · Answer

Let the first even integer be : x

Other will be : (x + 2)

And third will be: (x + 4)

$$(x+4)^2 = 76 +  (x+2)^2$$

$$(x+4)^2 - ( x+2 )^2 = 76$$

Solve for x now..

anonymous · Answer

You can use the formula:

$$(a+b)^2 = a^2 + b^2 + 2ab$$

anonymous · Answer

or we can call the middle one $x$ since the first one does not enter in to the question
solve 
$$x^2+76=(x+2)^2$$
$$x^2+76=x^2+4x+4$$
$$4x+4=76$$
$$4x=72$$
$$x=18$$ so the three are $\{16,18,20\}$ and the smallest one is 16

anonymous · Answer

Is this a factoring method

anonymous · Answer

no factoring needed for this problem

anonymous · Answer

$$x^2+76=(x+2)^2$$ looks like it might be a quadratic equation, but it is not one because there is an $x^2$ on both sides, subtract it from both sides and you have a linear equation

anonymous · Answer

No factoring method, x squared terms will automatically cancel each other you have to solve for x and constant terms...

anonymous · Answer

just learning the quadratic formula trying to understand each method not to be confused. Thank you