OpenStudy (anonymous):

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.

5 years ago
OpenStudy (anonymous):

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..

5 years ago
OpenStudy (anonymous):

You can use the formula: \[(a+b)^2 = a^2 + b^2 + 2ab\]

5 years ago
OpenStudy (anonymous):

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

5 years ago
OpenStudy (anonymous):

Is this a factoring method

5 years ago
OpenStudy (anonymous):

no factoring needed for this problem

5 years ago
OpenStudy (anonymous):

\[x^2+76=(x+2)^2\] 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

5 years ago
OpenStudy (anonymous):

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

5 years ago
OpenStudy (anonymous):

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

5 years ago