Question 148007
Remember, consecutive even integers follow the pattern: {{{2x}}}, {{{2x+2}}}, {{{2x+4}}}, etc.



So if the "product of two consecutive even intergers is 34 larger than their sum", then we have the equation {{{2x(2x+2)=(2x)+(2x+2)+34}}}


{{{2x(2x+2)=2x+2x+2+34}}} Start with the given equation.



{{{4x^2+4x=2x+2x+2+34}}} Distribute.



{{{4x^2+4x=4x+36}}} Combine like terms



{{{4x^2+4x-4x-36=0}}} Subtract {{{4x}}} from both sides. Subtract {{{36}}} from both sides.



{{{4x^2+0x-36=0}}} Combine like terms.



Notice we have a quadratic equation in the form of {{{ax^2+bx+c}}} where {{{a=4}}}, {{{b=0}}}, and {{{c=-36}}}



Let's use the quadratic formula to solve for x



{{{x = (-b +- sqrt( b^2-4ac ))/(2a)}}} Start with the quadratic formula



{{{x = (-(0) +- sqrt( (0)^2-4(4)(-36) ))/(2(4))}}} Plug in  {{{a=4}}}, {{{b=0}}}, and {{{c=-36}}}



{{{x = (-0 +- sqrt( 0-4(4)(-36) ))/(2(4))}}} Square {{{0}}} to get {{{0}}}. 



{{{x = (-0 +- sqrt( 0--576 ))/(2(4))}}} Multiply {{{4(4)(-36)}}} to get {{{-576}}}



{{{x = (-0 +- sqrt( 0+576 ))/(2(4))}}} Rewrite {{{sqrt(0--576)}}} as {{{sqrt(0+576)}}}



{{{x = (-0 +- sqrt( 576 ))/(2(4))}}} Add {{{0}}} to {{{576}}} to get {{{576}}}



{{{x = (-0 +- sqrt( 576 ))/(8)}}} Multiply {{{2}}} and {{{4}}} to get {{{8}}}. 



{{{x = (-0 +- 24)/(8)}}} Take the square root of {{{576}}} to get {{{24}}}. 



{{{x = (-0 + 24)/(8)}}} or {{{x = (-0 - 24)/(8)}}} Break up the expression. 



{{{x = (24)/(8)}}} or {{{x =  (-24)/(8)}}} Combine like terms. 



{{{x = 3}}} or {{{x = -3}}} Simplify. 





So the first number is either {{{2(3)=6}}} or {{{2(-3)=-6}}}


And the second number is either {{{2(3)+2=8}}} or {{{2(-3)+2=-4}}}

  
This means that the two numbers are 6 and 8


OR


the two numbers are -6 and -4