Question 225261
Find 2 consecutive integers such that three times the square of the first is equal to seven more than five times the second.

What is the answer?


Step 1.  Let {{{n}}} be the first integer.


Step 2.  Let {{{n+1}}} be the next consecutive integer


Step 3.  Let {{{3n^2}}} be three times the square of the first.


Step 4.  Let {{{7+5(n+1)}}} be seven more than five times the second.


Step 5.  Then {{{3n^2=7+5(n+1)}}} since three times the square of the first is equal to seven more than five times the second.


Step 6.  Solving the equation in Step 5 yields the following steps.


{{{3n^2=7+5(n+1)}}}


{{{3n^2=7+5n+5=12+5n}}}


Subtract 5n+12 from both sides of the equation


{{{3n^2-5n-12=12+5n-5n-12}}}


{{{3n^2-5n-12=0}}}


Step 7.  To solve use the quadratic equation given as


{{{x = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}}


where a=3, b=-5, and c=-12


*[invoke quadratic "x", 1, -2, -3 ]


With n=3 then n+1=4.  Check {{{3n^2=7+5(n+1)}}} if true so {{{3*3^2=7+5*4}}} or {{{27=27}}} which is a true statement.


With n=-1 then n+1=0  Check {{{3n^2=7+5(n+1)}}} if true so {{{3*(-1)^2=7+5*0}}} or {{{3=7}}} which is not a true statement.


Step 8.  ANSWER:  The numbers are 3 and 4


I hope the above steps were helpful.


For FREE Step-By-Step videos in Introduction to Algebra, please visit http://www.FreedomUniversity.TV/courses/IntroAlgebra and for Trigonometry visit http://www.FreedomUniversity.TV/courses/Trigonometry.


And good luck in your studies!


Respectfully,
Dr J

http://www.FreedomUniversity.TV