One positive integer is 3 more than 2 times the other positive integer. The difference of their squares is 144.

Question

Solve Applications of Quadratic Equations

Answer 1

Let x = 1st positive integer.

Then 2x+3 represents the 2nd positive integer (3 more than twice the 1st)

(2x+3)²- x²=144 (difference of their squares is 144).

Now solve:

4x²+12x+9 -x²=144 (square the binomial)

3x²+12x-135=0 (set equal to zero and combine similar terms)

x²+4x-45 = 0 (notice that every term of the equation is evenly divisible by 3 --might as well make the numbers smaller)

(x+9)(x-5) = 0 (factor the trinomial into the product of two binomials)

x+9 = 0 or x-5 = 0 (zero product property)

x= -9 or x = 5 (reject the -9 since x represents a positive integer)

If x = 5, then 2x+3 = 13.

13²-5²= 169 - 25 = 144 (check)