Question

If the sum of two positive integers is 24 and the difference of their squares is 48, what is the product of the two integers?(A) 108(B) 119(C) 128(D) 135(E) 143

Answer 1

Answer:

143

Step-by-step explanation:

Denote by x and y such integers. The hypotheses given can be written as:

$x+y=24, x^2-y^2=48$

Use the difference of squares factorization to solve for x-y

$48=x^2-y^2=(x-y)(x+y)=24(x-y)\text{ then }x-y=2$

Remember that

$(x+y)^2=x^2+2xy+y^2$

$(x-y)^2=x^2-2xy+y^2$

Substract the second equation from the first to obtain

$(x+y)^2-(x-y)^2=4xy$

Substituting the known values, we get

$4xy=24^2-2^2=572\text{ then }xy=\frac{572}{4}=143$