Question

State true or false:

Two natural numbers which differ by $4$ and the sum of their squares is $58$ are $7$ and $4$.

• A. True
• B. False

Answer 1

The correct option is B False

Let the numbers be $a$ and $b$
$a-b=4$
$\therefore a=b+4$........(1)
$a^2+b^2=58$
$(b+4{)}^2+b^2=58$......... From (1)
$b^2+8b+16+b^2=58$
$2b^2+8b-42=0$
$b^2+4b-21=0$
$b^2+7b-3b-21=0$
$(b+7)(b-3)=0$
$b=3,-7$
Neglect the negative number. Hence, the numbers are $3$ and $7$