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The product of two numbers is 45 and their difference is 4. The sum of squares of the two numbers is
As we know $$(a-b)^{2}$$ = $$a^{2} + b^{2} - 2ab$$
We assume that first number is a and second number is b hence ab = 45
and a - b = 4
after putiing values we will get $$a^{2} + b^{2}$$ = 106
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