Class 10 Math All topics Quadratic Equations

Find a two-digit number which exceeds by $12$ the sum of the squares of its digits and by $16$ the doubled product of its digits.

Solution: let the number be $10a+b$

From the given conditions

$10a+b=a_{2}+b_{2}+12......(1)$

$10a+b=2ab+16..........(2)$

$(1)−(2)⟹(a−b)_{2}=4⟹a−b=2$

$⟹10a+(a−2)=2(a)(a−2)+16$

$⟹2a_{2}−4a+16−10a−a+2=0$

$⟹a=6$

$⟹b=4$

So the number is $64$

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