Question

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

Answer 1

Answer:

The two digit number is 64

Step-by-step explanation:

Let the number be 10x + y

Now, According to the Question,

10x + y = x² + y² + 12 ------ 1

Next,

10x + y = 2xy + 16 ----- 2

When we subtract eq.1 and eq.2 we get,

(10x + y) - (10x + y) = x² + y² + 12 - (2xy + 16)

10x + y - 10x - y = x² + y² + 12 - 2xy - 16

0 = x² - 2xy + y² - 4

x² - 2xy + y² = 4

Now, If you observe closely,

(x - y)² = x² - 2xy + y²

Thus,

(x - y)² = 4

So,

x - y = √4

x - y = 2

We are not considering the -ve value because it will give an imaginary solution

Now,

x - y = 2

x = y + 2 ------ 3

Putting eq.3 in eq.2 we get,

10(y + 2) + y = 2y(y + 2) + 16

10y + 20 + y = 2y² + 4y + 16

11y + 20 = 2y² + 4y + 16

2y² + 4y + 16 - (11y + 20) = 0

2y² + 4y + 16 - 11y - 20 = 0

2y² - 7y - 4 = 0

ay² + by + c = 0

where a = 2, b = -7, c = -4

By splitting the middle term method,

Sum = b = -7

Product = a × c = -8

Factors are -1 and 8

2y² - 1y + 8y - 4 = 0

y(2y - 1) + 4(2y - 1) = 0

(2y - 1)(y + 4) = 0

y = 4 or y = 1/2

We know that,

y is a digit so it must be a whole number and can't be a fraction

Thus,

y = 4

Putting y = 4 in eq.3 we get,

x = 4 + 2

x = 6

Thus, the number is 10(6) + 4 = 60 + 4 = 64

We may also think that 46 can also be the two digit number, because it also satisfies the above conditions

But, we can't derive it, Remember I told that we are not considering the -ve answer of the Square root because it would then give us an imaginary solution and so can never reach 46, thus, 46 is not considered.

Hence, the two digit number is 64

Hope it helped and you understood it........All the best