Question

Divide 12 into 2 parts such that their sum is74

Answer 1

$\huge\pink{\underline{Good-Morning}}$$\huge\blue{\fbox{\fbox{\fbox{\red{\mathcal{Hlo}}}}}}$$plz \: thx \: my \: amswer$Let x = one of the numbers

Then the other number = 12 - x

Sum of the squares is 74:

x^2 + (12-x)^2 = 74

Solve for x:

x^2 + 144 - 24x + x^2 = 74

2x^2 - 24x + 70 = 0

x^2 - 12x + 35 = 0

Factor:

(x-7)(x-5) = 0

So the numbers are 5 and 7 

Answer 2

$\bf{\underline{Given}}:-$

• Divide 12 into two parts such that the sum of their square is 74.

To find :-

• The required parts

Solution :-

Let the required part be x and (12 - x)

According to the question,

x² + (12 - x)² = 74

⤇ x² + (12)² - 2 × 12 × x + (x)² = 74

⤇ x² + 144 - 24x + x² = 74

⤇ x² + x² - 24x + 144 = 74

⤇ 2x² - 24x = 74 - 144

⤇ 2x² - 24x = -70

⤇ 2x² - 24x + 70 = 0

⤇ 2(x² - 12x + 35) = 0

⤇ x² - 12x + 35 = 0

⤇ x² - 7x - 5x + 35 = 0

⤇ x(x - 7) - 5(x - 7) = 0

⤇ (x - 7) (x - 5) = 0

Now,

x - 7 = 0

⤇ x = 0 + 7

⤇ x = 7

Then,

x - 5 = 0

⤇ x = 0 + 5

⤇ x = 5

Hence,the required parts will be 5 and 7.