Question

Divide 20 into two parts such that twice the square of the smaller part is 16 less than the square
of the larger part.
only question no c​

Answer 1

Step-by-step explanation:

Let the smaller part be x

Hence the larger part is (20 – x)

Given,

2(20 – x)^2 = x^2 + 16

⇒ 2(400 – 40x + x^2) = x^2 + 16

⇒ 800 – 80x + 2x^2 – x^2 – 16 = 0

⇒ x^2– 80x + 784 = 0

Solve this to get the value of x

Answer 2

ANSWER:

To Do(+Given):

• Divide 20 into 2 parts,
• 2*Square of smaller part = square of larger part -16

Solution:

Let the bigger part be x.

So, the smaller part will be (20 - x).

We are given that,

⇒ Twice the (smaller side)² = (larger side)² - 16.

So,

⇒ 2 * (20 - x)² = (x)² - 16

We know that, (a - b)² = a² + b² - 2ab. So,

⇒ 2 * (400 + x² - 40x) = x² - 16

⇒ 2x² - 80x + 800 = x² - 16

Transposing RHS to LHS,

⇒ 2x² - 80x + 800 - x² + 16 = 0

⇒ x² - 80x + 816 = 0

Using Middle term splitting,

⇒ x² - 12x - 68x + 816 = 0

⇒ x(x - 12) - 68(x - 12) = 0

⇒ (x - 12)(x - 68) = 0

⇒ x = 12 or 68

As none of the parts can be greater than 20,

⇒ x = 12

⇒ 20 - x = 20 - 12 = 8

So, the numbers are 12 and 8 respectively.

Verification:

We are given that,

⇒ 2 * (20 - x)² = (x)² - 16

Putting value of x = 12, and (20 - x) as 8.

So,

⇒ 2 * (8)² = (12)² - 16

⇒ 2 * 64 = 144 - 16

⇒ 128 = 128

As, LHS = RHS,

HENCE VERIFIED!!!