Math, asked by Mudepakasamuel123, 11 months ago

Is the following situation.
if
possible? If so, determine
their present ages, the sum
of the ages of two
friends is 20 years.
Four years ago, the product
of their ages in years was 48
present age​

Answers

Answered by Anonymous
27

Answer:

Situation is not possible

Step-by-step explanation:

Let the present ages of two friends be 'x' and 'y' years

Sum of present ages of two friends = 20 years

⇒ x + y = 20

⇒ y = 20 - x

Age of one of the friend four years ago  = ( x - 4 ) years

Age of the other friend four years ago = ( y - 4 ) years = ( 20 - x - 4 ) = ( 16 - x ) years

Product of their ages four years ago = 48 years

⇒ ( x - 4 )( 16 - x ) = 48

⇒ x( 16 - x ) - 4( 16 - x ) = 48

⇒ 16x - x² - 64 + 4x = 48

⇒ - x² + 20x - 64 - 48 = 0

⇒ - x² + 20x - 112 = 0

⇒ x² - 20x + 112 = 0

Comparing the above equation with ax² + bx + c = 0 we get

  • a = 1
  • b = - 20
  • c = 112

Discriminant = D = b² - 4ac

                     = ( - 20 )² - 4( 1 )( 112 )  

                      = 400 - 448

                      = - 48

Since D < 0, the equation has no real roots

Therefore the situation is not possible.

Answered by Anonymous
3

Let’s say, the age of one friend be x years.

Then, the age of the other friend will be (20 – x) years.

Four years ago,

Age of First friend = (x – 4) years

Age of Second friend = (20 – x – 4) = (16 – x) years

As per the given question, we can write,

(x – 4) (16 – x) = 48

16x – x2 – 64 + 4x = 48

– x2 + 20x – 112 = 0

x2 – 20x + 112 = 0

Comparing the equation with ax2 + bx + c = 0, we get

a = 1, b = -20 and c = 112

Discriminant = b2 – 4ac

= (-20)2 – 4 × 112  

= 400 – 448 = -48

b2 – 4ac < 0

Therefore, there will be no real solution possible for the equations. Hence, condition doesn’t exist.

Similar questions