Math, asked by ishanewymbang, 11 months ago

the sum of the ages of two friend is 20. Four years ago, the product of their age in years was 48, Is it possible to determine their present age?


Answers

Answered by Anonymous
214
Solutions :-


Let age of first friend be x
Sum of age of two friends = 20 years
So, Age of second friend = (20 - x)

Four years ago,
Age of first friend = x - 4
Age of second friend = (20 - x) - 4 = 16 - x

Given : Four years ago, Product of the age of two friends = 48 years


A/q

=> (x - 4) (16 - x) = 48
=> x (16 - x) - 4 (16 - x) = 48
=> 16x - x² - 64 + 4x = 48
=> - x² + 16x + 4x - 64 - 48 = 0
=> - x² +20x - 112 = 0
=> x² - 20x + 112 = 0


Now,
Comparing the equation with ax² + bx + c

a = 1
b = -20
c = 112


We know,
D = b² - 4ac
= (-20)² - (4 × 1 × 112)
= 400 - 448
= - 48

Since,

D < 0 ( No real roots )



Hence,
It is not possible to determine their present age.

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Answered by SmãrtyMohït
98
❤Here is your solutions❤

Given :-

Sum of age of two friends is 20 years.

Let ,
Age of first friend be x

Age of second friend =20-x

Four years ago:-

age of 1st friend = x - 4✔

Age of 2nd friend = (20 - x) - 4 =16 - x✔

Product of the age of two friends = 48 years :-

According to question

=> (x - 4) (16 - x) = 48
=> x (16 - x) - 4 (16 - x) = 48
=> 16x - x² - 64 + 4x = 48
=>- x² + 16x + 4x - 64 - 48= 0
=> - x² +20x - 112 = 0
=> x² - 20x + 112 = 0

Comparing the equation with ax² + bx + c = x² - 20x + 112

Here

a = 1
b = -20
c = 112

Now we will used discriminant formula :-

D = b² - 4ac
=> (-20)² - (4 × 1 × 112)
=> 400 - 448
=> - 48

(D < 0 )i.e D is less than 0 its mean it have no real roots

Hence,

It is not possible to find the present age.

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