Question

Question 4 Is the following situation 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.

Class 10 - Math - Quadratic Equations Page 91

Answer 1

Let the age of one friend be x years.
then the age of the other friend will be (20 - x) years.
4 years ago,

Age of 1st friend = (x - 4) years
Age of 2nd friend = (20 - x - 4) = (16 - x) years
A/q we get that,
(x - 4) (16 - x) = 48

16x - x2 - 64 + 4x = 48
 - x2 + 20x - 112 = 0
x2 - 20x + 112 = 0
Comparing this 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. Such type of condition doesn't exist.

Answer 2

So a quadratic equation ax² + bx + c =0, has
  i) Two distinct real roots, if b² - 4ac >0 , then x= -b/2a + √D/2a &x= -b/2a - √D/2a
  ii) Two equal real roots, if b² - 4ac = 0 , then x= -b/2a or -b/2a
  iii) No real roots, if b² - 4ac <0
  _______________________________
  Solution:
 
Let the age of one of two friends be x yr.

Then the age of the other friend = (20 – x) yr.

[ The sum of the ages of two friends is 20 year]

4 years ago,

Age of one of two friends = (x – 4) yr

Age of other friend = (20 – x – 4) = (16 – x) yr

A.T.Q

(x – 4) (16 – x) = 48

16x – x² – 64 + 4x = 48
– x² + 20x – 112 = 0
x² – 20x + 112 = 0

On Comparing with ax² + bx + c = 0,
a = 1, b = -2 and c = 112

Discriminant = b2 – 4ac
= (-20)² – 4 × 112
= 400 – 448 = -48
D= - 48<0
b2 – 4ac < 0

Which implies that the real roots are not possible because this condition represent imaginary roots . So the solution does not exist and hence given situation is not possible.

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Hope this will help you.....