Question

2.29
The sum of the ages of two friends is 20 years. Four years ago, the product of their
ages in years was 48. Is the given situation possible? If so, determine their present
ages.
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Answer 1

Question:

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

Answer:

The given situation is not possible .

Note:

• The possible values of independent variable (say x) which satisfy the equation (in variable x ) are called its roots .

• A quadratic equation can have at most two roots.

• The discriminantnant D of a quadratic equation ,

ax² + bx + c = 0 , is given by; D = b² - 4ac .

• If D = 0 , then both the roots of the quadratic equation are real and equal.

• If D > 0 , then both the roots of the quadratic equation are real and distinct.

• If D < 0 , then both the roots of the quadratic equation are imaginary.

Solution:

Let's assume that ;

Present age of 1st friend = x years

Present age of 2nd friend = y years

Also,

Age of 1st friend 4 years ago = (x - 4) years

Age of 2nd friend 4 years ago = (y - 4) years

Now,

It is given that ;

The sum of the present ages of two friends is 20 years.

Thus,

=> x + y = 20

=> y = 20 - x --------(1)

Also,

It is given that ;

Four years ago, the product of their ages in years was 48.

Thus,

=> (x - 4)(y - 4) = 48

=> (x - 4)(20 - x - 4) = 48

=> (x - 4)(16 - x) = 48

=> 16x - x² - 64 + 4x = 48

=> x² - 16x - 4x + 64 + 48 = 0

=> x² - 20x + 112 = 0 ---------(2)

Now,

Let's find the discriminant of the above equation

{ ie: eq-(2) } ;

=> D = b² - 4•a•c

=> D = (-20)² - 4•1•112

=> D = 400 - 448

=> D = - 48

=> D < 0

Clearly,

The discriminant of the eq-(2) is negative ( less then zero ) thus , there exist no real value of x which satisfy the eq-(2).

Hence,

The given situation is not possible.

Answer 2

Given :---

• sum of ages of 2 friends = 20 years.
• 4 years ago both ages product = 48 .

To Find :--

• The situation is possible or not ?

Solution :---

it is given that, sum of ages of both friends is 20 .

Let one friend is x years old ,.

Than other friend = (20-x) years old.

Now, it is given that, 4 years ago, their ages product was 48.

→ 4 years before one friend was = (x-4) years old .

→ other friends was = (20-x)-4 = (16-x) years old.

A/q,

→ (x-4)(16-x) = 48

→ 16x -x² - 64 + 4x = 48

→ x² - 20x + 112 = 0

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Now, we use concept :----

If A*x^2 + B*x + C = 0 ,is any quadratic equation,

then its discriminant is given by;

D = B^2 - 4*A*C

• If D = 0 , then the given quadratic equation has real and equal roots.

• If D > 0 , then the given quadratic equation has real and distinct roots.

• If D < 0 , then the given quadratic equation has unreal (imaginary) roots...

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Here in Equation ,

x² - 20x + 112 = 0 , we have ,

→ A = 1

→ B = -20

→ C = 112 .

Putting values we get,

→ D = (-20)² - 4*1*112

→ D = 400 - 448

→ D = (-48)

Here , we get, D < 0 .

so, then the given quadratic equation has unreal (imaginary) roots...

_____________________________

Hence, we can say that, the given situation is not possible , as age of anyone never be an imaginary number ..