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.
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Answer 1

$\mathfrak{\huge \underline{Question:}}$

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.

$\mathfrak{\huge \underline{Solution:}}$

No, this situation is not possible as the roots ofthe equation is imaginary I.e not real

$\mathfrak{\huge \underline{Explanation:}}$

Let the present age of both the friends be x and y.

Their ages 4 yrs ago is x-4 and y-4.

Sum of ages= x+y=20___(i)

Product of their ages 4 years ago

= (x-4)(y-4)

=xy-4x-4y+16

=xy-4(x+y)+16

=xy-80+16=48

=xy=64+48

=xy= 112___(ii)

= x= 112/y

Putting x=112/y in x+y=20

$\sf \frac{112}{y}+y=20 \\\\\sf 112+y^2=20y \\\\\sf y^2-20y+112=0 \\\\\sf Using, quadratic~formula \\\\\sf\frac{-b \pm\sqrt{D}}{2a}\\\\\sf = \frac{-b \pm\sqrt{b^2-4ac}}{2a} \\\\\sf =\frac{20 \pm\sqrt{400-448}}{2}\\\\\sf= \frac{20 \pm \sqrt{-48}}{2} \\\\\sf Here~D= - 48 <0 \\\\\sf Hence, its~ roots ~are ~not~ real, the~ situation ~is ~not~ possible.$

Answer 2

Given :-

Sum of ages of two friends = 20 years

Product of the ages of two friends = 48 years

To find :-

Age of first friend

Age of second friend

Solution :-

Let the age of first friend be x

Then, the age of 2nd friend would be 20-x

According to the 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

Using the equation: ax² + bx + c

Substituting them, we get :-

(-20)² - (4 × 1 × 112)

= 400 - 448

= - 48

Hence, D < 0

$\therefore$ It is not possible to determine their ages.

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