Question

The of the ages of two friends is 20 years. Four years ago, the productofthér uges in
years was 48. Is the above situation possible? If so, determine their present ages.​

Answer 1

Appropriate 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 above situation please? If so, determine their present ages.

⠀⠀⠀⠀⠀⠀━━━━━━━━━━━━━━━━━━━⠀⠀⠀

⠀⠀⠀⠀⠀⠀

❍ Let's say, that the present age of first friend be x years. Then, the present age of second friend be (20 – x) years.

⠀⠀⠀⠀⠀⠀

$\underline{\bigstar\:\boldsymbol{According\: to \;the\; Question\; :}}$⠀⠀

• The sum of the ages of the both (first and second friend) friends is 20 years.

Also,

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

Therefore,

☆ Four years ago, their ages were —

• First friend = (x – 4) years.
• Second friend = (20 – x) – 4 = (16 – x) years.

⠀

★ ( First friend ) × ( Second friend ) = 48 ★

$:\implies\sf (x - 4) (16 - x) = 48 \\\\\\:\implies\sf 16x - x^2 - 64 + 4x = 48 \\\\\\:\implies\sf -x^2 + 20x - 112 = 0 \\\\\\:\implies\sf x^2 - 20x + 112 = 0$

⠀

Now, comparing this equation with the Quadratic equation (ax² + bx + c = 0).

$\frak{we\:have}\begin{cases}\frak{\:\;\; a = 1}\\\frak {\;\;\;b = -20}\\\frak{\;\;\;c = 112}\end{cases}$

$\rule{100px}{.3ex}$

Here If,

• (D < 0), then no real roots.
• (D = 0), then two equal real roots.
• (D > 0), then two distinct real roots.

$\bigstar\;\sf{{\pink{\underline{\boxed{\sf{\pmb{~D = (b^2 - 4ac)~}}}}}}}$

$\dashrightarrow\sf D = \Big(-20\Big)^2 - \Big(4 \times 1 \times 112\Big) \\\\\\\dashrightarrow\sf D = 400 - 448 \\\\\\\dashrightarrow{\underline{\pink{\boxed{\frak{\pmb{D = -48}}}}}}\;\bigstar$

Here we can see that (D < 0), it means equation hasn't real roots.

∴ Hence, the given situation isn't possible.

Answer 2

$olution:

Let the present age of first friend be x years. So,

present age of other be (20 - x).

4 years ago,

Age of first friend = (x - 4) yrs. and age of

second friend = (20 - x - 4) yrs = (16 - x) yrs.

According to question,

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

$= > 16 x - {x}^{2} - 64 + 4x = 48$

$= > - x2 + 20x - 112 = 0$

$= > {x}^{2} - 20x + 112 = 0$

$Here, a = 1, b = -20, c = 112$

$Now, D = {b}^{2} - 4 ac$

$= (-20)2 - 4(1)(112) = 400 - 448 = -48$

$D = -48$

$Since, D< 0$

So, the given quadratic equation has no real roots and hence it is not possible to calculate the present ages.