The present ages of two brothers are 15 years and 22 years respectively. After how many years will the product of their ages be 408? | Quadratic Equation | SciPiPupil

Question: The present ages of two brothers are 15 years and 22 years respectively. After how many years will the product of their ages be 408?

Solution:

Given,

The present ages of two brothers are 15 years and 22 years respectively.

Let the required years be represented by x years.

According to the question,

The product of their ages will be 408 after x years,

$or, (15 +x)(22 +x) = 408$

$or, 15(22+x) +x(22+x) = 408$

$or, 330 +15x +22x +x² = 408$

$or, x² +37x +330 -408 = 0$

$or, x² +37x -78 = 0$

$or, x² +(39-2)x -78 = 0$

$or, x² +39x -2x -78 = 0$

$or, x(x +39) -2(x +39) = 0$

$or, (x -2)(x +39) = 0$

Either,

$x -2 = 0$

$\therefore x = 2$

Or,

$x +39 = 9$

$\therefore x = -39$

Since, years can never be negative. We take value of x = 2.

Therefore, the product of the ages of the two brothers will be 408 in 2 years.

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