MAHARASHTRA Class 10: Word Problems in Quadratic Equations Videos | Word Problems in Quadratic Equations

Let Rohan’s age be x, so his mother’s age is x + 26.

After 3 years, their ages will be x + 3 and x + 29.

Product is 360.

(x + 3)(x + 29) = 360 x² + 32x - 273 = 0 (x - 7)(x + 39) = 0 x = 7 and x = −39 … age is not negative 

Length of the side of one square is x, so the perimeter is 4x.

So by the given condition, the other square’s perimeter will be 24 + 4x.

So, the sides will be (24 + 4x)/4 = 6 + x. Sum of the areas of squares is 468 m². 

x² + (x + 6)² = 468 2x² + 12x - 432 = 0 x² + 6x - 216 = 0 (x - 12)(x + 18) = 0 x = 12 or x = −18 … a side cannot be negative 

Let the height be x, so the base is x + 7. 

Applying Pythagoras theorem,

x² + (x + 7)² = 169 2x² + 14x + 49 - 169 = 0 x² + 7x - 60 = 0 (x + 12)(x − 5) = 0x = −12 or x = 5 … a side cannot be negative