Quadratic Equation Solution16





In this page quadratic equation solution16 we are going to see solution of the word problems of the topic quadratic equation.

Question 23:

The side of a square exceeds the side of another square by 4 cm and the sum of the area of two squares is 400 sq.cm. Find the dimensions of the squares.

Solution:

Let "x" be side length of one square

The side of a square exceeds the side of another square by 4 cm

So side length of another square (x + 4)

Area of one square with side length "x" = x²

Area of one square with side length "x + 4" = (x + 4)²

sum of the area of two squares = 400 sq.cm

x² + (x + 4)² =  400

x² + x² + 2 (x) (4) + 4² = 400

2 x² + 8 x + 16 - 400 = 0

2 x² + 8 x - 384 = 0

Now we are going to divide the whole equation by 2,so we get

x² + 4 x - 192 = 0

x² + 16 x - 12 x - 192 = 0

x (x + 16) - 12 (x + 16) = 0

(x + 16) (x - 12) = 0

x + 16 = 0             x - 12 = 0

x = -16                      x = 12

Here x represents side length of one square.

Therefore sides of one square is 12 cm

Side length of another square = (12 + 4) = 16 cm

Verification:

sum of the area of two squares = 400 sq.cm

12² + 16² = 400

144 + 256 = 400

 400 = 400

quadratic equation solution16