(Q) A rectangular floor area can be completely tiled with 200 square tiles. If the side length of each tile is increased by 1 unit, it would take only 128 tiles to cover the floor.
(i) Assuming the original length of each side of a tile be x units, make a quadratic equation from the above information.

(ii) Write the corresponding quadratic equation in standard form.

(iii) Find the value of x, the length of side of a tile by factorisation.


Ans:  Let’s consider the side of a tile is x units.

Therefore the area of 1 tile = x^2 sq. units

Hence, area of 200 tiles = 200x^2 sq. units

Next, The tile size is increased by 1 unit, the new side f tile = (x + 1) units

Hence, area of a new tile = (x + 1)^2sq. units

(i) Making the Quadratic equation:

By given condition, area of floor is covered by 200 old tiles or 128 new tiles,

Therefore, 200x^2 = 128 (x + 1)^2

(ii) Quadratic equation in standard form:

A quadratic equations standard form is ax^2 + bx + C = 0

we need to bring the above equation in above standard form.

By solving the equation, we get:

200 x^2 = 128 (x + 1)^2

\Rightarrow 200 x^2 = 128 (x^2 + 2 x +1)

\Rightarrow 200 x^2 = 128 x^2 + 256 x +128

\Rightarrow 200 x^2 - 128 x^2 - 256 x - 128 = 0

\Rightarrow 72x ^2 - 256 x - 128 = 0

\Rightarrow 9 x^2 - 32 x - 16 = 0

(iii) Find the value tile’s side:

Let’s solve this quadratic equation:

9 x^2 - 32 x - 16 = 0

\Rightarrow 9 x^2 - 36 x + 4 x - 16 = 0

\Rightarrow 9 x (x - 4) + 4 (x - 4) = 0

\Rightarrow (9 x + 4) (x - 4) = 0

Here, x =  \frac{-4}{9}~and~x = 4

Here, we reject x = \frac{-4}{9}, because the value of a tile’s side can not be negative number.

Hence x = 4

Therefore the side of the tile is 4 units

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