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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) (a) Find the value of $x$, the length of side of a tile by factorisation.
OR
(b) Solve the quadratic equation for $x$, using quadratic formula.
(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) (a) Find the value of $x$, the length of side of a tile by factorisation.
OR
(b) Solve the quadratic equation for $x$, using quadratic formula.
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Sol. (i) $200 x^2 = 128 (x + 1)^2$
(ii) $25x^2 = 16x^2 + 32x + 16$
$\Rightarrow 9x^2 - 32x - 16 = 0$
(iii) (a) $9x^2 - 32x - 16 = 0$
$\Rightarrow (9x + 4) (x - 4) = 0$
$x \neq -\frac{4}{9}$ so, $x = 4$
OR
(iii) (b) $x = \frac{32\pm\sqrt{1024+576}}{18} = \frac{32\pm 40}{18}$
$x \neq -\frac{4}{9}$ so, $x = 4$
(ii) $25x^2 = 16x^2 + 32x + 16$
$\Rightarrow 9x^2 - 32x - 16 = 0$
(iii) (a) $9x^2 - 32x - 16 = 0$
$\Rightarrow (9x + 4) (x - 4) = 0$
$x \neq -\frac{4}{9}$ so, $x = 4$
OR
(iii) (b) $x = \frac{32\pm\sqrt{1024+576}}{18} = \frac{32\pm 40}{18}$
$x \neq -\frac{4}{9}$ so, $x = 4$