Represent the following situations in the form of quadratic equations :
Question.

Represent the following situations in the form of quadratic equations :

(i) The area of a rectangular plot is $528 \mathrm{~m}^{2}$. The length of the plot (in metres) is one more than twice its breadth. We need to find the length and breadth of the plot.

(ii) The product of two consecutive positive integers is 306. We need to find the integers.

(iii)Rohan’s mother is 26 years older than him. The product of their ages (in years) 3 years from now will be 360. We would like to find Rohan’s present age.

(iv) A train travels a distance of 480 km at a uniform speed. If the speed had been 8 km/hr less, then it would have taken 3 hours more to cover the same distance. We need to find the speed of the train.


Solution:

(i) Let breadth be $=x$ meters

Then length $=(2 x+1)$ meters.

$x \times(2 x+1)=528($ Area of the plot $)$

or $\quad 2 x^{2}+x-528=0$

(ii) Let the consecutive integers be $x$ and $x+1$. It is given that their product is 306 .

$\therefore \quad x(x+1)=306$

$\Rightarrow x^{2}+x-306=0$

(iii) Let Rohan’s present age $=x$ years Then present age of Rohan’s mother $=(x+26)$ years

After 3 years, it is given that

$(x+3) \times\{(x+26)+3\}=360$

or $(x+3)(x+29)=360$

$\Rightarrow x^{2}+32 x+87=360$

$\Rightarrow x^{2}+32 x+87-360=0$

$\Rightarrow x^{2}+32 x-273=0$

(iv) Let the speed of train be $x \mathrm{~km} / \mathrm{h}$.

Time taken to travel $480 \mathrm{~km}=\frac{480}{\mathbf{x}} \mathrm{hrs}$ In second condition,

let the speed of train $=(x-8) \mathrm{km} / \mathrm{h}$

It is also given that the train will take 3 hours more to cover the same distance.

Therefore, time taken to travel $480 \mathrm{~km}=\left(\frac{480}{\mathbf{x}}+\mathbf{3}\right) \mathrm{hrs}$

Speed $\times$ Time $=$ Distance

$(x-8)\left(\frac{480}{x}+3\right)=480$

$\Rightarrow 480+3 x-\frac{3840}{x}-24=480$

$\Rightarrow 3 x-\frac{3840}{x}=24$

$\Rightarrow 3 x^{2}-24 x+3840=0$

$\Rightarrow x^{2}-8 x+1280=0$
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