# Quadratic Equation

## NCERT Exercise 4.1

### Part 2

Represent the following situation in the form of quadratic equation:

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

**Solution:** Let us assume breadth `= x`

Therefore; length `= 2x + 1`

Since area `= text(length) xx text(breadth)`

Hence; `x(2x + 1) = 528`

Or, `2x^2 + x = 528`

Or, `2x^2 + x – 528 = 0`

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

**Solution:** Let us assume the first integer `= x`

Hence; second integer `= x + 1`

As per question; `x(x + 1) = 306`

Or, `x^2 + x = 306`

Or, `x^2 + x – 306 = 0`

(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 the Rohan’s age.

**Solution:** Let us assume, Rohan’s present age `= x`

So, his mother’s present age `= x + 26`

Three years from now, Rohan’s age `= x + 3`

Three years from now, mother’s age `= x + 29`

As per question; `(x + 3)(x + 29) = 360`

Or, `x^2 + 29x + 3x + 87 = 360`

Or, `x^2 + 32x + 87 = 360`

Or, `x^2 + 32x + 87 – 360 = 0`

Or, `x^2 + 32x – 273 = 0`

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

**Solution:** Let us assume, speed of train `= x` km/h

Therefore; reduced speed `= x – 8` km/h

We know, `text(time) = text(distance)/(speed)`

Hence;

`t=(480)/(x)`-------(1)

In case of reduced speed;

`t+3=(480)/(x-8)`

Or, `t=(480)/(x-8)-3` -----------(2)

From equations (1) and (2);

`(480)/(x)=(480)/(x-8)-3`

Or, `(480)/(x)=(480-3(x-8))/(x-8)`

Or, `(480)/(x)=(480-3x+24)/(x-8)`

Or, `480(x-8)=x(504-3x)`

Or, `480x-3840=504x-3x^2`

Or, `480x-3840-504x+3x^2=0`

Or, `3x^2-24x-3840=0`

Or, `x^2-8x-1280=0`