Question

(ii) Construct a word problem on quadratic equation so that one answer will be 15
(age, metres, speed, rupees, etc). Solve the problem you have constructed.​

Answer 1

The original length was 15 miter

Step-by-step explanation:

Que: A business man constructs a square shape bath pool. But after sometimes, he wants to construct the walking path along its bank. So he decided to decrease its length by 2 miter and breadth be 3 miter. Now its area is found as 156 sq miter. What was the original length of its side, when it was square in shape?

Sol: Since, the pool was square in shape. but its length is decreased by 2 miter and breadth is decreased by 3 miter. Thus, its new shape is a rectangle.

Let the side length of square be x miter.

We know that;

The area of rectangle =$length \times breadth$

So, the new area of rectangular pool =$(x-2) \times (x-3)$

$156 = x^2 -2x -3x +6$

$x^2 -5x -150 = 0$

This is quadratic equation.

So, $x^2 - 15x + 10x - 150 = 0$

$x(x - 15) +10(x - 15) = 0$

$(x-15)(x+10) = 0$

x = 15, -10

Since, x is the length. so, it can't be negative.

Therefore, the original length was 15 miter.

Answer 2

Here is ur ans

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