Chapter 4: Quadratic Equations

A ball is thrown upwards from a rooftop. Its height $h$ (in meters) above the ground at time $t$ (in seconds) is given by the quadratic polynomial $h(t) = -5t^2 + 20t + 60$.

1. What is the height of the rooftop?
   (A) 20 m

   (B) 40 m

   (C) 60 m

   (D) 80 m
   Show Solution
   Solution: (C) 60 m
   Reason: At $t=0$ (start time), height is the rooftop height. $h(0) = -5(0)^2 + 20(0) + 60 = 60$ m.
2. When does the ball hit the ground?
   (A) 4 s

   (B) 6 s

   (C) 2 s

   (D) 5 s
   Show Solution
   Solution: (B) 6 s
   Step 1: Ball hits ground when $h(t) = 0$.
   Step 2: $-5t^2 + 20t + 60 = 0 \Rightarrow t^2 - 4t - 12 = 0$ (Divide by -5).
   Step 3: $(t-6)(t+2) = 0 \Rightarrow t=6$ or $t=-2$. Since time > 0, $t=6$ s.

A rectangular park is to be designed such that its length is 2 m more than its breadth. The area of the park is 195 sq m.

1. If breadth is $x$ m, which equation represents the area?
   (A) $x^2 + 2x - 195 = 0$

   (B) $x^2 - 2x - 195 = 0$

   (C) $x^2 + 2x + 195 = 0$

   (D) $2x^2 + x - 195 = 0$
   Show Solution
   Solution: (A) $x^2 + 2x - 195 = 0$
   Reason: Breadth $= x$, Length $= x+2$. Area $= x(x+2) = 195 \Rightarrow x^2 + 2x - 195 = 0$.
2. What is the breadth of the park?
   (A) 15 m

   (B) 13 m

   (C) 12 m

   (D) 10 m
   Show Solution
   Solution: (B) 13 m
   Step 1: Solve $x^2 + 2x - 195 = 0$.
   Step 2: Factorize: $x^2 + 15x - 13x - 195 = 0 \Rightarrow (x+15)(x-13) = 0$.
   Step 3: $x = 13$ or $x = -15$. Breadth cannot be negative, so $x = 13$ m.

A cottage industry produces a certain number of pottery articles in a day. It was observed on a particular day that the cost of production of each article (in rupees) was 3 more than twice the number of articles produced on that day. The total cost of production on that day was ₹90.

1. If $x$ is the number of articles, the quadratic equation is:
   (A) $x^2 + 3x - 90 = 0$

   (B) $2x^2 + 3x - 90 = 0$

   (C) $2x^2 - 3x - 90 = 0$

   (D) $x^2 + 2x + 90 = 0$
   Show Solution
   Solution: (B) $2x^2 + 3x - 90 = 0$
   Reason: Number of articles $= x$. Cost per article $= 2x+3$. Total cost $= x(2x+3) = 90 \Rightarrow 2x^2 + 3x - 90 = 0$.
2. Find the number of articles produced.
   (A) 10

   (B) 12

   (C) 6

   (D) 15
   Show Solution
   Solution: (C) 6
   Step 1: Solve $2x^2 + 3x - 90 = 0$.
   Step 2: $2x^2 + 15x - 12x - 90 = 0 \Rightarrow x(2x+15) - 6(2x+15) = 0$.
   Step 3: $(x-6)(2x+15) = 0 \Rightarrow x = 6$ (since articles cannot be negative/fraction).