application problem of quadratic equation
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Applications Of The Quadratic Equations
Many physical and mathematical problems are in the form of quadratic equations. In mathematics, the solution of the quadratic equation is of particular importance. As already discussed, a quadratic equation has no real solutions if D < 0. This case, as you will see in later classes is of prime importance. It helps develop a different field of mathematics known as the Complex Analysis.In other fields, we see quadratic equations in many forms. Here we will try to describe a few uses by considering a few examples. Let us start.
Application to Problems of Area
Example 1: There is a hall whose length is five times the width. The area of the floor is 45m2. Find the length and width of the hall.
Solution: Let us suppose that ‘w’ is the width of the hall. Then we see that w (5w) will give the area of the hall. Therefore, we can write:
5w2 = 45
w2 = 9
w2 – 9 = 0
(w+3)(w-3) = 0
w = -3 or w = 3. Therefore, the width is 3m and length is 5(3) = 15m.
Example 2: The three sides of a right-angled triangle are x, x+1 and 5. Find x and the area, if the longest side is 5.
Solution: The longest side will be the Hypotenuse. Therefore, we can write:
x2 + (x+1)2 = 52 (Pythagoras’ Theorem)
x2 + x2 + 2x + 1 = 25
2x2 + 2x – 24 = 0
Hence, x2 + x – 12 = 0
(x – 3)(x + 4) = 0
(x + 4) = 0 or (x – 3) = 0
x = -4 or x = 3
We can only take x = 3 here because the length can’t be negative. (Why?)
Hence, x = 3 and therefore, Area = 1/2 x 3 x 4 = 6
Application to Problems of Motion
Example 3: A ball is thrown upwards from a rooftop, 80 m above the ground. It will reach a maximum vertical height and then fall back to the ground. The height of the ball from the ground at time t is h, and is given by h = -16t2 + 64t + 80. Find:
1. The height reached by the ball after 1 second?
2. The maximum height reached by the ball?
3. The time it will take before hitting the ground?
Solution: 1) The given equation is h = -16t2 + 64t + 80. Let us find ‘h’ after 1 sec. For that we substitute t = 1. Therefore, we have:
h = -16(1)2 + 64(1) + 80 = 128m
2) To find the maximum height, let us rearrange the equation:
h = -16[t2 – 4t – 5]
Hence, h = -16[(t – 2)2 – 9]
h = -16(t – 2)2 + 144
Now for h to be maximum, the negative term should be minimum. Hence, for t = 2, the negative term vanishes and we get a maximum value for h.
In other words, when the height is maximum, t = 2; therefore, maximum height = 144m.
3) When the ball hits the ground, h = 0;
-16t2 + 64t + 80 = 0
Divide the equation by -16
t2 – 4t – 5 = 0
(t – 5)(t + 1) = 0
t = 5 or t = -1
The time cannot be negative; so, the time = 5 seconds.
Many physical and mathematical problems are in the form of quadratic equations. In mathematics, the solution of the quadratic equation is of particular importance. As already discussed, a quadratic equation has no real solutions if D < 0. This case, as you will see in later classes is of prime importance. It helps develop a different field of mathematics known as the Complex Analysis.In other fields, we see quadratic equations in many forms. Here we will try to describe a few uses by considering a few examples. Let us start.
Application to Problems of Area
Example 1: There is a hall whose length is five times the width. The area of the floor is 45m2. Find the length and width of the hall.
Solution: Let us suppose that ‘w’ is the width of the hall. Then we see that w (5w) will give the area of the hall. Therefore, we can write:
5w2 = 45
w2 = 9
w2 – 9 = 0
(w+3)(w-3) = 0
w = -3 or w = 3. Therefore, the width is 3m and length is 5(3) = 15m.
Example 2: The three sides of a right-angled triangle are x, x+1 and 5. Find x and the area, if the longest side is 5.
Solution: The longest side will be the Hypotenuse. Therefore, we can write:
x2 + (x+1)2 = 52 (Pythagoras’ Theorem)
x2 + x2 + 2x + 1 = 25
2x2 + 2x – 24 = 0
Hence, x2 + x – 12 = 0
(x – 3)(x + 4) = 0
(x + 4) = 0 or (x – 3) = 0
x = -4 or x = 3
We can only take x = 3 here because the length can’t be negative. (Why?)
Hence, x = 3 and therefore, Area = 1/2 x 3 x 4 = 6
Application to Problems of Motion
Example 3: A ball is thrown upwards from a rooftop, 80 m above the ground. It will reach a maximum vertical height and then fall back to the ground. The height of the ball from the ground at time t is h, and is given by h = -16t2 + 64t + 80. Find:
1. The height reached by the ball after 1 second?
2. The maximum height reached by the ball?
3. The time it will take before hitting the ground?
Solution: 1) The given equation is h = -16t2 + 64t + 80. Let us find ‘h’ after 1 sec. For that we substitute t = 1. Therefore, we have:
h = -16(1)2 + 64(1) + 80 = 128m
2) To find the maximum height, let us rearrange the equation:
h = -16[t2 – 4t – 5]
Hence, h = -16[(t – 2)2 – 9]
h = -16(t – 2)2 + 144
Now for h to be maximum, the negative term should be minimum. Hence, for t = 2, the negative term vanishes and we get a maximum value for h.
In other words, when the height is maximum, t = 2; therefore, maximum height = 144m.
3) When the ball hits the ground, h = 0;
-16t2 + 64t + 80 = 0
Divide the equation by -16
t2 – 4t – 5 = 0
(t – 5)(t + 1) = 0
t = 5 or t = -1
The time cannot be negative; so, the time = 5 seconds.
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Answer:
Quadratic equations are useful to solve problems arising in our day - to - day life.
The method of solving problems consists of the following three steps :
Step - by - step explanation:
Step 1 :
Convert the word problem, into symbolic language, i. e. form mathematical equation by identifying the relationship existing in the problem.
Step 2 :
Solve the quadratic equation thus formed.
Step 3 :
Interpret the solution of the equation into verbal language. The appropriate solution / solutions satisfying the given conditions is / are to be considered.
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