# Applications of Quadratic Equations

Picture a swing moving back and forth. When you look at it from the side, it draws an outline, a shape, perhaps? Is it some kind of arc or part of a circle? Yes, we call it a parabola. Do you know that a parabola is a graph drawn by quadratic equations? But first let’s understand what exactly do we mean by the applications of quadratic equations.

## 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 3 m and length is 5(3) = 15 m.

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

Download NCERT Solutions for Class 10 Mathematics

### 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 = 128 m

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.

## More Solved Examples for You

Question 4: A motorboat whose speed is 18 km/h in still water takes hour more to go 24 km upstream than to return downstream to the same spot. The speed of the stream is:
A) 6 km/h       B) 5 km/h      C) 3.5 km/h      D) 4.5 km/h

Answer : A) Let the speed of the stream be represented by x. Therefore, the speed of the motorboat upstream is (18 – x) km/h and the speed of the motorboat downstream is (18 + x) km/h.

Time taken by the boat to go upstream = $$\frac{Distance}{Velocity}$$ = $$\frac{24}{18 – x}$$

Similarly, the time taken by the boat to go downstream = $$\frac{24}{18 + x}$$

From the condition given here, we have: $$\frac{24}{18 – x}$$ – $$\frac{24}{18 + x}$$ = 1

24(18 + x) – 24(18 – x) =  (18 – x) (18 + x)

Simplification of the above equation gives: x2 + 48x -324 = 0. Hence, using the quadratic formula, we have x = 6 and x = -54. Speed can’t be negative, so we have x = 6 km/h.

Question 5: What is quadratic?

Answer: We can define this equation as an equation of second degree or degree of 2. This means that the highest exponent of the function is 2. In addition, the standard form of a quadratic equation is y = ax2 + bx + c, where a, b, and c are number and a is not equal to zero (a ≠ 0).

Question 6: What is the quadratic formula and what is it used for?

Answer: It refers to a formula that produces the zeros of any parabola. Furthermore, we can use the quadratic formula to identify the axis f symmetry of the parabola, and the number of real zeros the quadratic equation contains.

Question 7: What makes a problem quadratic?

Answer: It describes a problem that deals with a variable multiplied by itself, which we know as squaring. Moreover, this language originates from the area of a square is its side length multiplied by itself.

Question 8: State some application of the quadratic formula in real life?

Answer: In daily life we use quadratic formula as for calculating areas, determining a product’s profit or formulating the speed of an object. In addition, quadratic equations refer to an equation that has at least one squared variable.

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vee kwari

how do we state the nature of quadratics

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Karthik

A polynomial can have real numbers as zero(ie rational and irrational)to decide it’s nature we can use the relationship between the coefficients and zero (by comparing sum and product of zeroes)

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Vikas

HI

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HELLOOO

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Bhargav

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If d 0 – real & distinct roots
d =0 – real & equal roots ….