Examples on Quadratic Equations

We will discuss here about some examples on quadratic equations.

We know many word problems involving unknown quantities can be translated into quadratic equations in one unknown quantity.

1. Two pipes working together can fill a tank in 35 minutes. If the large pipe alone can fill the tank in 24 minutes less than the time taken by the smaller pipe then find the time taken by each pipe working alone to fill the tank.

Solution:

Let the large pipe and smaller pipe working alone fill the tank in x minutes and y minutes respectively.

Therefore, the large pipe fills \(\frac{1}{x}\) of the tank in 1 minute and the smaller pipe fills \(\frac{1}{y}\) of the tank in 1 minute.

Therefore, two pipes working together can fill (\(\frac{1}{x}\) + \(\frac{1}{y}\)) of the tank in 1 minute.

Therefore, two pipes working together can fill 35(\(\frac{1}{x}\) + \(\frac{1}{y}\)) of the tank in 35 minutes.

From the question, 35(\(\frac{1}{x}\) + \(\frac{1}{y}\)) = 1 (whole being 1). ......................... (i)

Also, x + 24 =y (from the question). ......................... (ii)

Putting y = x + 24 in (i), 35(\(\frac{1}{x}\) + \(\frac{1}{x + 24}\)) = 1

⟹ 35\(\frac{x + 24 + x}{x(x + 24)}\) = 1

⟹ \(\frac{35(2x + 24)}{x(x + 24)}\) = 1

⟹ 35(2x + 24) = x(x + 24)

⟹ 70x + 35 × 24 = x\(^{2}\) + 24x

⟹ x\(^{2}\) - 46x - 840 = 0

⟹ x\(^{2}\) – 60x + 14x – 840 = 0

⟹ x(x - 60) + 14(x - 60) = 0

⟹ (x - 60)(x + 14) = 0

⟹ x - 60 = 0 or, x + 14 = 0

⟹ x = 60 or x = -14

But x cannot be negative. So, x = 60 and then y = x + 24 = 60 + 24 = 84.

Therefore, when working alone, the large pipe takes 60 minutes and the smaller pipe takes 84 minutes to fill the tank.

  

2. Find a positive number, which is less than its square by 30.

Solution:

Let the number be x

By the condition, x\(^{2}\) - x = 30

⟹ x\(^{2}\) - x - 30 = 0

⟹ (x - 6)(x + 5) = 0

⟹ Therefore,  x = 6, -5

As the number is positive, x = - 5 is not acceptable, Thus the required number is 6.


3. The product of the digits of a two-digit number is 12. If 36 is added to the number, a number is obtained which is the same as the number obtained by reversing the digits of the original number.

Solution:

Let the digit at the units place be x and that at the tens place be y.

Then, the number = 10y + x.

The number obtained by reversing the digits = 10x + y

From the question, xy = 12 ................... (i)

10y + x + 36 = 10x + y ........................... (ii)

From (ii), 9y - 9x + 36 = 0

⟹ y – x + 4 =0

⟹ y = x – 4 .................................. (iiii)

Putting y = x- 4 in (i), x(x – 4) =12

⟹ x\(^{2}\) – 4x – 12 = 0

⟹ x\(^{2}\) – 6x + 2x – 12 = 0

⟹ x(x – 6) + 2(x – 6) = 0

⟹ (x – 6)(x + 2) = 0

⟹ x – 6 = 0 or x + 2 = 0

⟹ x = 6 or x = -2

But a digit in a number cannot be negative. So, x ≠ -2.

Therefore, x = 6.

Therefore, from (iii), y = x – 4 = 6 – 4 = 2.

Thus, the original number 10y + x = 10 × 2 + 6 = 20 + 6 = 26.

 

4. After completing a journey of 84 km. A cyclist noticed that he would take 5 hours less, if he could travel at a speed which is 5 km/hour more. What was the speed of cyclist in km/hour?

Solution:

Suppose, the cyclist has travelled with a speed of x km/hour

Therefore, by the condition \(\frac{84}{x}\) - \(\frac{84}{x + 5}\) = 5

⟹ \(\frac{84x + 420 - 84x}{x(x + 5)}\)= 5

⟹ \(\frac{420}{x^{2} + 5x}\) = 5

⟹ 5(x\(^{2}\) + 5x) = 420

⟹ x\(^{2}\) + 5x - 84 = 0

⟹ (x + 12)(x - 7) = 0

Therefore, x = -12, 7

But x ≠- 12, because speed cannot be negative

x = 7

Therefore, the cyclist has travelled with a speed of 7 km/hour.



9 Grade Math

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