Telangana SCERT Class 10 Maths Solution Chapter 5 Quadratic Equations Exercise 5.3

Telangana SCERT Solution Class X (10) Maths Chapter 5 Quadratic Equations Exercise 5.3

 

Quadratic Equations

Exercise – 5.3

 

(iv) x2 + 5 = – 6x

=> Solution:

x2 + 5 = – 6x

x2 + 6x + 5 = 0

x2 2(x)/2 X(6) + 5 = 0

x2 + 2 (x) X 6/2 = – 5

x2 + 2 (x) (3) = – 5

x2 + 2 (x) (3) + (3)2 = – 5 + (3)2 (by adding (3)2 on both sides)

=> (x + 3)2 = – 5 + 9

=> (x + 3)2 = 4

Square root on both sides,

=> x + 3 = ± 2

x + 3 = 2 or x + 3 = – 2

x = 2 – 3 or x = – 2 – 3

x = – 1 or x = – 5

∴ The roots of given equation = – 1 and – 5

 

(Q10) An express train takes 1 hour less than a passenger train to travel 132 km between Mysore and Bengaluru (without taking into consideration the time they stop at intermediate stations). If the average speed of the express train is 11 km/h more than that of the passenger train, find the average speed of the two trains

=> Solution:

The distance between Mysore and Bengaluru

= 132 km.

Let the speed of the passenger train be ‘x’ kmph.

Then the speed of the express train = (x + 11) kmph.

Time taken by the passenger train to travel 132 km, t2 = 132/x +11 hours —— (2)

But the difference between the two times is 1 hour.

 

Here is your solution of Telangana SCERT Class 10 Math Chapter 5 Quadratic Equations Exercise 5.3

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Updated: June 5, 2021 — 1:20 pm

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  1. Thank you very much for

  2. Identification frequencies identical Jack Kristina Rodamestri

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