Students can practice 10th Class Maths Solutions Telangana Chapter 5 Quadratic Equations Ex 5.1 to get the best methods of solving problems.
TS 10th Class Maths Solutions Chapter 5 Quadratic Equations Exercise 5.1
Question 1.
Check whether the following are quadratic equations or not.
i) (x + 1)2 = 2(x – 3)
Solution:
Given: (x + 1)2 = 2(x – 3)
⇒ x2 + 2x + 1 = 2(x – 3)
⇒ x2 + 2x + 1 – 2x + 6 = 0
⇒ x2 + 7 = 0 is a Q.E.
ii) x2 – 2x = -2(3 – x)
Solution:
Given : x2 – 2x = -2(3 – x)
⇒ x2 – 2x = -6 + 2x
⇒ x2 – 4x + 6 = 0 is a Q.E.
iii) (x – 2) (x + 1) = (x – 1) (x + 3)
Solution:
Given : (x – 2) (x + 1) = (x – 1) (x + 3)
⇒ x(x + 1) -2(x + 1)
= x(x + 3) – 1(x + 3)
Note : Compare the coefficients of x2 on both sides. If they are equal it is not a Q.E.
⇒ x2 + x – 2x – 2 = x2 + 3x – x – 3
⇒ x2 – x – 2 = x2 + 2x – 3
⇒ 3x – 1 = 0 is not a Q.E.
iv) (x – 3) (2x + 1) = x(x + 5)
Solution:
Given : (x – 3) (2x + 1) = x(x + 5)
⇒ x(2x + 1) – 3(2x + 1) = x. x + 5.x
⇒ 2x2 + x – 6x – 3 = x2 + 5x
⇒ x2 – 10x – 3 = 0 is a Q.E.
(or)
Comparing the coefficients of x2 on both sides, x . 2x and x. x ⇒ 2x2 and x2 2x2 ≠ x2
Hence it’s a Q.E.
v) (2x – 1)(x – 3) = (x + 5)(x – 1)
Solution:
Given : (2x – 1)(x – 3) = (x + 5)(x – 1)
⇒ 2x(x – 3) – 1(x – 3) = x(x – 1)+ 5(x – 1)
⇒ 2x2 – 6x – x + 3 = x2 – x + 5x – 5
⇒ 2x2 – 7x + 3 – x2 – 4x + 5 = 0
⇒ x2 – 11x + 8 = 0
Hence it’s a Q.E.
(or)
Co.eff. of x2 on L.H.S. = 2 × 1 = 2
Co.eff. of x2 on R.H.S = 1 × 1 = 1
LHS ≠ RHS
Hence it is a Q.E.
vi) x2 + 3x + 1 = (x – 2)2
Solution:
Given : x2 + 3x + 1 = (x – 2)2
⇒ x2 + 3x + 1 = (x – 2)2
⇒ x2 + 3x + 1 = x2 – 4x + 4
⇒ 7x – 3 = 0 is not a Q.E.
vii) (x + 2)3 = 2x (x2 – 1)
Solution:
Given : (x + 2)3 = 2x(x2 – 1)
⇒ x3 + 6x2 + 12x + 8 = 2x3 – 2x
[∵ (a + b)3 = a3 + 3a2b + 3ab2 + b3]
⇒ x3 + 6x2 + 14x + 8 = 0 is not a Q.E. [∵ degree = 3]
viii) x3 – 4x2 – x + 1 = (x – 2)3
Solution:
Given : x3 – 4x2 – x + 1 = (x – 2)3
⇒ x3 – 4x2 – x + 1 = (x – 2)3
= x3 – 6x2 + 12x – 8
⇒ 6x2 – 12x + 8 – 4x2 – x + 1 = 0
⇒ 2x2 – 13x + 9 = 0 is a Q.E.
Question 2.
Represent the following situations in the form-of quadratic equations :
i) The area of a rectangular plot is 528 m2. The length of the plot (in metres) is one more than twice its breadth. We need to find the length and breadth of the plot.
Solution:
Let the breadth of the rectangular plot be x m.
Then its length (by problem) = 2x + 1.
Area = l . b = (2x + 1). x = 2x2 + x but area = 528 m2 (∵ given)
∴ 2x2 + x = 528
⇒ 2x2 + x – 528 = 0 where x is the breadth of the rectangle.
ii) The product of the consecutive positive integers is 306. We need to find the integers.
Solution:
Let the consecutive integers be x and x + 1
Their product = x(x + 1) = x2 + x
By problem x2 + x = 306
⇒ x2 + x – 306 = 0
Where x is the smaller integer.
iii) Rohan’s mother is 26 years older than him. The product of their ages after 3 years will be 360 years. We need to find Rohan’s present age.
Solution:
Let the present age of Rohan be x years. Then age of Rohan’s mother = x + 26
After 3 years :
Age of Rohan would be = x + 3
Rohan’s mother’s age would be = (x + 26) + 3 = x + 29
By problem (x + 3) (x + 29) = 360
⇒ x(x + 29) + 3(x + 29) = 360
⇒ x2 + 29x + 3x + 87 = 360
⇒ x2 + 32x + 87 – 360 = 0
⇒ x2 + 32x – 273 = 0
Where x is Rohan’s present age.
iv) A train travels a distance of 480 km at a uniform speed. If the speed had been 8km/hr less, then it would have taken 3 hours more to cover the same distance. We need to find the speed of the train.
Solution:
Let the speed of the train be x kmph. Then time taken to travel a distance
It the speed is 8km/hr less, then time needed to cover the same distance
would be \(\frac{480}{x-8}\)
By problem = \(\frac{480}{x-8}\) – \(\frac{480}{x}\) = 3
⇒ x2 – 8x = 1280
⇒ x2 – 8x – 1280 = 0
Where x is the speed of the train.