Telangana SCERT Solution Class X (10) Maths Chapter 5 Quadratic Equations Exercise 5.1
Quadratic Equations
Exercise – 5.1
(Q1) Check whether the following are quadratic equations.
(i) (x + 1)2 = 2 (x – 3)
=> Solution: The general equation of quadratic equation is.
ax2 + bx + x = 0
∴ (x + 1)2 = 2(x – 3)
x2 + 2x + 1 = 2x – 6 [(a + b)2 = a2 + 2ab + b2]
x2 + 2x – 2x + 6 + 1 = 0
x2 + 7 = 0
∴ It is in the form ax2 + bx + c = 0
∴ The given equation is a quadratic equation.
(ii) x2 – 2x = (-2) (3 – x)
=> Solution: Given equation is,
x2 – 2x = (-2) (3-x)
x2 – 2x = – 6 + 2x
x2 – 2x – 2x + 6 = 0
x2 – 4x + 6 = 0
∴ It is in the form ax2 + bx + c = 0
∴ The given equation is a quadratic equation.
(iii) (x – 2) (x + 1) = (x – 1) (x + 3)
=> Solution: Given equation is,
(x – 2) (x + 1) = (x – 1) (x + 3)
x2 + x – 2x – 2 = x2 + 3x – x – 3
x2 – x2 + x – 3x + x – 2x – 2 + 3 = 0
2x – 2x – 3x + 1 = 0
– 3x + 1 = 0
∴ It is not in the form ax2 + bx + c = 0
∴ The given equation is not a quadratic equation.
(iv) (x – 3) (2x + 1) = x(x + 5)
=> Solution: The given equation is,
(x – 3) (2x + 1) = x (x + 5)
2x2 + x – 6x – 3 = x2 + 5x
2x2 – x2 – 5x – 5x – 3 = 0
x2 – 10x – 3 = 0
∴ It is in the form ax2 + bx + c = 0
∴ The given equation is a quadratic equation.
(v) (2x – 1) (x – 3) = (x + 5) (x – 1)
=> Solution: The given equation is,
(2x – 1) (x – 3) = (x + 5) (x – 1)
2x2 – 6x – x + 3 = x2 – x + 5x – 5
2x2 – x2 – 7x + 3 + x – 5x + 5 = 0
x2 – 11x + 8 = 0
∴ It is in the form ax2 + bx + c = 0
∴ The given equation is a quadratic equation.
(vi) x2 + 3x + 1 = (x – 2)2
=> Solution: The given equation is,
x2 + 3x + 1 = (x – 2)2
x2+ 3x + 1 = x2 – 2 X 2 X x + 4 [(a – b)2= a2 – 2ab + b2]
x2 + 3x + 1 = x2 – 4x + 4
3x + 4x + 1 – 4 = 0
7x – 3 = 0
∴ It is not in the form ax2 + bx + c = 0
∴ The given equation is not a quadratic equation.
(vii) (x + 2)2 = 2x (x2 – 1)
=> Solution: The given equation is,
(x + 2)2 = 2x (x2 – 1)
x2 + 4x + 4 = 2x3 – 2x [(a – b)2 = a2 – 2ab + b2]
– 2x3 + x2 + 4x + 2x + 4 = 0
– 2x3 + x2 + 6x + 4 = 0
∴ The highest degree is 3.
But in quadratic equation the highest degree is 2.
So, it is not in the form ax2 + bx + c = 0
∴ The given equation is not quadratic equation.
(viii) x3 – 4x2 – x + 1 = (x – 2)3
=> Solution: The given equation is,
X3– 4x2 – x + 1 = (x – 2)3
x3 – 4x2 – x + 1 = x3 – 3x2 (2) + 3x (2)3 – 23
[(a – b)3 = a3 – 3a2b + 3ab2 – b3]
– 4x2 – x + 1 = – 6x2 + 3x (8) – 8
– 4x2 – x + 1 = – 6x2 + 24x – 8
– 4x2 + 6x2 – x – 24x + 1 + 8 = 0
2x2– 25x + 9 = 0
∴ It is in the form ax2 + bx + c = 0
∴ The given equation is a quadratic equation.
(Q2)
(ii) The product of two consecutive positive integers in 306 we need to find the integers
=> Solution:
Let, the consecutive positive integers be x, x + 1
∴ The product of two consecutive positive integers is 306.
x (x + 1) = 306
x2 + x = 306
x2 + x – 306 = 0
(iii) Rohan’s mother is 26 years older than him, the product of their after 3 years will be 360 years, we need to find Rohan’s present age.
=> Solution: At present,
Let, the age of Rohan be ‘x’ years then the age of his mother becomes (x + 26) years.
After 3 years,
The age of Rohan = (x + 3) years and
The age of his mother = (x + 26 + 3)
= (x + 29) Years.
∴ The product of their ages after 3 years will be 360 years.
∴ (x + 3) (x + 29) = 360
x2 + 29x + 3x + 87 = 360
x2 + 32x + 87 – 360 = 0
x2 + 32x – 273 = 0
(iv) A train travels a distance of 480 km at a uniform speed. If the speed had been 8 km/h less, then it would have taken 3 hours more to cover the same distance we used to find the speed of the train.
=> Solution:
Distance to be travelled = 480 km.
Let the speed of the train be x km/h
we have, by formula
Time = distance/speed
Time taken by the train to travel 480 km at a speed of ‘x’ km/h = t1 = 480/x —— (1)
If the speed is 8 km/h less, i.e. speed = (x – 8) km/h
Time taken by the train to travel 480 km at a speed of (x – 8) km/h = t2 = 480/x-8 —–(2)
But the difference in time = 3hr.
t2 – t1 = 3
From equations (1) and (2),
480/x-8 – 480/x = 8
480x – 480(x – 8)/x (x – 8) = 3
480x – 480x + 3840 = 3x (x – 8)
3840 = 3 (x2 – 8x)
3840 = 3x2 – 24x
3x2 – 24x – 3840 = 0
3 (x2 – 8x – 1280) = 0
X2 – 8x – 1280 = 0
∴ The required quadratic equation is x2 – 8x – 1280 = 0
Here is your solution of Telangana SCERT Class 10 Math Chapter 5 Quadratic Equations Exercise 5.1 .
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