Class10 NCRT Quadratic Equations Exercise – 4.1 pdf || UP Board
Quadratic Equations
Exercise – 4.1
1. Check whether the following are
quadratic equations:
(i) (x + 1)2 = 2(x – 3)
on simplifying the equation,
(x + 1)2 = 2(x – 3)
x2 + 2x + 1 = 2x – 6
x2 + 7 = 0
Or x2 + 0x + 7 = 0
This equation is of the type ax2
+ bx + c = 0.
Hence, the given equation is a
quadratic equation.
(ii) x2
– 2x = (-2)(3 – x)
on simplifying the equation,
x2 – 2x = (-2)(3 - x)
x2 – 2x = -6 + 2x
x2 – 4x + 6 = 0
Or x2 – 4x + 6 = 0
This equation is of the type ax2
+ bx + c = 0.
Hence, the given equation is a
quadratic equation.
(iii) (x –
2)(x + 1) = (x – 1)(x + 3)
on simplifying the equation,
(x – 2)(x + 1) = (x – 1)(x + 3)
x2 – 2x + x – 2 = x2
– x + 3x – 3
-3x + 1 = 0
Or 3x – 1 = 0
This equation is not of the type ax2
+ bx + c = 0.
Hence, the given equation is not a
quadratic equation.
(iv) (x –
3)(2x + 1) = x(x + 5)
on simplifying the equation,
2x2 – 6x + x – 3 = x2
+ 5x
x2 – 10x – 3 = 0
Or x2 +(-10)x + (-3) = 0
This equation is of the type ax2
+ bx + c = 0.
Hence, the given equation is a
quadratic equation.
(v) (2x –
1)(x – 3) = (x + 5)(x – 1)
on simplifying the equation,
2x2 – x – 6x + 3 = x2
+ 5x – x – 5
x2 – 11x + 8 = 0
This equation is of the type ax2
+ bx + c = 0.
Hence, the given equation is a
quadratic equation.
(vi) x2
+ 3x + 1 = (x – 2)2
on simplifying the equation,
x2 + 3x + 1 = x2
– 4x + 4
7x – 3 =0
This equation is not of the type ax2
+ bx + c = 0.
Hence, the given equation is not a
quadratic equation.
(vii) (x + 2)3
= 2x(x2 – 1)
on simplifying the equation,
x3 + 6x2 + 12x
+ 8 = 2x3 – 2x
-x3 + 6x2 +
14x + 8 = 0
अर्थात x3 – 6x2 – 14x
– 8 = 0
This equation is not of the type ax2 + bx + c = 0.
Hence, the given equation is not a
quadratic equation.
(viii) x3
– 4x2 – x + 1 = (x – 2)3
on simplifying the equation,
x3 – 4x2 – x +
1 = x3 – 6x2 + 12x – 8
2x2 – 13x + 9 = 0
2x2 + (– 13)x + 9 = 0
This equation is of the type ax2
+ bx + c = 0.
Hence, the given equation is a
quadratic equation.
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 plot = x m
Hence, length of plot = 2x + 1 m
Hence, area = x(2x + 1) m2
According to Question,
x(2x + 1) = 528
2x2 + x = 528
2x2 + x – 528 = 0
Hence, the quadratic equation 2x2
+ x – 528 = 0 represents the length and breadth of the plot.
(ii) The product of two consecutive
positive integers is 306. we need to find the integers.
Let first integer = x
Therefore, second integer = x + 1
So, product = x(x + 1)
According to Question,
x(x + 1) = 306
x2 + x = 306
x2 + x – 306 = 0
Hence, the quadratic equation x2
+ x – 306 = 0 represents the product of two consecutive positive integers.
(iii) Rohan’s mother is 26 years older
than him. The product of their ages (in years) 3 years from now will be 360. We
would like to find Rohon’s present age.
Let Rohan's age = x years
Therefore, age of Rohan's mother = x
+ 26 years
after three years,
Let Rohan's age = x + 3 years
Therefore, age of Rohan's mother = x
+ 29 years
Therefore, product of ages = (x + 3)(x + 29) years
According to Question,
(x + 3)(x + 29) = 360
x2 + 3x + 29x + 87 = 360
x2 + 32x – 273 = 0
Hence, the quadratic equation x2
+ 32x – 273 = 0 represents the Rohon’s present age.
(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 need to find the speed of the
train.
Let speed of train = x km/h
Total distance = 480 km
Hence, time taken by the train = 480/x hours
If its speed had been 8 km/m less, then time taken = 480/(x – 8) hours
According to Question,
480x – 480x + 3640 = 3(x – 8)x
3640 = 3x2 – 24x
3x2 – 24x – 3640 = 0
Hence, the
quadratic equation 3x2 – 24x – 3640 = 0 represents the speed of the
train.
Download Class10 NCRT Quadratic Equations Exercise – 4.1 pdf
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