NCERT Class 10th Math chapter 4 Exercise 4.1 Question Answer. NCERT Solution for Class 10 Maths Chapter 4 Exercise 4.1 of Quadratic Equations NCERT Solution for CBSE, HBSE, RBSE and Up Board and all other chapters of class 10 math.
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NCERT Class 10 Math chapter 4 Quadratic Equations Exercise 4.1 Question Answer Solution.
Class 10 Math Chapter 4 Ex. 4.1 Question Answer
1. Check whether the following are quadratic equations :
(i) (x + 1)2 =2(x – 3)
Ans. LHS = (x + 1)2 =x2 + 1 + 2x
RHS = 2(x – 3) = 2x – 3
As LHS = RHS
x2 + 1 + 2x = 2x – 3
x2 + 1 + 2x – 2x + 3 = 0
x2 + 4 = 0
It is of the form ax2 + bx + c = 0, where a is not equal to zero.
Therefore, the given equation is a quadratic equation.
(ii) x2 – 2x = (-2)(3 – x)
Ans. LHS = x2 – 2x
RHS = (-2)(3 – x) = – 6 + x
As LHS = RHS
x2 – 2x = – 6 + x
x2 – 2x + 6 – x = 0
x2 – 3x + 6 = 0
It is of the form ax2 + bx + c = 0
Therefore, the given equation is a quadratic equation.
(iii) (x -2)(x + 1) = (x-1)(x + 3)
Ans. LHS = (x -2)(x + 1) =x2 + x – 2x – 2 = x2 – x – 2
RHS = (x-1)(x + 3) = x2 + 3x – x – 3 = x2 + 2x – 3
As LHS = RHS
x2 – x – 2 = x2 + 2x – 3
x2 – x – 2 – x2 – 2x + 3= 0
-3x + 1 = 0
It is not in the form of ax2 + bx + c = 0
Therefore, the given equation is not a quadratic equation.
(iv) (x – 3)(2x + 1) = x(x + 5)
Ans. LHS = (x – 3)(2x + 1) = 2x2 + x – 6x – 3 = 2x2 – 5x – 3
RHS = x(x + 5)= x2 + 5x
As LHS = RHS
2x2 – 5x – 3= x2 + 5x
2x2 – 5x – 3 – x2 – 5x= 0
x2 – 10x – 3 = 0
It is of the form ax2 + bx + c = 0
Therefore, the given equation is a quadratic equation.
(v) (2x – 1)(x – 3) = (x + 5)(x – 1)
Ans. LHS = (2x – 1)(x – 3) =2x2 – 6x – x + 3 = 2x2 – 7x + 3
RHS = (x + 5)(x – 1) = x2 – x + 5x – 5 = x2 + 4x – 5
As LHS = RHS
2x2 – 7x + 3 = x2 + 4x – 5
x2 + 1 + 2x – x2 – 4x + 5= 0
– 2x + 6 = 0
It is not in the form of ax2 + bx + c = 0
Therefore, the given equation is not a quadratic equation.
(vi) x2 + 3x + 1 = (x – 2)2
Ans. LHS = x2 + 3x + 1
RHS =(x – 2)2 = x2 + 4 – 4x
As LHS = RHS
x2 + 3x + 1 =x2 + 4 – 4x
x2 + 3x + 1 – x2 – 4 + 4x = 0
7x – 3 = 0
It is not in the form of ax2 + bx + c = 0
Therefore, the given equation is not a quadratic equation.
(vii) (x + 2)3 =2x(x2 -1)
Ans. LHS = (x + 2)3 = x3 + 8 + 6x(x + 2) = x3 + 8 + 6x2 +12x
RHS = 2x(x2 -1) = 2x3 – 2x
As LHS = RHS
x3 + 8 + 6x2 +12x = 2x3 – 2x
x3 + 8 + 6x2 +12x – 2x3 + 2x = 0
– x3 + 6x2 + 14x + 8 = 0
It is not in the form of ax2 + bx + c = 0
Therefore, the given equation is not a quadratic equation..
(viii) x3 – 4x2 – x + 1 = (x – 2)3
Ans. LHS = x3 – 4x2 – x + 1
RHS = (x – 2)3 = x3 – 8 – 6x(x – 2) = x3 – 8 – 6x2 +12x
As LHS = RHS
x3 – 4x2 – x + 1 = x3 – 8 – 6x2 +12x
x3 – 4x2 – x + 1 – x3 + 8 + 6x2 – 12x = 0
2x2 – 13x + 9 = 0
It is of the form ax2 +bx + c = 0
Therefore, 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.
Ans. Let x be the breadth of rectangular plot.
According to question, length = 2x + 1
Area of rectangle = length × breath
Therefore,
528 = (2x + 1) × x
528 = 2x2 + x
2x2 + x – 528 = 0
(ii) The product of two consecutive positive integers is 306. We need to find the integers.
Ans. Let x be the first positive integer.
Then second consecutive positive integer will be x + 1
According to question,
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 ages ( in years ) 3 years from now will be 360. We would like to find Rohan’s present age.
Ans. Let x be the present age of Rohan.
Then Present age of Rohan’s mother will be x + 26.
Three years from now, their product will be 360.
(x + 3) × (x + 26 + 3) = 360
(x + 3) × (x + 29) = 360
x2 + 29x + 3x + 69 = 360
x2 + 32x + 69 – 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 need to find the speed of the train.
Ans. As we know that Distance = Speed × Time
Let x be the speed of train.
According to question,
480/(x – 8) – 480/x = 3
{480x – 480(x – 8)} / x(x – 8) = 3 ( cross multiply )
480x – 480x + 3840 = 3x(x – 8)
3840 = 3x2 – 24x
3x2 – 24x – 3840 = 0 ( dividing equation by 3)
x2 – 8x – 1280 = 0