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NCERT New Syllabus Mathematics
Class: 10
Exercise 4.1
Topic: Quadratic Equation
Understanding Quadratic Equations in the New NCERT Syllabus for Class 10
Quadratic equations form a fundamental concept in algebra, crucial for building a strong mathematical foundation. In the new Class 10 syllabus of NCERT, quadratic equations take center stage, preparing students to solve real-world problems efficiently using algebraic techniques.
A quadratic equation is a polynomial equation of degree two, expressed in the form:
ax2 + bx + c = 0 where a, b, and c are real numbers, with a ≠ 0.
In this blog, we will explore the various methods of solving quadratic equations, including factorization, completing the square, and the quadratic formula. We’ll also dive into the geometric significance of these equations and how they apply to practical scenarios.
Let’s start by breaking down the key components of quadratic equations and how to solve them step by step!
EXERCISE 4.1
1. Check whether the following are quadratic equations :
(i) (x + 1)2 = 2(x – 3) (ii) x2 – 2x = (–2) (3 – x)
(iii) (x – 2)(x + 1) = (x – 1)(x + 3) (iv) (x – 3)(2x +1) = x(x + 5)
(v) (2x – 1)(x – 3) = (x + 5)(x – 1) (vi) x2 + 3x + 1 = (x – 2)2
(vii) (x + 2)3 = 2x (x2 – 1) (viii) x3 – 4x2 – x + 1 = (x – 2)3
Explanation:
1) A quadratic polynomial is expressed as ax2 + bx + c, where a ≠ 0.
2) A quadratic equation takes the form ax2 + bx + c = 0, where a ≠ 0.
3) Any equation of the form p(x) = 0, where p(x) is a polynomial of degree 2, is
called a quadratic equation.
4) The standard form of a quadratic equation is ax2 + bx + c = 0, where a ≠ 0.
The degree of this equation is 2.
Solution:
(i) (x + 1)2 = 2(x – 3)
1) The given equation is:
(x + 1)2 = 2(x – 3)x2 + 2x +1 = 2x – 6
x2 + 2x +1 – 2x + 6 = 0
x2 + 7 = 0 -------------------equation 1
2) Since the highest power of the variable x is 2, it confirms that this is a
quadratic equation.
(ii) x2 – 2x = (– 2) (3 – x)
1) The given equation is:
x2 – 2x = (–2) (3 – x)x2 – 2x = – 6 + 2x
x2 – 2x + 6 – 2x = 0
x2 – 4x + 6 = 0 -------------------equation 1
2) Since the highest power of the variable x is 2, it confirms that this is a
quadratic equation.
(iii) (x – 2)(x + 1) = (x – 1)(x + 3)
1) The given equation is:
(x – 2)(x + 1) = (x – 1)(x + 3)x(x + 1) – 2(x + 1) = x(x + 3) – (x + 3)
x2 + x – 2x – 2 = x2 + 3x – x – 3
x2 – x – 2 = x2 + 2x – 3
x2 – x – 2 – x2 – 2x + 3 = 0
– 3x + 1 = 0
3x – 1 = 0 -------------------equation 1
2) Since the highest power of the variable x is 1, it is not the quadratic equation.
(iv) (x – 3)(2x + 1) = x(x + 5)
1) The given equation is:
(x – 3)(2x +1) = x(x + 5)x(2x + 1) – 3(2x + 1) = x(x + 5)
2x2 + x – 6x – 3 = x2 + 5x
2x2 + x – 6x – 3 – x2 – 5x = 0
x2 – 10x – 3 = 0 -------------------equation 1
2) Since the highest power of the variable x is 2, it confirms that this is a
quadratic equation.
(v) (2x – 1)(x – 3) = (x + 5)(x – 1)
1) The given equation is:
(2x – 1)(x – 3) = (x + 5)(x – 1)2x(x – 3) – (x – 3) = x(x – 1) + 5(x – 1)2x2 – 6x – x + 3 = x2 – x + 5x – 5
2x2 – 6x – x + 3 – x2 + x – 5x + 5 = 0
2x2 – x2 – 6x – x + x – 5x + 3 + 5 = 0
x2 – 11x + 8 = 0 -------------------equation 1
2) Since the highest power of the variable x is 2, it confirms that this is a
quadratic equation.
(vi) x2 + 3x + 1 = (x – 2)2
1) The given equation is:
x2 + 3x + 1 = (x – 2)2x2 + 3x + 1 = x2 – 4x + 4
x2 + 3x + 1 – x2 + 4x – 4 = 0
x2 – x2 + 3x + 4x + 1 – 4 = 0
7x – 3 = 0-------------------equation 1
2) Since the highest power of the variable x is 1, it is not the quadratic equation.
(vii) (x + 2)3 = 2x (x2 – 1)
1) We know that (a + b)3 = a3 + 3a2 b + 3a b2 + b3
(x + 2)3 = 2x (x2 – 1)x3 + 3x2 (2) + 3x (2)2 + 23 = 2x3 – 2x
x3 + 6x2 + 12x + 8 = 2x3 – 2x
x3 + 6x2 + 12x + 8 – 2x3 + 2x = 0
x3 – 2x3 + 6x2 + 12x + 2x + 8 = 0
– x3 + 6x2 + 14x + 8 = 0 -------------------equation 1
2) Since the highest power of the variable x is 3, it is not the quadratic equation.
(viii) x3 – 4x2 – x + 1 = (x – 2)3
1) We know that (a – b)3 = a3 – 3a2 b + 3a b2 – b3
x3 – 4x2 – x + 1 = (x – 2)3x3 – 4x2 – x + 1 = x3 – 3x2 (2) + 3x (2)2 – 23
x3 – 4x2 – x + 1 = x3 – 6x2 + 12x – 8
x3 – 4x2 – x + 1 – x3 + 6x2 – 12x + 8 = 0
x3 – x3 – 4x2 + 6x2 – x – 12x + 1 + 8 = 0
2x2 – 13x + 9 = 0 -------------------equation 1
2) Since the highest power of the variable x is 2, it confirms that this is indeed a
quadratic equation.
Q2. 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.(ii) The product of two consecutive positive integers is 306. We need to find the 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 Rohan’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.
Explanation:
1) Let x be any suitable variable representing the unknown value.
2) Based on the given conditions, frame the quadratic equation in the standard
form ax2 + bx + c = 0, where a ≠ 0.
3) Solve this quadratic equation using an appropriate method (such as factorization,
completing the square, or the quadratic formula) to find the required solutions.
Solution:
(i) The area of a rectangular plot is 528 m2. The length of the plot (in meters)
is one more than twice its breadth. We need to find the length and breadth of the plot.
1) Let the breadth of the plot be x meters, as the length depends on it.
2) According to the problem, the length is more than twice the breadth,
so the length is (2x + 1) meters.
3) Given that the area of the plot is 528 m2, we can write the equation for the area
as:
x(2x + 1) = 528
2x2 + x = 528
2x2 + x – 528 = 0 ------------ equation 1
4) Therefore, the required quadratic equation is:
2x2 + x – 528 = 0, where x is breadth (in metres) of the plot.
(ii) The product of two consecutive positive integers is 306. We need to find the integers.
1) Let the first integer be x.
2) The next consecutive integer will be (x + 1).
3) Since the product of the two integers is 306, we can write the equation as:
x(x + 1) = 306
x2 + x = 306
x2 + x – 306 = 0 ------------ equation 1
4) Therefore, the required quadratic equation is: 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
(in years) 3 years from now will be 360. We would like to find Rohan’s present age.
1) Let Rohan's current age be x.
2) According to the problem, his mother's age is (x + 26).
3) Three years later, Rohan's age will be (x + 3), and his mother's age will be
(x + 29).
4) From the given condition, the product of their ages after 3 years is 360:
(x + 3)(x + 29) = 360
x(x + 29) + 3(x + 29) = 360
x2 + 29x + 3x + 87 = 360
x2 + 32x + 87 = 360
x2 + 32x + 87 – 360 = 0
x2 + 32x – 273 = 0 ------------ equation 1
4) Therefore, the required quadratic equation is x2 + 32x – 273 = 0.
where x (in years) is the present age of Rohan.
(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.
1) Let the uniform speed of a train be x km/h.
2) To cover 480 km, the time taken will be (480/x) hrs.
3) If the speed is reduced by 8 km/h, then time will be increased by 3 hrs, so
new speed = (x – 8) km/h, and new time = [(480/x) + 3] hrs.
4) According to the condition,
(x – 8)[(480/x) + 3] = 480
(x – 8)[(480 + 3x)/x] = 480
(x – 8)(480 + 3x) = 480x
x(480 + 3x) – 8(480 + 3x) = 480x
3x2 + 480x – 3840 – 24x = 480x
3x2 – 24x – 3840 = 0
x2 – 8x – 1280 = 0 ------------ equation 1
5) The required quadratic equation is x2 – 8x – 1280 = 0.
where x (in km/h) is the speed of the train.
Conclusion: Unlocking the Power of Quadratic Equations
As we conclude our exploration of Quadratic Equations, we’ve unlocked a crucial mathematical tool that helps us solve a variety of complex problems, from calculating areas to predicting trajectories. Mastering the factorization method, completing the square, and applying the quadratic formula equips you with versatile techniques to tackle challenges beyond the classroom. Embrace these concepts, as they lay the groundwork for higher-level math and real-world applications. Keep pushing your limits, and remember, every problem has a solution—just like every quadratic equation has its roots!
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