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NCERT New Syllabus Mathematics
Class: 10
Exercise 4.3
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 4.3
Q1. Find the nature of the roots of the following quadratic equations.
If the real roots exist, find them:
(i) 2x2 – 3x + 5 = 0 (ii) 3x2 – 4√3 x + 4 = 0 (iii) 2x2 – 6x + 3 = 0
Explanation:
1) In a quadratic equation of the form ax2 + bx + c = 0, where a ≠ 0, the expression
(b2 – 4ac) is known as discriminant.
2) Based on the discriminant, the nature of the roots of the quadratic equation
ax2 + bx + c = 0 can be determined as follows:
(a) If b2 – 4ac > 0, the equation has two distinct real roots.
(b) If b2 – 4ac = 0, the equation has two equal real roots.
(c) If b2 – 4ac < 0, the equation has no real roots.
Solution:
(i) 2x2 – 3x + 5 = 0
1) The given equation is 2x2 – 3x + 5 = 0 ------------------ equation 1.
2) Equate the coefficient of equation 2x2 – 3x + 5 = 0 with ax2 + bx + c = 0,
we have,
a = 2, b = – 3, c = 5.
3) First we will find:
b2 – 4ac = (– 3)2 – 4(2)(5)
b2 – 4ac = 9 – 40
b2 – 4ac = – 31 ------------------ equation 2.
4) Since b2 – 4ac < 0, the quadratic equation 2x2 – 3x + 5 = 0 has no real roots.
(ii) 3x2 – 4√3 x + 4 = 0
1) The given equation is 3x2 – 4√3 x + 4 = 0 ------------------ equation 1.
2) Equate the coefficient of equation 3x2 – 4√3 x + 4 = 0 with ax2 + bx + c = 0,
we have,
a = 3, b = – 4√3, c = 4.
3) First we will find:
b2 – 4ac = (– 4√3)2 – 4(3)(4)
b2 – 4ac = 48 – 48
b2 – 4ac = 0 ------------------ equation 2.
4) As, b2 – 4ac = 0, it has two equal real roots, so from equation 2 and equation 3,
we have,
x = [– b ± √(b2 – 4ac)]/2a
x = [– (– 4√3) ± √0]/2(3)
x = (4√3 ± 0)/2(3)
x = 2(√3)/3
x = 2(√3)/(√3√3)
x = 2/√3
x = (2√3)/3
5) Since, b2 – 4ac = 0, the equation has two equal real roots.
So, the roots are x = (2√3)/3 or x = (2√3)/3.
(iii) 2x2 – 6x + 3 = 0
1) The given equation is 2x2 – 6x + 3 = 0 ------------------ equation 1.
2) Equate the coefficient of equation 2x2 – 6x + 3 = 0 with ax2 + bx + c = 0,
we have,
a = 2, b = – 6, c = 3.
3) First we will find:
b2 – 4ac = (– 6)2 – 4(2)(3)
b2 – 4ac = 36 – 24
b2 – 4ac = 12 ------------------ equation 2.
4) As, b2 – 4ac ≥ 0, it has two distinct real roots, so from equation 2 and equation 3,
we have,
x = [– b ± √(b2 – 4ac)]/2a
x = [– (– 6) ± √12]/2(2)
x = (6 ± √12)/4
x = (6 ± 2√3)/4
x = (3 ± √3)/25) Since b2 – 4ac > 0, the equation has two distinct real roots.
So, the roots are x = (3 + √3)/2 or x = (3 – √3)/2.
Q2. Find the values of k for each of the following quadratic equations, so that
they have two equal roots.
(i) 2x2 + kx + 3 = 0 (ii) kx (x – 2) + 6 = 0
Explanation:
1) In a quadratic equation of the form ax2 + bx + c = 0, where a ≠ 0, the expression
(b2 – 4ac) is known as discriminant.
2) Based on the discriminant, the nature of the roots of the quadratic equation
ax2 + bx + c = 0 can be determined as follows:
(a) If b2 – 4ac = 0, the equation has two equal real roots.
Solution:
(i) 2x2 + kx + 3 = 0
1) The given equation is 2x2 + kx + 3 = 0 ------------------ equation 1.
2) Equate the coefficient of equation 2x2 + kx + 3 = 0 with ax2 + bx + c = 0,
we have,
a = 2, b = k, c = 3.
3) First we will find:
b2 – 4ac = (k)2 – 4(2)(3)
b2 – 4ac = k2 – 24 ------------------ equation 2.
4) As the quadratic equation has two equal roots,
b2 – 4ac = 0
k2 – 24 = 0
k2 = 24
k = ± √24 = ± 2√6
(ii) kx (x – 2) + 6 = 0
1) The given equation is
kx (x – 2) + 6 = 0
kx2 – 2kx + 6 = 0 ------------------ equation 1.
2) Equate the coefficient of equation kx2 – 2kx + 6 = 0 with ax2 + bx + c = 0,
we have,
a = k, b = – 2k, c = 6.
3) First we will find:
b2 – 4ac = (– 2k)2 – 4(k)(6)
b2 – 4ac = 4k2 – 24k ------------------ equation 2.
4) As the quadratic equation has two equal roots,
b2 – 4ac = 0
4k2 – 24k = 0
4k(k – 6) = 0
k(k – 6) = 0
5) So, k = 0 or k = 6.
Q3. Is it possible to design a rectangular mango grove whose length is twice
its breadth and the area is 800 m2? If so, find its length and breadth.
1) Let the breadth of a rectangular mango grove be x m.
2) So, the length of a rectangular mango grove will be 2x m
3) According to the problem, the area is 800, so
2x(x) = 800
2x2 = 800
x2 = 400
x2 – 400 = 0 ------------------ equation 1.
4) Equate the coefficient of equation x2 – 400 = 0 with ax2 + bx + c = 0,
we have,
a = 1, b = 0, c = – 400.
5) First we will find:
b2 – 4ac = (0)2 – 4(1)(– 400)
b2 – 4ac = 1600 ------------------ equation 2.
6) As b2 – 4ac = 1600 > 0, it has real roots, so from equation 1 and equation 2,
we have,
x = [– b ± √(b2 – 4ac)]/2a
x = [– (0) ± √1600]/2x = (0 ± 40)/2
7) So, x = (40)/2 or x = (– 40)/2, i.e. x = 20, or x = – 20.
8) As length is always positive, ignore x = – 20.
9) So, the breadth of the rectangular mango grove is 20 m and
its length is 40 m.
Q4. Is the following situation possible? If so, determine their present ages.
The sum of the ages of two friends is 20 years. Four years ago, the product of their age in years was 48.
1) Let the present age of the first friend be x.
2) So, the present age of the second friend will be (20 – x).
3) 4 years ago, their ages will be (x – 4) and (20 – x – 4).
4) According to the problem,
(x – 4)(16 – x) = 48
x(16 – x) – 4(16 – x) = 48
16x – x2 – 64 + 4x = 4820x – x2 – 64 – 48 = 0
20x – x2 – 112 = 0
x2 – 20x + 112 = 0 ------------------ equation 1.
5) Equate the coefficient of equation x2 – 20x + 112 = 0 with ax2 + bx + c = 0, so
a = 1, b = – 20, c = 112
6) First we will find:
b2 – 4ac = (– 20)2 – 4(1)(112)
b2 – 4ac = 400 – 448
b2 – 4ac = – 48 ------------------ equation 2.
7) Since b2 – 4ac = – 48 < 0, it has no real roots, so the given situation is
impossible.
Q5. Is it possible to design a rectangular park of perimeter 80 m and an area
of 400 m2? If so, find its length and breadth.
1) Let the breadth of a rectangular park be x m.
2) As, the perimeter of a rectangular park = 80 m.
3) So, length = [(perimeter/2) – breadth]
length = [(80/2) – x]
length = (40 – x)
4) According to the problem,
x(40 – x) = 400
40x – x2 = 400
x2 – 40x + 400 = 0 ------------------ equation 1.
5) Equate the coefficient of equation x2 – 40x + 400 = 0 with ax2 + bx + c = 0, so
a = 1, b = – 40, c = 400
6) First we will find:
b2 – 4ac = (– 40)2 – 4(1)(400)
b2 – 4ac = 1600 – 1600
b2 – 4ac = 0 ------------------ equation 2.
7) As, b2 – 4ac = 0, it has two equal real roots, so from equation 1 and equation 2,
we have,
x = [– b ± √(b2 – 4ac)]/2a
x = [– (– 40) ± √0]/2(1)
x = (40 ± 0)/2
x = (20 ± 0)8) Therefore, x = 20, the breadth = 20 m and the length = 40 – 20 = 20 m.
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 various 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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