Master Exercise 2.5 Class 10 with Ease
In this post, we will explore the process of forming a quadratic equation when the roots are given. Exercise 2.5 class 10 is based on the concept of constructing a quadratic equation from its roots, using the formula:
x² − (sum of the roots)x + (product of the roots) = 0.
We will solve various problems that help reinforce this concept, ensuring a solid understanding of how to apply this formula. These solutions of exercise 2.5 Class 10 provide clear steps to make learning easier and more effective.
Question 1 of Exercise 2.5 Class 10. Write the quadratic equation having the following roots
(a) Roots: 1, 5
Since 1 and 5 are the roots of the required quadratic equation, therefore:
Sum of roots S = 1 + 5 = 6
Product of roots P = 1 × 5 = 5
As x² – Sx + P = 0, so the required equation is:
x² – 6x + 5 = 0
(b) Roots: 4, 9
Since 4 and 9 are the roots of the required quadratic equation, therefore:
Sum of roots S = 4 + 9 = 13
Product of roots P = 4 × 9 = 36
As x² – Sx + P = 0, so the required equation is:
x² – 13x + 36 = 0
(c) Roots: -2, 3
Since -2 and 3 are the roots of the required quadratic equation, therefore:
Sum of roots S = -2 + 3 = 1
Product of roots P = -2 × 3 = -6
As x² – Sx + P = 0, so the required equation is:
x² – x – 6 = 0
(d) Roots: 0, -3
Since 0 and -3 are the roots of the required quadratic equation, therefore:
Sum of roots S = 0 + (-3) = -3
Product of roots P = 0 × (-3) = 0
As x² – Sx + P = 0, so the required equation is:
x² – 3x = 0
(e) Roots: 2, -6
Since 2 and -6 are the roots of the required quadratic equation, therefore:
Sum of roots S = 2 + (-6) = -4
Product of roots P = 2 × (-6) = -12
As x² – Sx + P = 0, so the required equation is:
x² + 4x – 12 = 0
(f) Roots: -1, -7
Since -1 and -7 are the roots of the required quadratic equation, therefore:
Sum of roots S = -1 + (-7) = -8
Product of roots P = -1 × -7 = 7
As x² – Sx + P = 0, so the required equation is:
x² + 8x + 7 = 0
(g) Roots: 1 + i, 1 – i
Since 1 + i and 1 – i are the roots of the required quadratic equation, therefore:
Sum of roots S = (1 + i) + (1 – i) = 2
Product of roots P = (1 + i)(1 – i) = 1² – i² = 1 – (-1) = 2
As x² – Sx + P = 0, so the required equation is:
x² – 2x + 2 = 0
(h) Roots: 3 + √2, 3 – √2
Since 3 + √2 and 3 – √2 are the roots of the required quadratic equation, therefore:
Sum of roots S = (3 + √2) + (3 – √2) = 6
Product of roots P = (3 + √2)(3 – √2) = 3² – (√2)² = 9 – 2 = 7
As x² – Sx + P = 0, so the required equation is:
x² – 6x + 7 = 0
Question 2 of Exercise 2.5 Class 10. If α, β are the roots of the equation x² – 3x + 6 = 0
form an equation whose roots are:
(a) 2α + 1, 2β + 1 (b) -α², -β² (c) 1/α, 1/β (d) α/β, β/α (e) α + 1/α, β + 1/β
Solution:
As α, β are the roots of the equation x² – 3x + 6 = 0
a = 1, b = -3, c = 6
∴
α + β = -b/a = -(-3)/1 = 3 ⇒ α + β = 3
(a) 2α + 1, 2β + 1
Sum of roots:
S = 2α + 1 + 2β + 1
S = 2(α + β) + 2
S = 2(3) + 2 = 6 + 2 = 8
Product of roots:
P = (2α + 1)(2β + 1)
P = 4αβ + 2α + 2β + 1
P = 4αβ + 2(α + β) + 1
P = 4(6) + 2(3) + 1 = 24 + 6 + 1 = 31
Using x² – Sx + P = 0, we have:
x² – 8x + 31 = 0
(b) -α², -β²
Sum of roots:
S = α² + β²
S = (α + β)² – 2αβ
S = (3)² – 2(6) = 9 – 12 = -3
Product of roots:
P = α²β² = (αβ)²
P = (6)² = 36
Using x² – Sx + P = 0, we have:
x² + 3x + 36 = 0
(c) 1/α, 1/β
Sum of roots:
S = 1/α + 1/β = (α + β)/αβ
S = 3/6 = 1/2
Product of roots:
P = 1/α × 1/β = 1/αβ
P = 1/6
Using x² – Sx + P = 0, we have:
x² – (1/2)x + (1/6) = 0
Multiplying by 6 on both sides, we have:
6x² – 3x + 1 = 0
(d) α/β, β/α
Sum of roots:
S = α/β + β/α
S = (α² + β²)/αβ
S = ((α + β)² – 2αβ)/αβ
S = (3² – 2(6))/6
S = (9 – 12)/6 = -3/6
S = -1/2
Product of roots:
P = (α/β)(β/α) = 1
Using x² – Sx + P = 0, we have:
x² – (1/2)x + 1 = 0
Multiplying both sides by 2, we have:
2x² – x + 2 = 0
(e) α + 1/α, β + 1/β
Sum of roots:
S = (α + β) + (1/α + 1/β)
S = (α + β) + (α + β)/αβ
S = 3 + 3/6
S = 3 + 1/2
S = (6 + 1)/2
S = 7/2
Product of roots:
P = (α + β)(1/α + 1/β)
P = (α + β)(α + β)/αβ
P = 3(3/6)
P = 9/6
P = 3/2
Using x² – Sx + P = 0, we have:
x² – (7/2)x + (3/2) = 0
Multiplying both sides by 2, we have:
2x² – 7x + 3 = 0
Question 3 of Exercise 2.5 Class 10. If α, β are the roots of the equation x² + px + q = 0, form the equation whose roots are
(a) α², β² (b) α/β, β/α
Solution:
Since α, β are the roots of the equation x² + px + q = 0
ax² + bx + c = 0
By comparing the coefficients of these equations, we have:
a = 1, b = p, c = q
α + β = -b/a = -p/1 = -p ⇒ α + β = -p
αβ = c/a = q/1 = q ⇒ αβ = q
(a) α², β²
Sum of roots:
S = α² + β²
S = (α + β)² – 2αβ
S = (-p)² – 2q
S = p² – 2q
Product of roots:
P = α²β²
P = (αβ)²
P = q²
As x² – Sx + P = 0 so the required equation is:
x² – (p² – 2q)x + q² = 0
(b) α/β, β/α
Since α, β are the roots of the equation x² + px + q = 0
ax² + bx + c = 0
By comparing the coefficients of these equations, we have:
a = 1, b = p, c = q
α + β = -b/a = -p/1 = -p ⇒ α + β = -p
αβ = c/a = q/1 = q ⇒ αβ = q
Sum of roots:
S = α/β + β/α
S = (α² + β²)/αβ
S = ((α + β)² – 2αβ)/αβ
S = ((-p)² – 2q)/q
S = (p² – 2q)/q
Product of roots:
P = (α/β)(β/α) = 1
As x² – Sx + P = 0 so the required equation is:
x² – ((p² – 2q)/q)x + 1 = 0
Multiplying by q on both sides, we have:
qx² – (p² – 2q)x + q = 0
We hope this post has helped you understand how to form a quadratic equation when the roots are given, as covered in Exercise 2.5 class 10. For students looking to deepen their understanding of quadratic equations, we recommend checking out this introduction to quadratic equations on Math is Fun. This resource provides easy-to-follow explanations, examples, and interactive tools to help make learning quadratic equations more engaging and accessible. With consistent practice, this concept will become second nature.
If you’d like to review the concepts from the previous exercise, you can check out Exercise 2.4 solutions. For those ready to move forward, the next step is Exercise 2.6 solutions, where we will delve into solving quadratic equations by factoring.
Here’s a video lecture of this exercise 2.5 class 10 to help you understand the concepts better.
