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Class 10 (SSC Board)Mathematics & Statistics2026-27 Syllabus

Chapter 2 Quadratic Equations Practice Set 2.1 Solutions

Complete Maharashtra State Board Balbharati & Yuvakbharati textbook solutions for Chapter 2 Quadratic Equations Practice Set 2.1. Step-by-step solved exercises, numerical problems, and digest answers.

10 Solved Questions• 6 Diagrams• 1523 words

Practice Set 2.1 Algebra 10th Std Maths Part 1 Answers Chapter 2 Quadratic Equations

Question 1 Maharashtra Board Solution
Write any two quadratic equations.
Solution & Step-by-Step Answer:
i. y2 – 7y + 12 = 0 ii. x2 – 8 = 0
Question 2 Maharashtra Board Solution
Decide which of the following are quadratic i. x2 – 7y + 2 = 0 ii. y2 = 5y – 10 iii. y2 + = 2 iv. x + = -2 v. (m + 2) (m – 5) = 03 vi. m3 + 3m2 – 2 = 3m3
Solution & Step-by-Step Answer:
i. The given equation is x2 + 5x – 2 = 0 Here, x is the only variable and maximum index of the variable is 2. a = 1, b = 5, c = -2 are real numbers and a ≠ 0. ∴ The given equation is a quadratic equation.

ii. The given equation is
y2= 5y – 10
∴ y2– 5y + 10 = 0
Here, y is the only variable and maximum index of the variable is 2.
a = 1, b = -5, c = 10 are real numbers and a ≠ 0.
∴ The given equation is a quadratic equation.

iii. The given equation is
y2+ = 2
∴ y3+ 1 = 2y …[Multiplying both sides by y]
∴ y3– 2y + 1 = 0
Here, y is the only variable and maximum index of the variable is not 2.
∴ The given equation is not a quadratic equation.

iv. The given equation is
x + = -2
∴ x2+ 1 = -2x …[Multiplying both sides by x]
∴ x2+ 2x+ 1 = 0
Here, x is the only variable and maximum index of the variable is 2.
a = 1, b = 2, c = 1 are real numbers and a ≠ 0.
∴ The given equation is a quadratic equation.

v. The given equation is
(m + 2) (m – 5) = 0
∴ m(m – 5) + 2(m – 5) = 0
∴ m2– 5m + 2m – 10 = 0
∴ m2– 3m – 10 = 0
Here, m is the only variable and maximum index of the variable is 2.
a = 1, b = -3, c = -10 are real numbers and a ≠ 0.
∴ The given equation is a quadratic equation.

vi. The given equation is
m3+ 3m2– 2 = 3m3
∴ 3m3– m3– 3m2+ 2 = 0
∴ 2m3– 3m2+ 2 = 0
Here, m is the only variable and maximum
index of the variable is not 2.
∴ The given equation is not a quadratic equation.

Question 3 Maharashtra Board Solution
Write the following equations in the form ax2 + bx + c = 0, then write the values of a, b, c for each equation. i. 2y = 10 – y2 ii. (x – 1)2 = 2x + 3 iii. x2 + 5x = – (3 – x) iv. 3m2 = 2m2 – 9 v. P (3 + 6p) = – 5 vi. x2 – 9 = 13
Solution & Step-by-Step Answer:
i. 2y – 10 – y2 ∴ y2 + 2y – 10 = 0 Comparing the above equation with ay2 + by + c = 0, we get a = 1, b = 2, c = -10

ii. (x – 1)2= 2x + 3
∴ x2– 2x + 12x + 3
x2– 2x + 1 – 2x – 30
∴ x2– 4x – 2 = 0
Comparing the above equation with
ax2+ bx + c = 0, we get
a = 1, b = -4, c = -2

iii. x2+ 5x = – (3 – x)
∴ x2+ 5x = -3 + x
∴ x2+ 5x – x + 3 = 0
∴ x2+ 4x + 3 = 0
Comparing the above equation with
ax2+ bx + c = 0, we get
a = 1, b = 4, c = 3

iv. 3m2= 2m2– 9
∴ 3m2– 2m2+ 9 = 0
∴ m2+ 9 = 0
∴ m2+ 0m + 9 = 0
Comparing the above equation with
am2+ bm + c = 0, we get
a = 1, b = 0, c = 9

v. p (3 + 6p) = – 5
∴ 3p + 6p2= -5
∴ 6p2+ 3p + 5 = 0
Comparing the above equation with
ap2+ bp + c = 0, we get
a = 6, b = 3, c = 5

vi. x2– 9 = 13
∴ x2– 9 – 13 = 0
∴ x2– 22 = 0
∴ x2+ 0x – 22 = 0
Comparing the above equation with
ax2+ bx + c = 0, we get
a = 1, b = 0, c = -22

Question 4 Maharashtra Board Solution
Determine whether the values given against each of the quadratic equation are the roots of the equation. i. x2 + 4x – 5 = 0; x = 1,-1 ii. 2m2 – 5m = 0; m = 2,
Solution & Step-by-Step Answer:
i. The given equation is x2 + 4x – 5 = 0 …(i) Putting x = 1 in L.H.S. of equation (i), we get L.H.S. = (1)2 + 4(1) – 5 = 1 + 4 – 5 = 0 ∴ L.H.S. = R.H.S. ∴ x = 1 is the root of the given quadratic equation. Putting x = -1 in L.H.S. of equation (i), we get L.H.S. = (-1)2 + 4(-1) – 5 = 1 – 4 – 5 = -8 ∴ LH.S. ≠ R.H.S. ∴ x = -1 ¡s not the root of the given quadratic equation.

ii. The given equation is
2m2– 5m = 0 …(i)
Putting m = 2 in L.H.S. of equation (i), we get
L.H.S. = 2(2)2– 5(2) = 2(4) -10 = 8 – 10 = -2
∴ L.H.S. ≠ R.H.S.
∴ m = 2 is not the root of the given quadratic equation.
Putting m = in L.H.S. of equation (i), we get

Question 5 Maharashtra Board Solution
Find k if x = 3 is a root of equation kx2 – 10x + 3 = 0.
Solution & Step-by-Step Answer:
x = 3 is the root of the equation kx2 – 10x + 3 = 0. Putting x = 3 in the given equation, we get k(3)2 – 10(3) + 3 = 0 ∴ 9k – 30 +3 = 0 ∴ 9k – 27 = 0 ∴ 9k = 27 ∴ k = ∴ k = 3
Question 6 Maharashtra Board Solution
One of the roots of equation 5m2 + 2m + k = 0 is Complete the following activity to find the value of ‘k’.
Solution & Step-by-Step Answer:

Question 1 Maharashtra Board Solution
x2 + 3x – 5, 3x2 – 5x, 5x2; Write the polynomials In the index form. Observe the coefficients and fill in the boxes. (Textbook p. no. 31)
Solution & Step-by-Step Answer:
Index form of the given polynomials: x2 + 3x – 5, 3x2 – 5x + 0, 5x2 + 0x + 0 i. Coefficients of x2 are [1], [3] and [5] respectively, and these coefficients are non zero. ii. Coefficients of x are 3, [-5] and [0] respectively. iii. Constant terms are [-5], [0] and [0] respectively. Here, constant terms of second and third polynomial is zero.
Question 2 Maharashtra Board Solution
Complete the following table (Textbook p. no. 31)
Solution & Step-by-Step Answer:

Question 3 Maharashtra Board Solution
Decide which of the following are quadratic equations? (Textbook pg. no. 31) i. 9y2 + 5 = 0 ii. m3 – 5m2 + 4 = 0 iii. (l + 2)(l – 5) = 0
Solution & Step-by-Step Answer:
i. In the equation 9y2 + 5 = 0, [y] is the only variable and maximum index of the variable is [2]. ∴ It [is] a quadratic equation.

ii. In the equation m3– 5m2+ 4 = 0, [m] is the only variable and maximum index of the variable is not 2.
∴ It [is not] a quadratic equation.

iii. (l + 2)(l – 5) = 0
∴ l(l – 5) + 2(l – 5) = 0
∴ l2– 5l + 2l – 10 = 0
∴ l2– 3l – 10 = 0.
In this equation [l] is the only variable and maximum index of the variable is [2]
∴ it [is] a quadratic equation.

Question 4 Maharashtra Board Solution
If x = 5 is a root of equation kx2 – 14x – 5 = 0, then find the value of k by completing the following activity. (Textbook pg, no. 33)
Solution & Step-by-Step Answer:

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Chapter 1 Linear Equations in Two Variables Problem Set 1
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Chapter 2 Quadratic Equations Practice Set 2.2