Class 10 Mathematic 1 Quadratic Equations Practice Set 2.1 Solution
Class 10 Maths 1 – Practice Set 2.1 Solutions
Introduction
In this article, you will find complete step-by-step solutions to Practice Set 2.1 from Chapter 2 – Quadratic Equations. These solutions are prepared in a simple exam-oriented format for Maharashtra Board Class 10 students.
1. Write any two quadratic equations.
Answer:
Any two quadratic equations are:
1. x² + 5x − 2 = 0
2. y² − 5y + 10 = 0
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2. Decide which of the following are quadratic equations.
(1) x² + 5x − 2 = 0
Solution:
In the equation x² + 5x − 2 = 0
x is the only variable and maximum
index of the variable is 2
Therefore, x² + 5x − 2 = 0 is a quadratic equation.
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(2) y² = 5y − 10
Solution:
Write the equation in standard form:
y² − 5y + 10 = 0
In the equation. y² − 5y + 10 = 0
y is the only variable and maximum
index of the variable is 2
Therefore, y² = 5y − 10 is a quadratic equation.
Answer: Quadratic Equation ✔
(3) y² + 1/y = 2
Solution:
The variable y appears in the denominator (1/y). Hence, it is not a polynomial equation.
Answer: Not a quadratic equation.
(4) x + 1/x = -2
Solution:
The variable x appears in the denominator (1/x). Hence, it is not a polynomial equation.
Answer: Not a quadratic equation.
(5) (m + 2)(m - 5) = 0
Solution:
Expand the equation:
(m + 2)(m - 5) = 0
m² - 5m + 2m - 10 = 0
m² - 3m - 10 = 0
In the equation m is the only variable and maximum index of the variable is 2
Therefore, (m + 2)(m - 5) = 0 is a quadratic equation.
Answer: Quadratic equation.
(6) m³ + 3m² - 2 = 3m³
Solution:
Bring all terms to one side:
m³ + 3m² - 2 - 3m³ = 0
-2m³ + 3m² - 2 = 0
The highest power of m is 3.
Answer: Not a quadratic equation
3. Write in the form ax² + bx + c = 0 and write a, b, c
(1) 2y = 10 - y²
Solution
y² + 2y - 10 = 0
a = 1, b = 2, c = -10
(2) (x - 1)² = 2x + 3
Solution
x² - 2x + 1 = 2x + 3
x² - 4x - 2 = 0
a = 1, b = -4, c = -2
(3) x² + 5x = -(3 - x)
Solution
x² + 5x = -3 + x
x² + 4x + 3 = 0
a = 1, b = 4, c = 3
(4) 3m² = 2m² - 9
Solution
3m²-2m²+9=0
m² + 9 = 0
a = 1, b = 0, c = 9
(5) p(3 + 6p) = -5
Solution
3p + 6p² + 5 = 0
6p² + 3p + 5 = 0
a = 6, b = 3, c = 5
(6) x² - 9 = 13
Solution
x²-9-13=0
x² - 22 = 0
a = 1, b = 0, c = -22
4. Determine whether the given values are roots of the equation.
(1) x² + 4x − 5 = 0 ; x = 1, −1
Solution:
To check whether a given value is a root, substitute it into the equation.
For x = 1:
LHS=x² + 4x − 5
= (1)² + 4(1) − 5
= 1 + 4 − 5
= 0
Since the LHS becomes 0, x = 1 is a root.
For x = −1:
LHS= x² + 4x − 5
= (−1)² + 4(−1) − 5
= 1 − 4 − 5
= −8
Since the LHS is not equal to 0, x = −1 is not a root.
Answer:
✓ x = 1 is a root.
✗ x = −1 is not a root.
(2) 2m² − 5m = 0 ; m = 2, 5/2
Solution:
For m = 2:
LHS= 2m² − 5m
= 2(2)² − 5(2)
= 8 − 10
= −2
Therefore, m = 2 is not a root.
For m = 5/2:
LHS= 2m² − 5m
= 2(5/2)² − 5(5/2)
= 2 × 25/4 − 25/2
= 25/2 − 25/2
= 0
Therefore, m = 5/2 is a root.
Answer:
✗ m = 2 is not a root.
✓ m = 5/2 is a root.
5. Find k if x = 3 is a root of the equation
kx² − 10x + 3 = 0.
Solution:
Given equation:
kx² − 10x + 3 = 0
Since x = 3 is a root, substitute x = 3.
k(3)² − 10(3) + 3 = 0
9k − 30 + 3 = 0
9k − 27 = 0
9k = 27
k = 27 ÷ 9
k = 3
Final Answer:
✓ k = 3
6. One of the roots of the equation 5m² + 2m + k = 0 is −7/5. Find the value of k by completing the activity.
Solution
Given:
Quadratic equation: 5m² + 2m + k = 0
One root = −7/5
Since −7/5 is a root of the equation, substitute m = −7/5 into the equation.
5(−7/5)² + 2(−7/5) + k = 0
5 × 49/25 − 14/5 + k = 0
49/5 − 14/5 + k = 0
35/5 + k = 0
7 + k = 0
k = −7
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