Exercise 1.1 Class 10 Expert Solutions
Welcome to the Solutions of Exercise 1.1 Class 10 Math! In this exercise, we dive into the foundational concepts of quadratic equations, focusing on solving equations, simplifying expressions, and understanding the properties of quadratics. These detailed solutions are designed to guide you step-by-step through each problem, helping you build confidence and clarity as you work through the material. Whether you’re studying for exams or simply aiming to strengthen your math skills, these solutions will serve as a valuable resource for mastering the essentials of quadratic equations. Get ready to deepen your understanding and tackle each question with ease!
Question 1. Question 2. Question 3. Video Lecture
Question 1. Write the following quadratic equations in the standard form and point out pure quadratic equations.
(i) (x + 7)(x – 3) = −7
Solution:
(x + 7)(x – 3) = −7
x² + 4x − 21 = −7
x² + 4x − 21 + 7 = 0
x² + 4x − 14 = 0 (standard form)
(ii) (x² + 4) / 3 − x / 7 = 1
Solution:
(x² + 4) / 3 − x / 7 = 1
Multiplying both sides by 21 (L.C.M. of 3, 7)
21 * ((x² + 4) / 3) − 21 * (x / 7) = 21 * 1
7(x² + 4) − 3x = 21
7x² + 28 − 3x = 21
7x² + 28 − 3x − 21 = 0
7x² − 3x + 7 = 0 (standard form)
(iii) (x / (x + 1)) + ((x + 1) / x) = 6
Solution:
(x / (x + 1)) + ((x + 1) / x) = 6
Multiplying both sides by (x + 1)(x):
(x + 1)(x)((x) / (x + 1)) + (x + 1)(x)((x + 1) / x) = 6(x + 1)(x)
x² + (x + 1)(x + 1) = 6(x² + x)
x² + x² + 2x + 1 = 6x² + 6x
x² + x² + 2x + 1 – 6x² – 6x = 0
-4x² – 4x + 1 = 0
4x² + 4x – 1 = 0 (standard form)
(IV)(x + 4) / (x – 2) – (x – 2) / x + 4 = 0
Solution:
(x + 4) / (x – 2) – (x – 2) / x + 4 = 0
Multiplying both sides by ((x – 2)(x)):
(x – 2)(x) * ((x + 4) / (x – 2)) – (x – 2)(x) * ((x – 2) / x) + 4(x – 2)(x) = 0
x(x + 4) – (x – 2)(x – 2) + 4(x^2 – 2x) = 0
x² + 4x – (x² – 4x + 4) + 4x² – 8x = 0
x² + 4x – x² + 4x – 4 + 4x² – 8x = 0
4x² – 4 = 0 (dividing by 4)
x² – 1 = 0 (pure quadratic equation)
(V) (x + 3) / (x + 4) – (x – 5) / x = 1
Solution:
(x + 3) / (x + 4) – (x – 5) / x = 1
Multiplying by (x(x + 4)) both sides:
x(x + 4) * ((x + 3) / (x + 4)) – x(x + 4) * ((x – 5) / x) = x(x + 4)(1)
x(x + 3) – (x + 4)(x – 5) = x² + 4x
x² + 3x – x² + x + 20 = x² + 4x
x² + 3x – x² + x + 20 – x² – 4x = 0
4x + 20 – x² – 4x = 0
-x² + 20 = 0
=> x² – 20 = 0 (pure quadratic equation)
(VI) (x + 1) / (x + 2) + (x + 2) / (x + 3) = 25 / 12
Solution:
Question 2 of Exercise 1.1 Class 10. Solve by factorization:
(i) x² – x – 20 = 0
Solution:
x² – x – 20 = 0
x² – 5x + 4x – 20 = 0
x(x – 5) + 4(x – 5) = 0
(x – 5)(x + 4) = 0
=> x – 5 = 0 or x + 4 = 0
x = 5 or x = -4
Solution set = ({5, -4})
(ii) 3y² = y(y – 5)
Solution:
3y² = y(y – 5)
3y² = y² – 5y
3y² – y² + 5y = 0
2y² + 5y = 0
y(2y + 5) = 0
=> y = 0 or 2y + 5 = 0
Thus:
y = 0 and 2y = -5
y = -5 / 2
Solution set is {0, -5 / 2}
(iii) 4 – 32x = 17x²
Starting with the equation: 4 – 32x = 17x²
Rewrite it in standard form: 4 – 32x – 17x² = 0
Rearrange to: 17x² + 32x – 4 = 0
Now factor by grouping: 17x² + 34x – 2x – 4 = 0
Factor out common terms: 17x(x + 2) – 2(x + 2) = 0
Factor (x + 2): (x + 2)(17x – 2) = 0
Set each factor to zero: x + 2 = 0 or 17x – 2 = 0
Solving these gives: x = -2 or x = 2 / 17
(iv)x² – 11x = 152
First, rewrite the equation in standard form: x² – 11x – 152 = 0
Next, factorize the expression by finding two numbers that multiply to (-152) and add to (-11). These numbers are (-19) and (8).
So, the equation factors to: (x – 19)(x + 8) = 0
Now, set each factor equal to zero:
x – 19 = 0 or x + 8 = 0
This gives the solutions:
x = 19 and x = -8
(V) (x + 1) / x + x / (x + 1) = 25 / 12
Solution:
(x + 1) / x + x / (x + 1) = 25 / 12
Multiplying both sides by (12x(x + 1)):
12x(x + 1) * ((x + 1) / x) + 12x(x + 1) * (x / (x + 1)) = 25 / 12 * 12x(x + 1)
Simplifying:
12(x + 1)(x + 1) + 12x(x) = 25x(x + 1)
12(x² + 2x + 1) + 12x² = 25(x² + x)
12x² + 24x + 12 + 12x² = 25x² + 25x
12x² + 24x + 12 + 12x² – 25x² – 25x = 0
12x² + 12x² – 25x² + 24x – 25x + 12 = 0
-x² – x + 12 = 0
or
x² + x – 12 = 0
x² + 4x – 3x – 12 = 0
x(x + 4) – 3(x + 4) = 0
(x + 4)(x – 3) = 0
Thus, x + 4 = 0 or x – 3 = 0
Rightarrow x = -4, x = 3
Solution Set {-4, 3}
(VI) (2 / (x – 9)) = (1 / (x – 3)) – (1 / (x – 4))
Solution:
2 / (x – 9) = (1 / (x – 3)) – (1 / (x – 4))
Simplifying the right side:
2 / (x – 9) = ((x – 4) – (x – 3)) / ((x – 3)(x – 4))
2 / (x – 9) = (x – 4 – x + 3) / ((x – 3)(x – 4))
2 / (x – 9) = -1 / ((x – 3)(x – 4))
Multiplying both sides by (x – 9)(x – 3)(x – 4):
2 / (x – 9) = -1 / ((x – 3)(x – 4))
2 * (x – 3)(x – 4) = -(x – 9)
2(x² – 7x + 12) = -x + 9
2x² – 14x + 24 = -x + 9
2x² – 13x + 15 = 0
2x² – 10x – 3x + 15 = 0
2x(x – 5) – 3(x – 5) = 0
(2x – 3)(x – 5) = 0
2x – 3 = 0 ⇒ x = 1.5
x – 5 = 0 ⇒ x = 5
Solutions: (x = 5) and (x = 1.5).
Question 3 of Exercise 1.1 Class 10. Solve the following equations completing square.
(i) 7x² + 2x – 1 = 0
Solution: 7x² + 2x – 1 = 0
Dividing each term of the equation by 7
7/7 x² + 2/7x – 1/7 = 0
x² + (2/7)x (2/7)(1) – 1/7 = 0
x² + 2(x)(1/7) = 1/7
Adding (1/7)² on both sides:
x² + 2(x)(1/7) + (1/7)² = 1/7 + (1/7)²
(x + 1/7)² = 1/7 + 1/49
(x + 1/7)² = (7 + 1)/49
(x + 1/7)² = 8/49
Taking the square root of both sides:
√((x + 1/7)²) = ±√(8/49)
√((x + 1/7)²) = ±√(4 × 2/49)
x + 1/7 = ±(2√2/7)
x = -1/7 ± 2√2/7
∴ x = (-1 ± 2√2)/7
Solution set is { (-1 ± 2√2)/7 }
(ii) ax² + 4x – a = 0, a ≠ 0
Solution: ax² + 4x – a = 0, a ≠ 0
Dividing each term of the equation by ‘a’:
(ax²)/a + (4x)/a – a/a = 0/a
x² + (4x)/a – 1 = 0
Add and subtract (4/a²):
x² + (4x)/a + (4/a²) – (4/a²) – 1 = 0
Rewrite as a perfect square:
(x + 2/a)² – (4/a² + 1) = 0
Isolate the perfect square:
(x + 2/a)² = 4/a² + 1
Solve for x:
x + 2/a = ±√(4/a² + 1)
x = -2/a ± √(4/a² + 1)
Solution set is [(-2 ± √(a² + 4))/a]
(iii)11x² – 34x + 3 = 0
Rewrite the equation:
11x² – 34x + 3 = 0
11x² – 34x = -3
Divide by 11:
x² – (34/11)x = -3/11
Complete the square:
- Take (34/11), divide by 2, and square it:
((34/11)/2)² = (17/11)² = 289/121 - Add and subtract 289/121:
x² – (34/11)x + (289/121) = -3/11 + (289/121)
Simplify the right side:
-3/11 = -33/121
-33/121 + 289/121 = 256/121
Write as a square:
(x – 17/11)² = 256/121
Take square roots:
x – 17/11 = ±16/11
Solve for x:
- For +16/11:
x = 17/11 + 16/11 = 33/11 = 3 - For -16/11:
x = 17/11 – 16/11 = 1/11
Final Solutions:
x = 3
x = 1/11
(IV)
To solve the quadratic equation lx² + mx + n = 0 by completing the square:
Divide the whole equation by l to make the coefficient of x² equal to 1:
x² + (m/l)x + (n/l) = 0
Move the constant term to the right-hand side:
x² + (m/l)x = -(n/l)
To complete the square, add (m/2l)² to both sides:
x² + (m/l)x + (m/2l)² = -(n/l) + (m/2l)²
The left-hand side is now a perfect square:
(x + m/2l)² = (m²/4l²) – (n/l)
Simplify the right-hand side:
(x + m/2l)² = (m² – 4ln) / 4l²
Take the square root of both sides:
x + m/2l = ±√(m² – 4ln) / 2l
Solve for x:
x = -(m/2l) ± √(m² – 4ln) / 2l
The solution is:
x = (-m ± √(m² – 4ln)) / 2l
(V)
To solve 3x² + 7x = 0 by completing the square:
First, divide the equation by 3:
x² + (7/3)x = 0
To complete the square, add (7/6)² to both sides:
x² + (7/3)x + (7/6)² = (7/6)²
This becomes a perfect square:
(x + 7/6)² = 49/36
Take the square root of both sides:
x + 7/6 = ±7/6
Solve for x:
x = 0 or x = -7/3
(VI)
To solve x² – 2x – 195 = 0:
First, move the constant to the other side:
x² – 2x = 195
To complete the square, add (-2/2)² = 1 to both sides:
x² – 2x + 1 = 195 + 1
This becomes a perfect square:
(x – 1)² = 196
Take the square root of both sides:
x – 1 = ±14
Solve for x:
x = 1 ± 14
Thus, x = 15 or x = -13.
(VII)
To solve -x² + 15/2 = 7/2x:
First, rearrange the terms:
-x² – (7/2)x + 15/2 = 0
Multiply through by -1 to simplify:
x² + (7/2)x – 15/2 = 0
Move the constant to the other side:
x² + (7/2)x = 15/2
To complete the square, add (7/4)² = 49/16 to both sides:
x² + (7/2)x + 49/16 = 15/2 + 49/16
Find a common denominator and simplify:
x² + (7/2)x + 49/16 = 120/16 + 49/16
x² + (7/2)x + 49/16 = 169/16
This becomes a perfect square:
(x + 7/4)² = 169/16
Take the square root of both sides:
x + 7/4 = ±13/4
Solve for x:
x = -7/4 ± 13/4
Thus:
x = (6/4) = 3/2 or x = -5
(VIII)
To solve x² + 17x + 33/4 = 0:
Move the constant to the other side:
x² + 17x = -33/4
To complete the square, add (17/2)² = 289/4 to both sides:
x² + 17x + 289/4 = -33/4 + 289/4
Simplify the right-hand side:
x² + 17x + 289/4 = 256/4
This becomes a perfect square:
(x + 17/2)² = 64
Take the square root of both sides:
x + 17/2 = ±8
Solve for x:
x = -17/2 ± 8
Thus:
x = (-17 + 16)/2 = -1/2 or x = (-17 – 16)/2 = -33/2
(IX)
To solve 4 – (8 / (3x + 1)) = (3x² + 5) / (3x + 1):
Multiply through by (3x + 1) to eliminate the denominators:
(3x + 1) * 4 – 8 = 3x² + 5
12x + 4 – 8 = 3x² + 5
12x – 4 = 3x² + 5
Rearrange the terms:
3x² – 12x + 9 = 0
Divide by 3:
x² – 4x + 3 = 0
To complete the square, add (-4/2)² = 4 to both sides:
x² – 4x + 4 = -3 + 4
(x – 2)² = 1
Take the square root of both sides:
x – 2 = ±1
Solve for x:
x = 2 ± 1
Thus:
x = 3 or x = 1
(X)
To solve 7(x + 2a)² + 3a² = 5a(7x + 23a):
Expand both sides:
7(x² + 4ax + 4a²) + 3a² = 35ax + 115a²
7x² + 28ax + 28a² + 3a² = 35ax + 115a²
Simplify:
7x² + 28ax + 31a² = 35ax + 115a²
Rearrange the terms:
7x² + 28ax – 35ax + 31a² – 115a² = 0
7x² – 7ax – 84a² = 0
Divide by 7:
x² – ax – 12a² = 0
To complete the square, add (-a/2)² = a²/4 to both sides:
x² – ax + a²/4 = 12a² + a²/4
(x – a/2)² = 12a² + a²/4
Simplify the right-hand side:
(x – a/2)² = (49a²) / 4
Take the square root of both sides:
x – a/2 = ±7a/2
Solve for x:
x = a/2 ± 7a/2
Thus:
x = 4a or x = -3a
We hope these Solutions of Exercise 1.1 Class 10 Math have made the concepts of quadratic equations clearer and easier to understand. Working through these solutions will reinforce your understanding, improve your problem-solving skills, and prepare you well for exams. Remember, consistent practice is key to mastering math—so revisit these solutions and try solving similar questions on your own. If you need more guidance, explore other exercises and resources to build a strong foundation in math. Keep practicing, stay focused, and success will follow. Good luck with your studies!
Ready for more practice? Check out the solutions for Exercise 1.2 to continue mastering quadratic equations.
Still having issues understanding the questions? have a look at this video lecture.
Video Lecture of Exercise 1.1 Class 10
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