Balaji Class 10 Maths Solutions Chapter 4 Quadratic Equations Ex 4.1

Balaji Class 10 Maths Solutions Chapter 4 Quadratic Equations Ex 4.1 द्विघात समीकरण

Ex 4.1 Quadratic Equations गुणनखण्ड विधि (Factorization Method)

निम्नलिखित समीकरणों को गुणनखण्ड विधि द्वारा हल करें।
प्रश्न 1.
(i) 9x2 – 3x – 2 = 0
(ii) 8x2 – 22x – 21 = 0
हल:
(i) 9x2 – 3x – 2 = 0
9x2 – 6x + 3x – 2 = 0
3x(3x – 2) + 1(3x – 2) = (UPBoardSolutions.com) 0
(3x – 2)(3x + 1) = 0
यदि 3x – 2 = 0, तब x = [latex]\frac{2}{3}[/latex]
यदि 3x + 1 = 0, तब x = [latex]-\frac{1}{3}[/latex]
अतः x = [latex]\frac{2}{3}[/latex] व [latex]-\frac{1}{3}[/latex]

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(ii) 8x2 – 22x – 21 = 0
8x2 – 28x + 6x – 21 = 0
4x(2x – 7) + 3(2x – 7) = 0
(2x – 7)(4x + 3) = (UPBoardSolutions.com) 0
यदि 2x – 7 = 0, तब x = [latex]\frac{7}{2}[/latex]
यदि 4x + 3 = 0, तब x = [latex]-\frac{3}{4}[/latex]
अत: x = [latex]\frac{7}{2}[/latex] व [latex]-\frac{3}{4}[/latex]

प्रश्न 2.
Balaji Class 10 Maths Solutions Chapter 4 Quadratic Equations Ex 4.1 1
हल:
Balaji Class 10 Maths Solutions Chapter 4 Quadratic Equations Ex 4.1 2
4x2 + 6x – x – 3 + 3x + 9 = 0
4x2 + 10x + 6 = 0
4x2 + 6x + 4x + 6 = 0
2x(2x + 3) + 2(2x + 3) = 0
(2x + 3)(2x + 2) = (UPBoardSolutions.com) 0
यदि 2x + 3 = 0 तब x = [latex]-\frac{3}{2}[/latex]
यदि 2x + 2 = 0 तब x = [latex]-\frac{2}{2}[/latex] = -1
अतः x = [latex]-\frac{3}{2}[/latex] व -1

UP Board Solutions

प्रश्न 3.
4x2 – 2(a2 + b2)x + a2b2 = 0
हलः
4x2 – 2(a2 + b2)x + a2b2 = 0
4x2 – 2a2x – 2b2x + a2b2 = 0
2x(2x – a2) – b2(2x – a2) = 0
(2x – a2)(2x – b2)
यदि 2x – a2 = 0 तब x = [latex]\frac{a^{2}}{2}[/latex]
यदि 2x – b2 = 0 तब x = [latex]\frac{b^{2}}{2}[/latex]
अतः x = [latex]\frac{a^{2}}{2}[/latex] व [latex]\frac{b^{2}}{2}[/latex]

प्रश्न 4.
(i) a2b2x2 + b2x – a2x – 1 = 0
(ii) x2 + [latex]\left(\frac{\boldsymbol{a}}{\boldsymbol{a}+\boldsymbol{b}}+\frac{\boldsymbol{a}+\boldsymbol{b}}{\boldsymbol{a}}\right)[/latex]x + 1 = 0
हलः
(i) a2b2x2 + b2x – ax – 1 = 0
b2x(a2x + 1) – 1(a2x + 1) = 0
(a2x + 1)(b2x – 1) = 0
Balaji Class 10 Maths Solutions Chapter 4 Quadratic Equations Ex 4.1 3

UP Board Solutions

प्रश्न 5.
(i) abx2 + (b2 – ac)x – bc = 0 (UPBoardSolutions.com)
(ii) x2 + [latex]\left(\boldsymbol{a}+\frac{\mathbf{1}}{\boldsymbol{a}}\right)[/latex]x + 1 = 0
(iii) [latex]\frac{x-1}{x-2}+\frac{x-3}{x-4}=\frac{10}{3}[/latex], x ≠ 2, x ≠ 4
(iv) 3x2 – 2[latex] \sqrt{{6}} [/latex]x + 2 = 0 (NCERT)
(v) [latex]\frac{1}{x-1}-\frac{1}{x+5}=\frac{6}{7}[/latex], x ≠ 1, -5
(vi) [latex]x+\frac{2}{x}[/latex] = 3, x ≠ 0 (NCERT)
(vii) [latex]\frac{1}{x+4}-\frac{1}{x-7}=\frac{11}{30}[/latex], x ≠ 4, 7 (NCERT)
हलः
(i) abx2 + (b2 – ac)x – bc = 0
abx2 + b2x – acx – bc = 0
bx(ax + b) – c(ax + b) = 0
(ax + b) (bx – c) = 0
ax + b = 0 तथा bx – c = 0
x = [latex]-\frac{b}{a}[/latex], (UPBoardSolutions.com) x = [latex]\frac{c}{b}[/latex]
अतः [latex]-\frac{b}{a}[/latex], व x = [latex]\frac{c}{b}[/latex]
Balaji Class 10 Maths Solutions Chapter 4 Quadratic Equations Ex 4.1 4
Balaji Class 10 Maths Solutions Chapter 4 Quadratic Equations Ex 4.1 5

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प्रश्न 6.
Balaji Class 10 Maths Solutions Chapter 4 Quadratic Equations Ex 4.1 6
हलः
(i)
Balaji Class 10 Maths Solutions Chapter 4 Quadratic Equations Ex 4.1 7
a(x2 – cx – bx + bc) + b(x2 – cx + ax + ac) – 2c(x2 – bx – ax + ab) = 0
ax2 – acx – abx + abc + bx2 – bcx – abx + abc – 2cx2 + 2bcx + 2acx – 2abc = 0
ax2 + bx2 – 2cx2 – 2abx + bcx + acx = 0
x[ax + bx – 2cx – 2ab + bc + ac] = 0
∴ x = 0

UP Board Solutions
Balaji Class 10 Maths Solutions Chapter 4 Quadratic Equations Ex 4.1 8
8x2 – 40x + 48 = 3x2 – 8x
8x2 – 40x + 48 – 3x2 + 8x = 0
5x2 – 32x + 48 = 0
5x2 – 20 x – 12x + 48 = 0
x(x – 4) – 12(x – 4) = 0
(x – 4)(5x – 12) = 0
(x – 4) = 0 तथा 5x – 12 = 0
x = 4 , x = [latex]\frac{12}{5}[/latex]
अतः x = 4 व [latex]\frac{12}{5}[/latex]

(iii)
Balaji Class 10 Maths Solutions Chapter 4 Quadratic Equations Ex 4.1 9
5x – 8x – 12 + 6x2 + 9x = 0
6x2 + 6x – 12 = 0
6(x2 + x – 2) = 0
x2 + x – = 0
x2 + 2x – x – 2 = 0
x(x + 2) – 1(x + 2) = 0
(x + 2)(x – 1) = 0
x + 2 = 0 तथा x – 1 = 0
x = – 2 , x = 1
अतः x = -2 व 1

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(iv)
Balaji Class 10 Maths Solutions Chapter 4 Quadratic Equations Ex 4.1 10
या 16x = 75 – 3x2
या 16x – 75 + 3x2 = 0
3x2 + 16 x – 75 = 0
या 3x2 + 25x – 9x – 75 = 0
या x(3x + 25) – 3(3x + 25) = 0
या (3x + 25)(x – 3) = 0
3x + 25 = 0 तथा x – 3 = 0 (UPBoardSolutions.com)
x = [latex]-\frac{25}{3}[/latex], x = 3
अतः x = [latex]-\frac{25}{3}[/latex], 3

UP Board Solutions

(v)
Balaji Class 10 Maths Solutions Chapter 4 Quadratic Equations Ex 4.1 11
30x2 + 35x – 15 – 19x2 + 42x + 15 = 0
11x2 + 77x = (UPBoardSolutions.com) 0
11x(x + 7) = 0
11x = 0 तथा x + 7 = 0
x = 0 , x = -1
अतः x = 0, – 7

(vi)
Balaji Class 10 Maths Solutions Chapter 4 Quadratic Equations Ex 4.1 12
29x2 – 29 = 28x2 + 4x – 7x – 1
29x2 – 29 = 28x2 – 3x – 1
29x2 – 29 – 28x2 + 3x + 1 = 0
x2 + 3x – 28 = 0
x2 + 7x – 4x – 28 = 0
x(x + 7) – 4(x + 7) = 0
(x + 7)(x – 4) = 0
x + 7 = 0 तथा x – 4 = 0
x = – 7 , x = 4
अतः x = – 7, 4

UP Board Solutions

प्रश्न 7.
Balaji Class 10 Maths Solutions Chapter 4 Quadratic Equations Ex 4.1 13
हलः
Balaji Class 10 Maths Solutions Chapter 4 Quadratic Equations Ex 4.1 14
3y2 – 11y – 4 = 0
3y2 – 12y + y – 4 = 0
3y(y – 4) + 1(y – 4) = 0
(y – 4)(3y + 1) = (UPBoardSolutions.com) 0
y – 4 = 0 तथा 3y + 1 = 0
y = 4 , y = [latex]-\frac{1}{3}[/latex]
Balaji Class 10 Maths Solutions Chapter 4 Quadratic Equations Ex 4.1 15

UP Board Solutions
10x2 – 80x + 150 = 6x2 – 42x + 66
10x2 – 80x + 150 – 6x2 + 42x – 66 = 0
4x2 – 38x + 84 = 0
2(2x2 – 19x + 42) = 0
2x2 – 19x + 42 = 0
2x2 – 12x – 7x + 42 = 0
2x(x – 6) – 7(x – 6) = 0
(x – 6)(2x – 7) = 0
x – 6 = 0 तथा 2x – 7 = 0
x = 6 , x = [latex]\frac{7}{2}[/latex]
अत: x = 6, [latex]\frac{7}{2}[/latex]

(iii)
Balaji Class 10 Maths Solutions Chapter 4 Quadratic Equations Ex 4.1 16
(2a + b + 2x)(bx + 2ax + ab) = 2abx
2abx + 4a2x + 2a2b + b2x + 2abx + ab2 + 2bx2 + 4ax2 + 2abx – 2abx = 0
4ax2 + 2bx2 + 4a2x + b2x + 4abx + 2a2b + ab2 = 0
2x2(2a + b) + x(4a2 + b2 + 4ab) + ab(2a + b) = 0
2x2(2a + b) + x(2a + b)2 + ab(2a + b) = 0
(2a + b)[2x2 + x(2a + b) + ab] = 0
2x2 + 2ax + bx + ab = 0
2x(x + a) + b(x + a) = 0
(x + a)(2x + b) = 0
x + a = 0 तथा 2x + b = 0
x = – a , x = [latex]-\frac{b}{2}[/latex]
अतः x = -a, [latex]-\frac{b}{2}[/latex]

UP Board Solutions

Ex 4.1 Quadratic Equations पूर्ण वर्ग द्वारा हल (Solution by Completing the Square)

पूर्ण वर्ग बनाकर हल करने की विधि से निम्नलिखित (UPBoardSolutions.com) द्विघात समीकरणों को हल कीजिए

प्रश्न 8.
(i) 2x22 – 5x + 3 = 0 (NCERT)
(ii) 5x2 – 6x – 2 = 0 (NCERT)
(iii) 4x2 + 4bx – (a2 – b2) = 0
(iv) x2 – ([latex] \sqrt{{3}} [/latex] + 1)x + [latex] \sqrt{{3}} [/latex] = 0
(v) a2x2 – 3abx + 2b2 = 0 (NCERT)
हलः
(i) 2x22 – 5x + 3 = 0
x2 का गुणांक = 2 से समीकरण को भाग करने पर,
Balaji Class 10 Maths Solutions Chapter 4 Quadratic Equations Ex 4.1 17
अब x के गुणांक के आधे का वर्ग करके दोनों पक्षों में जोड़ने पर,
Balaji Class 10 Maths Solutions Chapter 4 Quadratic Equations Ex 4.1 18
Balaji Class 10 Maths Solutions Chapter 4 Quadratic Equations Ex 4.1 19
Balaji Class 10 Maths Solutions Chapter 4 Quadratic Equations Ex 4.1 20

UP Board Solutions
Balaji Class 10 Maths Solutions Chapter 4 Quadratic Equations Ex 4.1 21
UP Board Solutions

प्रश्न 9.
पूर्ण वर्ग बनाने वाली विधि का प्रयोग करके, (UPBoardSolutions.com) सिद्ध कीजिए कि समीकरण 4x2 + 3x + 5 = 0 के मूल वास्तविक नहीं हैं। (NCERT)
हलः
4x2 + 3x + 5 = 0
4 से भाग देने पर,
Balaji Class 10 Maths Solutions Chapter 4 Quadratic Equations Ex 4.1 22
∵ दायाँ पक्ष ऋणात्मक है। [latex]\left(x+\frac{3}{8}\right)^{2}[/latex], x के किसी भी वास्तविक मान के लिए ऋणात्मक नहीं हो सकता है।
∴. दी गई समीकरण के मूल वास्तविक नहीं है।
यही सिद्ध करना था।

UP Board Solutions

Ex 4.1 Quadratic Equations द्विघात सूत्र के प्रयोग द्वारा (By using the Quadratic Formula)

द्विघात सूत्र का प्रयोग करके निम्नलिखित (UPBoardSolutions.com) समीकरणों को हल कीजिए-

प्रश्न 10.
Balaji Class 10 Maths Solutions Chapter 4 Quadratic Equations Ex 4.1 23
हलः
(i) Balaji Class 10 Maths Solutions Chapter 4 Quadratic Equations Ex 4.1 24
Balaji Class 10 Maths Solutions Chapter 4 Quadratic Equations Ex 4.1 25
4(x2 + 3x + 2) = (3x + 4)(x + 4) (UPBoardSolutions.com)
4x2 + 12x + 8 = 3x2 + 12x + 4x + 16
4x2 + 12x + 8 – 3x2 – 12x – 4x – 16 = 0
x2 – 4x – 8 = 0
द्विघात समीकरण ax2 + bx + c = 0 से तुलना करने पर,
a = 1, b = – 4, c = – 8
द्विघात सूत्र (श्रीधराचार्य सूत्र) से,
Balaji Class 10 Maths Solutions Chapter 4 Quadratic Equations Ex 4.1 26

(ii)
2x2 – 2[latex] \sqrt{{2}} [/latex]x + 1 = 0
द्विघात समीकरण ax2 + bx + c = 0 से तुलना करने पर,
a = 2, b = – 2[latex] \sqrt{{2}} [/latex], c = 1
Balaji Class 10 Maths Solutions Chapter 4 Quadratic Equations Ex 4.1 27

UP Board Solutions

(iii) [latex] \sqrt{{2}} [/latex]. x2 + 7x + 5[latex] \sqrt{{2}} [/latex] = 0
द्विघात समीकरण ax2 + bx + c = 0 से (UPBoardSolutions.com) तुलना करने पर,
a = [latex] \sqrt{{2}} [/latex], b = 7, c = 5[latex] \sqrt{{2}} [/latex]
Balaji Class 10 Maths Solutions Chapter 4 Quadratic Equations Ex 4.1 28
Ex 4.1 Quadratic Equations विविध प्रश्न (Miscellaneous Problems)

प्रश्न 11.
निम्नलिखित. समीकरणों को गुणनखण्ड (UPBoardSolutions.com) विधि द्वारा हल करें
(i) 2x2 + x – 6 = 0 (NCERT)
(ii) 100x2 – 20x + 1 = 0 (NCERT)
(iii) 2x2 – x + [latex]\frac{1}{8}[/latex] = 0 (NCERT)
हलः
(i)
2x2 + x – 6 = 0
2x2 + 4x – 3x – 6 = 0
2x(x + 2) – 3(x + 2) = 0
(x + 2)(2x – 3) = 0
x + 2 = 0 तथा 2x – 3 = 0
x = – 2, x = [latex]\frac{3}{2}[/latex]
अतः x= – 2, [latex]\frac{3}{2}[/latex]

(ii) 100x2 – 20x + 1 = 0
100x2 – (10 + 10)x + 1 = 0
100x2 – 10x – 10x + 1 = 0
10x (10x – 1) – 1(10x – 1) = 0
(10x – 1)(10x – 1) = 0
10x – 1 = 0, तथा 10x – 1 = 0
x = [latex]\frac{1}{10}[/latex], x = [latex]\frac{1}{10}[/latex]
अतः x = [latex]\frac{1}{10}[/latex], [latex]\frac{1}{10}[/latex]

Balaji Class 10 Maths Solutions Chapter 4 Quadratic Equations Ex 4.1 29
⇒ 16x2 – 8x + 1 = 0
⇒ 16x2 – 4x – 4x + 1 = 0
⇒ 4x (4x – 1) – 1(4x – 1) = 0
⇒ (4x – 1)(4x – 1) = 0
4x – 1 = 0 तथा 4x – 1 = 0
x = [latex]\frac{1}{4}[/latex], x = [latex]\frac{1}{4}[/latex]
अत: x = [latex]\frac{1}{4}[/latex], [latex]\frac{1}{4}[/latex]

प्रश्न 12.
x के लिए हल कीजिए – 12abx2 – (9a2 – 8b2)x – 6ab = 0
हलः
12abx2 – (9a2 – 8b2)x – 6ab = 0
12abx2 – 9a2x + 8b2x – 6ab = 0
3ax(4bx – 3a) + 2b (4bx – 3a) = 0
(4bx – 3a) (3ax + 2b) = 0
4bx – 3a = 0 तथा 3ax + 2b = 0
x = [latex]\frac{3a}{4b}[/latex], x = [latex]-\frac{2b}{3a}[/latex]
अतः x [latex]\frac{3a}{4b}[/latex], [latex]-\frac{2b}{3a}[/latex]

UP Board Solutions

प्रश्न 13.
समीकरण 4x2 – 2x + [latex]\frac{1}{4}[/latex] = 0 (UPBoardSolutions.com) के मूल, पूर्ण वर्ग बनाने वाली विधि द्वारा ज्ञात कीजिए। (NCERT)
हलः
Balaji Class 10 Maths Solutions Chapter 4 Quadratic Equations Ex 4.1 30

प्रश्न 14.
x के लिए हल कीजिए – 4x2 – 4a2x + a4 – b4 = 0
हलः
4x2 – 4a2x + a4 – b4 = 0
4x2 – (2a2 + 2b2)x – (2a2 – 2b2)x + (a2)2 – (b2)2 = 0
4x2 – 2(a2 + b2)x – 2(a2 – b2)x + (a + b)(a2 – b2)= 0.
2x{2x – (a2 + b2)} – (a2 – b2){2x – (a2 + b2)} = 0
{2x – (a2 + b2)} {2x – (a2 – b2)} = 0
2x – (a2 + b2) = 0 तथा 2x – (a2 – b2) = 0
x = [latex]\frac{a^{2}+b^{2}}{2}[/latex], x = [latex]\frac{a^{2}-b^{2}}{2}[/latex]
अत: x = [latex]\frac{a^{2}+b^{2}}{2}[/latex], [latex]\frac{a^{2}-b^{2}}{2}[/latex]

UP Board Solutions

प्रश्न 15.
x के लिए हल कीजिए – 9x2 – 9(a + b)x + (2a2 + 5ab + 2b2) = 0
हलः
9x2 – 9(a + b)x + (2a2 + 5ab + 2b2) = 0
द्विघात समीकरण ax2 + bx + c = 0 से तुलना करने (UPBoardSolutions.com) पर,
a = 9, b = – 9(a + b), c = 2a2 + 5ab + 2b2
द्विघात सूत्र (श्रीधराचार्य सूत्र) से,
Balaji Class 10 Maths Solutions Chapter 4 Quadratic Equations Ex 4.1 31
Balaji Publications Mathematics Class 10 Solutions

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