Balaji Class 10 Maths Solutions Chapter 4 Quadratic Equations Ex 4.1

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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 = $\frac{2}{3}$
यदि 3x + 1 = 0, तब x = $-\frac{1}{3}$
अतः x = $\frac{2}{3}$ व $-\frac{1}{3}$

(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 = $\frac{7}{2}$
यदि 4x + 3 = 0, तब x = $-\frac{3}{4}$
अत: x = $\frac{7}{2}$ व $-\frac{3}{4}$

प्रश्न 2.

हल:

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 = $-\frac{3}{2}$
यदि 2x + 2 = 0 तब x = $-\frac{2}{2}$ = -1
अतः x = $-\frac{3}{2}$ व -1

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 = $\frac{a^{2}}{2}$
यदि 2x – b² = 0 तब x = $\frac{b^{2}}{2}$
अतः x = $\frac{a^{2}}{2}$ व $\frac{b^{2}}{2}$

प्रश्न 4.
(i) a²b²x² + b²x – a²x – 1 = 0
(ii) x² + $\left(\frac{\boldsymbol{a}}{\boldsymbol{a}+\boldsymbol{b}}+\frac{\boldsymbol{a}+\boldsymbol{b}}{\boldsymbol{a}}\right)$ x + 1 = 0
हलः
(i) a²b²x² + b²x – ax – 1 = 0
b²x(a²x + 1) – 1(a²x + 1) = 0
(a²x + 1)(b²x – 1) = 0

प्रश्न 5.
(i) abx² + (b² – ac)x – bc = 0 (UPBoardSolutions.com)
(ii) x² + $\left(\boldsymbol{a}+\frac{\mathbf{1}}{\boldsymbol{a}}\right)$ x + 1 = 0
(iii) $\frac{x-1}{x-2}+\frac{x-3}{x-4}=\frac{10}{3}$ , x ≠ 2, x ≠ 4
(iv) 3x² – 2 $\sqrt{{6}}$ x + 2 = 0 (NCERT)
(v) $\frac{1}{x-1}-\frac{1}{x+5}=\frac{6}{7}$ , x ≠ 1, -5
(vi) $x+\frac{2}{x}$ = 3, x ≠ 0 (NCERT)
(vii) $\frac{1}{x+4}-\frac{1}{x-7}=\frac{11}{30}$ , 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 = $-\frac{b}{a}$ , (UPBoardSolutions.com) x = $\frac{c}{b}$
अतः $-\frac{b}{a}$ , व x = $\frac{c}{b}$

प्रश्न 6.

हलः
(i)

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

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 = $\frac{12}{5}$
अतः x = 4 व $\frac{12}{5}$

(iii)

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

(iv)

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 = $-\frac{25}{3}$ , x = 3
अतः x = $-\frac{25}{3}$ , 3

(v)

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)

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

प्रश्न 7.

हलः

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 = $-\frac{1}{3}$

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 = $\frac{7}{2}$
अत: x = 6, $\frac{7}{2}$

(iii)

(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 = $-\frac{b}{2}$
अतः x = -a, $-\frac{b}{2}$

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 – ( $\sqrt{{3}}$ + 1)x + $\sqrt{{3}}$ = 0
(v) a²x² – 3abx + 2b² = 0 (NCERT)
हलः
(i) 2x²2 – 5x + 3 = 0
x² का गुणांक = 2 से समीकरण को भाग करने पर,

अब x के गुणांक के आधे का वर्ग करके दोनों पक्षों में जोड़ने पर,

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,

∵ दायाँ पक्ष ऋणात्मक है। $\left(x+\frac{3}{8}\right)^{2}$ , 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।

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.

हलः
(i)

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,

(ii)
2x² – 2 $\sqrt{{2}}$ x + 1 = 0
द्विघात समीकरण ax² + bx + c = 0 से तुलना करने पर,
a = 2, b = – 2 $\sqrt{{2}}$ , c = 1

(iii) $\sqrt{{2}}$ . x² + 7x + 5 $\sqrt{{2}}$ = 0
द्विघात समीकरण ax² + bx + c = 0 से (UPBoardSolutions.com) तुलना करने पर,
a = $\sqrt{{2}}$ , b = 7, c = 5 $\sqrt{{2}}$

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 + $\frac{1}{8}$ = 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 = $\frac{3}{2}$
अतः x= – 2, $\frac{3}{2}$

(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 = $\frac{1}{10}$ , x = $\frac{1}{10}$
अतः x = $\frac{1}{10}$ , $\frac{1}{10}$

⇒ 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 = $\frac{1}{4}$ , x = $\frac{1}{4}$
अत: x = $\frac{1}{4}$ , $\frac{1}{4}$

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 = $\frac{3a}{4b}$ , x = $-\frac{2b}{3a}$
अतः x $\frac{3a}{4b}$ , $-\frac{2b}{3a}$

प्रश्न 13.
समीकरण 4x² – 2x + $\frac{1}{4}$ = 0 (UPBoardSolutions.com) के मूल, पूर्ण वर्ग बनाने वाली विधि द्वारा ज्ञात कीजिए। (NCERT)
हलः

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 = $\frac{a^{2}+b^{2}}{2}$ , x = $\frac{a^{2}-b^{2}}{2}$
अत: x = $\frac{a^{2}+b^{2}}{2}$ , $\frac{a^{2}-b^{2}}{2}$

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 द्विघात समीकरण

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