Balaji Class 10 Maths Solutions Chapter 4 Quadratic Equations Ex 4.3

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Balaji Class 10 Maths Solutions Chapter 4 Quadratic Equations Ex 4.3 द्विघात समीकरण

निम्नलिखित समीकरणों को हल कीजिए –
प्रश्न 1.
(i) x(x + 1)(x + 3)(x + 4) = 180
(ii) (2x + 3)(2x + 5)(x – 1)(x – 2) = 30
(iii) (x – 5)(x – 7)(x + 4) (UPBoardSolutions.com) (x + 6) = 504
(iv) x(2x + 1)(x – 2)(2x – 3) = 63
(v) (x² – 3x – 10)(x² – 5x – 6) = 144
(vi) (x + 2)(3x + 4)(3x + 7)(x + 3) = 2400
हलः
(i) दिया गया समीकरण
{x(x + 1)(x + 3)(x + 4)} = 180
{x(x + 4)}{(x + 1)(x + 3)} = 180
(x² + 4x)(x² + x + 3x + 3) = 180
(x² + 4x)(x² + 4x + 3) = 180
माना x² + 4x = y
∴ दिया हुआ समीकरण, y(y + 3) = 180
y² + 3y – 180 = 0

अब y = 12 लेने पर, x² + 4x = 12
x² + 4x – 12 = 0
x² + 6x (UPBoardSolutions.com) – 2x – 12 = 0
x(x + 6) – 2(x + 6) = 0
(x + 6)(x – 2) = 0
x = 2, – 6
y = 15 लेने पर, x² + 4x = – 15
x² + 4x + 15 = 0

(ii) दिया गया समीकरण
(2x + 3) (2x + 5)(x – 1)(x – 2) = 30
{(2x + 3)(x – 1)}{(2x + 5)(x – 2)} = 30
(2x² – 2x + 3x – 3)(2x² – 4x + 5x – 10) = 30
(2x² + x – 3)(2x² + x – 10) = 30
माना 2x² + x = y
∴ दिया गया समीकरण
(y – 3)(y – 10) = 30
y² – 10y – 3y + 30 = 30
y² – 13y = 0
y(y – 13) = 0
y= 0 व y = 13
जब y = 0, 2x² (UPBoardSolutions.com) + x = 0
x(2x + 1) = 0
⇒ x = 0, x = $-\frac{1}{2}$
y = 13 लेने पर, 2x² + x = 13
2x² + x – 13 = 0

(iii) दिया गया समीकरण
{(x – 5)(x + 4)}{(x – 7)(x + 6)} = 504
(x² – 5x + 4x – 20)(x² – 7x + 6x – 42) = 504
(x² – x – 20)(x² – x – 42) = 504
माना x² – x = y
∴ दिया गया समीकरण
(y – 20)(y – 42) = 504
y² – 20y – 42y + 840 – 504 = 0
y² – 62y + 336 = 0
y² – 6y – 56y + 336 (UPBoardSolutions.com) = 0
y(y – 6) – 56(y – 6) = 0
(y – 6)(y – 56) = 0
जब y – 6 = 0 तब y = 6
जब y – 56 = 0
तब y = 56
y = 6 लेने पर, x² – x = 6
x² – x – 6 = 0
x² + 2x – 3x – 6 = 0
x(x + 2) – 3(x + 2) = 0
(x + 2)(x – 3) = 0
x = – 2, 3
y = 56 लेने पर, x² – x = 56
x² – x – 56 = 0
x² – 8x + 7x – 56 = 0
x(x – 8) + 7(x – 8) = 0
(x + 7)(x – 8) = 0
जब x + 7 = 0 तब x = – 7
जब x – 8 = 0 तब x = 8

(iv) दिया गया समीकरण
x(2x + 1)(x – 2)(2x – 3) = 63
{x(2x – 3)} {(2x + 1)(x – 2)} = 63
(2x² – 3x)(2x² – 4x + x – 2) = 63
(2x² – 3x)(2x² – 3x – 2) = 63
माना 2x² – 3x (UPBoardSolutions.com) = y
y(y – 2) = 63
y² – 2y – 63 = 0
y² – 9y + 7y – 63 = 0
y(y – 9) + 7(y – 9) = 0
(y + 7)(y – 9) = 0
जब y + 7 = 0 तब y = – 7
जब y – 9 = 0 तब y = 9
y = – 7 लेने पर, 2x² – 3x = – 7
2x² – 3x + 7 = 0

y = 9 लेने पर, 2x² – 3x = 9
2x² – 3x – 9 = 0
2x² + 3x – 6x – 9 = 0
x(2x + 3) – 3(2x + 3) = 0
(x – 3)(2x + 3) = 0
जब x – 3 = 0 तब (UPBoardSolutions.com) x = 3
जब 2x + 3 = 0 तब x = $\frac{-3}{2}$

(v) दिया गया समीकरण
(x² – 3x – 10)(x² – 5x – 6) = 144
(x² – 5x + 2x – 10)(x² – 6x + x – 6) = 144
[x(x – 5) + 2(x – 5)][x(x – 6) + 1(x – 6)] = 144
(x – 5)(x + 2)(x + 1)(x – 6) = 144
{(x – 5)(x + 1)}{(x + 2)(x – 6)} = 144
(x² – 4x – 5) (x² – 4x – 12) = 144
माना x² – 4x = y
∴ दिया गया समीकरण,
(y – 5)(y – 12) = 144
y² – 5y – 12y + 60 – 144 = 0
y² – 17y – 84 = 0
y² + 4y – 21y – 84 = 0
y(y + 4) – 21 (y + 4) = 0
(y + 4)(y – 21) = 0
जब y + 4 = 0, (UPBoardSolutions.com) तब y = -4
जब y – 21 = 0, तब y = 21
y = – 4 लेने पर, x² – 4x + 4 = 0
x² – 2x – 2x + 4 = 0
x(x – 2) – 2(x – 2) = 0
(x – 2)(x – 2) = 0
x = 2, 2
y = 21 लेने पर, x² – 4x = 21
x² – 4x – 21 = 0
x² – 7x + 3x – 21 = 0
x(x – 7) + 3(x – 7) = 0
x = – 3, 7

(vi) दिया गया समीकरण,
{(x + 2)(3x + 7)} {(3x + 4)(x + 3)} = 2400
(3x² + 7x + 6x + 14)(3x² + 9x + 4x + 12) = 2400
(3x² + 13x + 14) (UPBoardSolutions.com) (3x² + 13x + 12) = 2400
माना 3x² + 13x = y
∴ दिया गया समीकरण (y + 14)(y + 12) = 2400
y² + 14y + 12y + 168 = 2400
y² + 26y + 168 – 2400 = 0
y² + 26y – 2232 = 0
y² – 36y + 62y – 2232 = 0
y(y – 36) + 62(y – 36) = 0
(y – 36)(y + 62) = 0
जब y – 36 = 0 तब y = 36
जब y + 62 = 0 तब y = – 62
y = 36 लेने पर, 3x² + 13x = 36
3x² + 13x – 36 = 0

प्रश्न 2.

हल:
(i) दिया गया समीकरण
$\sqrt{3 x+1}-\sqrt{x-1}$ = 2
$\sqrt{3 x+1}$ = 2 + $\sqrt{x-1}$
3x + 1 = 4 + (x – 1) + 4 $\sqrt{x-1}$ (दोनों पक्षों का वर्ग करने पर)
3x – x + 1 – 4 + 1 = 4 $\sqrt{x-1}$
2x – 2 = 4 $\sqrt{x-1}$ (UPBoardSolutions.com)
2(x – 1) = 2.2 $\sqrt{x-1}$
(x – 1) = 2 $\sqrt{x-1}$
x² – 2x + 1 = 4(x – 1) (दोनों पक्षों का पुनः वर्ग करने पर)
x² – 2x – 4x + 1 + 4 = 0
x² – 6x + 5 = 0
x² –x – 5x + 5 = 0
x(x – 1) – 5(x – 1)
(x – 1)(x – 5) = 0
जब (x – 1) = 0 तब x = 1
तथा जब (x – 5) = 0 तब x = 5

(ii) दिया गया समीकरण.
$\sqrt{2 x+8}$ + $\sqrt{x+5}$ = 7
$\sqrt{2 x+8}$ = 7 – $\sqrt{x+5}$
2x + 8 = 49 + (x + 5) (UPBoardSolutions.com) – 14 $\sqrt{x+5}$ (दोनों पक्षों का वर्ग करने पर)
2x + 8 – x – 49 – 5 = – 14 $\sqrt{x+5}$
x – 46 = – 14 $\sqrt{x+5}$
(x – 46)² = ( – 14 $\sqrt{x+5}$ )² (दोनों पक्षों का पुनः वर्ग करने पर)
x² – 92x + (46)² = 196(x + 5)
x² – 92x – 196x + 2116 – 980 = 0
x² – 288x + 1136 = 0
x² – 4x – 284x + 1136 = 0
x(x – 4) – 284(x – 4) = 0
यदि x – 4 = 0 तब x = 4
तथा यदि x – 284 = 0 तब x = 284
परन्तु x = 284 समीकरण को सन्तुष्ट नहीं करता।
अतः x = 4

(iii) दिया गया समीकरण,
$\sqrt{x+4}$ + $\sqrt{x+20}$ = 2 $\sqrt{x+11}$
x + 4 + x + 20 + 2/ $\sqrt{(x+4)} \sqrt{x+20}$ = 4(x + 11) (दोनों पक्षों का वर्ग करने पर)
2x + 24 + 21 $\sqrt{x^{2}+24 x+80}$ = 4x + 44
2x + 24 – 4x – 44 = $\sqrt{x^{2}+24 x+80}$
– 2x – 20 = – 2 $\sqrt{x^{2}+24 x+80}$
– 2(x + 10) = -2 $\sqrt{x^{2}+24 x+80}$
(x + 10) = $\sqrt{x^{2}+24 x+80}$
x² + 20x + 100 = x² + 24x + 80 (दोनों पक्षों का पुन: वर्ग करने पर)
x² + 20x + 100 – x² – 24x – 80 = 0
– 4x + 20 = 0
– 4x = – 20
x = 5

(iv) दिया गया समीकरण,
$\sqrt{x+1}$ – $\sqrt{x-1}$ = $\sqrt{4x-1}$
x + 1 + x – 1 – 2 $\sqrt{x^{2}-1}$ = 4x – 1 (दोनों पक्षों का वर्ग करने पर)
2x – 4x + 1 = 20 $\sqrt{x^{2}-1}$ (UPBoardSolutions.com)
– 2x + 1 = 2 $\sqrt{x^{2}-1}$
4x² + 1 – 4x = 4(x² – 1) (दोनों पक्षों का पुनः वर्ग करने पर)
4x² – 4x + 1 – 4x² + 4 = 0
– 4x + 5 = 0
– 4x = – 5
x = $\frac{5}{4}$

Balaji Class 10 Maths Solutions Chapter 4 Quadratic Equations Ex 4.3 द्विघात समीकरण

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