Balaji Class 10 Maths Solutions Chapter 4 Quadratic Equations Ex 4.2

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Balaji Class 10 Maths Solutions Chapter 4 Quadratic Equations Ex 4.2 द्विघात समीकरण

निम्नलिखित समीकरणों को हल कीजिए
प्रश्न 1.
x⁴ – 8x² – 9 = 0
हलः
दिया हुआ समीकरण
x⁴ – 8x² – 9 = 0 …(1)
x² = y समीकरण (1) में रखने पर
y² – 8y – 9 = 0
⇒ y² – 9y + y – 9 = 0
⇒ y(y – 9) + 1(y – 9) = 0
⇒ (y – 9)(y + 1) = 0
⇒ y = 9, – 1
अब y = 9 ⇒ x² = 9
⇒ x = ±3
तथा y = – 1 = i²
⇒ x² = i²
⇒ x = ±i
अतः समीकरण के (UPBoardSolutions.com) हल = (±3, ±i)

प्रश्न 2.
4x⁴ – 5x² + 1 = 0
हलः
दिया गया समीकरण
4x⁴ – 5x² + 1 = 0
4(x²)² – 5x² + 1 = 0 …(1)
x² = y, समीकरण (1) में रखने पर
4y² – 5y + 1 = 0
⇒ 4y² – 4y – y + 1 = 0
⇒ 4y(y – 1) – 1(y – 1) = 0

प्रश्न 3.
$\left[\frac{\boldsymbol{x} - \boldsymbol{a}}{\boldsymbol{x} + \boldsymbol{a}}\right]^{2} - 5\left[\frac{\boldsymbol{x} - \boldsymbol{a}}{\boldsymbol{x} + \boldsymbol{a}}\right]$ + 6 = 0 (UPBoardSolutions.com)
हलः
दिया गया समीकरण
$\left[\frac{\boldsymbol{x} - \boldsymbol{a}}{\boldsymbol{x} + \boldsymbol{a}}\right]^{2} - 5\left[\frac{\boldsymbol{x} - \boldsymbol{a}}{\boldsymbol{x} + \boldsymbol{a}}\right]$ + 6 = 0
समीकरण (1) में $\frac{x - a}{x + a}$ = y रखने पर
y² – 5y + 6 = 0
⇒ y² – 2y – 3y + 6 = 0
⇒ y(y – 2) – 3(y – 2) = 0
⇒ (y – 2) (y – 3) = 0
⇒ y = 2, 3
जब y = 2 ⇒ $\frac{x - a}{x + a}$ = 2
x – a = 2(x + a)
2x + 2a = x – a
2x – x = – a – 2a = – 3a
x = – 3a

जब y = 3 ⇒ $\frac{x - a}{x + a}$ = 3
x – a = 3(x + a)
3x + 3a = x – a
3x – x = – a – 3a
2x = – 4a
x = – 2a
अतः समीकरण के हल क्रमशः – 2a, – 3a हैं।

प्रश्न 4.
x^{– 2} – 12 = – x^{– 1}
हलः

∴ दिया गया समीकरण y ² + y – 12 = 0
y² + 4y – 3y – 12 = 0
y(y + 4) – 3(y + 4) = 0
(y – 3)(y + 4) = 0
यदि y – 3 = 0, तब y = 3
और यदि y + 4 = 0, तब y = – 4
y = 3 लेने पर, $\frac{1}{x}$ = 3
x = $\frac{1}{3}$
y = – 4 लेने पर, $\frac{1}{x}$ = – 4
x = $- \frac{1}{4}$

प्रश्न 5.
(x² – 3x + 3)² – (x – 1)(x – 2) = 7
हलः
दिया गया समीकरण (UPBoardSolutions.com)
(x² – 3x + 3)² – (x – 1)(x – 2) = 7
(x² – 3x + 2 + 1)² – (x² – 3x + 2) – 7 = 0 …(1)
समीकरण (1) में x² – 3x + 2 = y रखने पर
(y + 1)² – y – 7 = 0
y² + 1 + 2y – y – 7 = 0
y² + y – 6 = 0
y² + 3y – 2y – 6 = 0
y(y + 3) – 2(y + 3) = 0
(y – 2)(y + 3) = 0
y = 2, – 3
जब y = 2 ⇒ x² – 3x + 2 = 2
x² – 3x = 0
x(x – 3) = 0
x= 0, 3
जब y = – 3 ⇒ x² – 3x + 2 = – 3
x² – 3x + 2 + 3 = 0
x² – 3x + 5 = 0

प्रश्न 6.
(x² – 5x)² – 30(x² – 5x) – 216 = 0
हलः
दिया गया समीकरण
(x² – 5x)² – 30(x² – 5x) – 216 = 0 …(1)
समीकरण (1) में x² – 5x = y रखने पर
y² – 30y – 216 = 0
y² – 36y + 6y – 216 = 0
y(y – 36) + 6(y – 36) = 0
(y + 6)(y – 36) = 0
y = – 6, 36
जब y = – 6 ⇒ x² – 5x = – 6
x² – 5x + 6 = 0
x² – 3x – 2x + 6 = 0
x(x – 3) – 2(x – 3) = 0
(x – 2)(x – 3) = 0
x = 2, 3
जब y = 36 ⇒ x² – 5x = 36
x² – 5x – 36 = 0
x² – 9x + 4x – 36 = 0
x(x – 9) + 4(x – 9) = 0
(x + 4)(x – 9) = 0
x = – 4, 9

प्रश्न 7.
(x² – 5x + 7)² – (x – 2)(x – 3) = 1
हलः
दिया गया (UPBoardSolutions.com) समीकरण
(x² – 5x + 7)² – (x – 2)(x – 3) = 1
{(x² – 5x + 6) + 1}² – (x² – 5x + 6) = 1 …(1)
समीकरण (1) में x² – 5x + 6 = y रखने पर
(y + 1)² – y = 1
y² + 2y + 1 – y – 1 = 0
y² + y = 0
y(y + 1) = 0
y = 0, – 1
जब y = 0 ⇒ x² – 5x + 6 = 0
x² – 2x – 3x + 6 = 0
x(x – 2) – 3(x – 2) = 0
x = 2, 3
जब y = – 1 ⇒ x² – 5x + 6 = – 1
x² – 5x + 7 = 0

प्रश्न 8.
12x⁴ – 56x³ + 89x² – 56x + 12 = 0
हलः
दिया गया समीकरण
12x⁴ – 56x³ + 89x² – 56x + 12 = 0
दोनों पक्षों में x² से भाग करने पर,

x² + $\frac{1}{x^{2}}$ + 2 = y²
x² + $\frac{1}{x^{2}}$ = y² – 2
समीकरण (1) में x² + (UPBoardSolutions.com) $\frac{1}{x^{2}}$ = y² – 2 तथा x + $\frac{1}{x}$ = y
12(y² – 2) – 56y + 89 = 0
12y² – 24 – 56y + 89 = 0
12y² – 56y + 65 = 0

2(x² + 1) = 5x
2x² – 5x + 2 = 0
2x² – x – 4x + 2 = 0
x(2x – 1) – 2(2x – 1) = 0
(x – 2)(2x – 1) = 0

6(x² + 1) = 13x
6x² – 13x + 6 = 0
6x² – 9x – 4x + (UPBoardSolutions.com) 6 = 0
3x(2x – 3) – 2(2x – 3) = 0
(3x – 2)(2x – 3) = 0
x = $\frac{2}{3}, \frac{3}{2}$

प्रश्न 9.

हलः

2(y² + 1) = 5y
2y² – 5y + 2 = 0
2y² – 4y – y + (UPBoardSolutions.com) 2 = 0
2y(y – 2) – 1(y – 2) = 0
(y – 2)(2y – 1) = 0
जब y – 2 = 0 तब जब y = 2
जब 2y – 1 = 0 तब y = $\frac{1}{2}$
y = 2 लेने पर, $\frac{3 x+1}{x+1}$ = 2
3x + 1 = 2(x + 1)
3x + 1 = 2x + 2
3x – 2x = 2 – 1
x = 1
y = $\frac{1}{2}$ लेने पर, $\frac{3 x+1}{x+1}=\frac{1}{2}$
2(3x + 1) = x + 1
6x + 2 = x + 1
6x – x = 1 – 2
5x = -1
x = $-\frac{1}{5}$

प्रश्न 10.

हलः

6(y² + 1) = 13y
6y² – 13y + 6 = 0
6y² – 4y – 9y + 6 = 0
2y(3y – 2) – 3(3y – 2) (UPBoardSolutions.com) = 0
(2y – 3)(3y – 2) = 0
जब 2y – 3 = 0 तब y = $\frac{3}{2}$
जब 3y – 2 = 0 तब y = $\frac{2}{3}$
y = $\frac{3}{2}$ लेने पर, $\frac{x}{1+x}=\frac{3}{2}$
3(1 + x) = 2x
3 + 3x = 2x
3x – 2x = -3
x = – 3
y = $\frac{2}{3}$ लेने पर, $\frac{x}{1+x}=\frac{2}{3}$
3x = 2(1 + x)
3x = 2 + 2x
3x – 2x = 2
x = 2

प्रश्न 11.
$\frac{4 x-1}{4 x+1}+\frac{4 x+1}{4 x-1}=\frac{10}{3}$
हलः

3(y² + 1) = 10y
3y² – 10y + 3 = 0
3y² – 9y – y + 3 = 0
3y(y – 3) – 1(y – 3) = 0
(y – 3)(3y – 1) = 0
जब y – 3 = 0, तब y = 3
जब 3y – 1 = 0, तब y = 1/3 (UPBoardSolutions.com)
y = 3 लेने पर, $\frac{4 x-1}{4 x+1}=\frac{3}{1}$
3(4x + 1)= 4x – 1
12x + 3 = 4x – 1
12x – 4x = – 1 – 3
8x = – 4
x = $-\frac{4}{8}=-\frac{1}{2}$
y = $\frac{1}{3}$ लेने पर, $\frac{4 x-1}{4 x+1}=\frac{1}{3}$
⇒ 3(4x – 1) = 4x + 1
12x – 3 = 4x + 1
12x – 4x = 1 + 3
8x = 4
⇒ x = $\frac{4}{8}=\frac{1}{2}$
अतः x = $\pm \frac{1}{2}$

प्रश्न 12.
5^1+x + 5^1-x = 26
हलः
दिया गया समीकरण 5·5^x + 5·5^-x = 26
5(5^x + 5^-x) = 26
5^x + 5^-x = $\frac{26}{5}$
माना 5^x = y तब 5^-x = $\frac{1}{y}$
∴ दिया गया समीकरण y + $\frac{1}{y}$ = $\frac{26}{5}$
$\frac{y^{2}+1}{y}=\frac{26}{5}$
5(y² + 1) = 26y
5y² + 5 – 26y = 0
5y² – 26y + 5 = 0
5y² – 25y – y + 5 = 0
5y(y – 5) – 1(y – 5) = 0
(y – 5)(5y – 1) = 0
जब y – 5 = 0 तब y = 5
तथा जब 5y – 1 = 0 तब y = $\frac{1}{5}$
y = 5 लेने पर, 5^x = 5 = 5¹
x = 1
y = $\frac{1}{5}$ लेने पर, 5^x = 5^-1
x = -1

प्रश्न 13.
5^x+1 + 5^2-x = 5³ + 1
हलः
दिया गया समीकरण
5^x+1 + 5^2-x = 5³ + 1
5^x ·5 + $\frac{5^{2}}{5^{x}}$ = 5³ + 1
5·5^x · 5^x + 5² = 5³ · 5^x + 5^x
माना 5^x = y 5y · y + 5² = 125y + y
5y² – 126y + 5² = 0 (UPBoardSolutions.com)
5y² – 126y + 25 = 0
5y² – 1y – 125y + 25 = 0
y(5y – 1) – 25(5y – 1) = 0
(y – 25)(5y – 1) = 0
y = 25 या y = $\frac{1}{5}$
यदि y = 25, तब 5^x = y = 25
5^x = 5²
तब x = 2 (घातों की तुलना करने पर)
तथा यदि y = $\frac{1}{5}$ = 5^-1
तब 5^x = y
5^x = 5^-1
x = – 1 (घातों की तुलना करने पर)
अतः x = – 1, 2

प्रश्न 14.
3^x + 3^-x – 2 = 0
हलः

y² + 1 – 2y = 0
(y – 1)² = 0
y = 1, 1
y = 1 लेने पर, 3^x = 1
3^x = 3⁰
x = 0

प्रश्न 15.
2^2x+8 – 8·2^x+2 + 1 = 0
हलः
दिया गया (UPBoardSolutions.com) समीकरण
2^2x+4-4 – 8·2^x+2+1 = 0
2^2x+4 · 2⁴ – 8·2^x+2 + 1 = 0
16·2^2(x+2) – 8·2^x+2 + 1 = 0
माना 2^x+2 = y
∴ दिया गया समीकरण
16y^x – 8y + 1 = 0
16y^x – 4y – 4y + 1 = 0
4y(4y – 1) – 1(4y – 1) = 0
(4y – 1)(4y – 1) = 0

दोनों पक्षों में घातों की तुलना करने पर
x + 2 = – 2
x = – 2 – 2
x = – 4

प्रश्न 16.

हलः

y² + 4 = 5y
y² – 5y + 4 = 0
y² – y – 4y + 4 = 0
y(y – 1) – 4(y – 1) = 0
(y – 1)(y – 4) = 0
जब y – 1 = 0 तब y = 1
जब y – 4 = 0 तब y = 4
y = 1 लेने पर, $\sqrt{3 x^{2}+1}$ = 1
3x² + 1 = 1
3x² = 0
x² = 0 ⇒ x = 0, 0
y = 4 लेने पर, $\sqrt{3 x^{2}+1}$ = 4 ⇒ 3x² + 1 = 16
3x² = 16 – 1 = 15
⇒ 3x² = 15
⇒ x² = ± $\sqrt{{5}}$
⇒ x = + 15

प्रश्न 17.
2^x = 4^2x-1
हलः
दिया गया समीकरण (UPBoardSolutions.com) 2² = 4^2x-1
2^x = $\frac{4^{2 x}}{4}$
4·2² = 4^2x
4·2^x = (2²)^2x
4·2^x = (2^2x)²
4·2^x = 2^2x2^2x = (2^x)²(2^x)²
माना 2^x = y तब, 4y = y^x·y^x
4y = y⁴
(4y – y⁴) = 0
y(4 – y³) = 0
y = 0 या 4 – y³ = 0
4 = y³
∴ y³ = 4
y = 4^1/3 = (2²)^1/3
यदि y = 0 तब 2^x = 0, x = 0 (अमान्य)
y = 4^1/3 तब 2^x = 4^1/3
2^x = 2^2/3
दोनों ओर घातों की तुलना करने पर
x = $\frac{2}{3}$

प्रश्न 18.

हलः
दिया गया समीकरण

8y² – 1 = 2y
8y² – 2y – 1 = 0
8y² – 4y + 2y – 1 = 0
4y(2y – 1) + 1(2y – 1) = 0
(2y – 1)(4y + 1) = 0
जब 2y – 1 = 0 तब y = $\frac{1}{2}$
जब 4y + 1 = 0 तब y = $-\frac{1}{4}$

4x = x + 3
4x – x = 3
3x = 3
x = 1

16x = x + 3
16x – x = 3
15x = 3 ⇒ x = $\frac{3}{15}=\frac{1}{5}$
परन x= $\frac{1}{5}$ समीकरण को सन्तुष्ट नहीं करती है अतः x = 1

Balaji Class 10 Maths Solutions Chapter 4 Quadratic Equations Ex 4.2 द्विघात समीकरण

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