Question

निम्न समीकरणों के मूल ज्ञात कीजिए:
(i) x-$\frac{1}{x}=3, x\neq0$
(ii) $\frac{1}{x+4}-\frac{1}{x-7}= \frac{11}{30}, x\neq-4,7.$

Answer 1

(i) x - 1/x = 3 , x ≠ 0

(x² - 1)/x = 3

x² - 1 = 3x

x² - 3x - 1 = 0

हम जानते हैं कि किसी द्विघात समीकरण ax² + bx + c = 0 का मूल ज्ञात करने के लिए उपयोग में आने वाले सूत्र हैं
$x=\frac{(-b\pm\sqrt{b^2-4ac})}{2a}$

इसीलिए,

x = {3 ± √(3² - 4×-1)}/2

= (3 ± √13)/2


(ii) 1/(x + 4) - 1/(x - 7) = 11/30 , x ≠ -4, 7

{(x - 7) - (x + 4)}/(x + 4)(x - 7) = 11/30

-11/(x² - 7x + 4x - 28) = 11/30

-1/(x² - 3x - 28) = 1/30

x² - 3x - 28 = -30

x² - 3x - 28 + 30 = 0

x² - 3x + 2 = 0

x² - x - 2x + 2 = 0

x(x - 1) - 2(x - 1) = 0

(x - 1)(x - 2) = 0

x = 1 , 2

Answer 2

Solution :

i ) Given x - 1/x = 3 [ x ≠ 0 ]----( 1 )

________________________
We know that ,

( a + b )² = ( a - b )² + 4ab
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=> ( x + 1/x )² = ( x - 1/x )² + 4

=> ( x + 1/x )² = 3² + 4 [ from ( 1 ) ]

=> ( x + 1/x )² = 9 + 4 = 13

=> x + 1/x = ± √13 ----( 2 )

add equations ( 1 ) & ( 2 ), we get

2x = 3 ± √13

x = [ 3±√13 ]/2

ii ) 1/(x+4) - 1/(x-7) = 11/30

=> [x-7 -(x+4)]/[(x+4)(x-7)] = 11/30

=> (x-7-x-4)/(x²-7x+4x-28) = 11/30

=> (-11)/(x²-3x-28) = 11/30

=> (-11 × 30 )/11 = x² - 3x - 28

=> -30 = x² - 3x - 28

=> x² - 3x - 28 + 30 = 0

=> x² - 3x + 2 = 0

=> x² -1x - 2x + 2 = 0

=> x( x - 1 ) - 2( x - 1 ) = 0

=> ( x - 1 )( x - 2 ) = 0

x - 1 = 0 or x - 2 = 0

x = 1 or x = 2

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