हिंदी

Solve the following quadratic equation: 1/(2x – 3) + 1/(x – 5) = 1 1/9, x ≠ 3/2, 5

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प्रश्न

Solve the following quadratic equation:

`1/(2x - 3) + 1/(x - 5) = 1 1/9, x ≠ 3/2, 5`

योग
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उत्तर

Given: `1/(2x - 3) + 1/(x - 5) = 1 1/9, x ≠ 3/2, 5`

Step-wise calculation:

1. Convert the mixed number:

`1 1/9 = 10/9`

2. Combine the left-hand side over the common denominator (2x – 3)(x – 5):

`(x - 5 + 2x - 3)/((2x - 3)(x - 5)) = (3x - 8)/((2x - 3)(x - 5))`

3. Set equal to `10/9`:

`(3x - 8)/((2x - 3)(x - 5)) = 10/9`

4. Cross-multiply:

9(3x – 8) = 10(2x – 3)(x – 5)

5. Expand both sides:

27x – 72 = 10(2x2 – 13x + 15) 

= 20x2 – 130x + 150

6. Bring all terms to one side:

0 = 20x2 – 130x + 150 – 27x + 72

= 20x2 – 157x + 222

7. Solve the quadratic 20x2 – 157x + 222 = 0.

Compute discriminant: D = 1572 – 4 × 20 × 222 

= 24649 – 17760

= 6889

= 832

8. Roots: `x = (157 ± 83)/(2 xx 20)` 

= `(157 ± 83)/40` 

`x_1 = (157 + 83)/40`

= `240/40` 

= 6

`x_2 = (157 - 83)/40` 

= `74/40`

= `37/20`

9. Check against excluded values `x ≠ 3/2 (1.5)` and 5:

Neither 6 nor `37/20` equals 1.5 or 5, so both are admissible.

 x = 6 or `x = 37/20`

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अध्याय 4: Quadratic Equations - EXERCISE 4A [पृष्ठ १८४]

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आर.एस. अग्रवाल Mathematics [English] Class 10
अध्याय 4 Quadratic Equations
EXERCISE 4A | Q 55. (ii) | पृष्ठ १८४
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