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`        Show that the equation 1/x-a +1/x-b +1/x-c =0 can have a pair of equal roots if a=b=c`
7 months ago

Vikas TU
10470 Points
```							Dear student The numerator becomes: (x-b)(x-c) + (x-a)(x-c) + (x-b)(x-c) We don't care about the bottom because if a fraction = 0, it's the numerator that must = 0 (ie, you can't divide by any number to make it 0) x^2 - cx - bx + bc + x^2 - ax - cx + ac + x^2 - bx - cx + bc = 0 3x^2 - 2cx - 2bx - 2ax + ac + bc + ab= 0 Quad formula x = (2c + 2b + 2a) +/- sqrt((2a+2b+2c)^2 - (4)(3)(ac+bc+ab)) / 6 x = (2a + 2b + 2c) +/- 2sqrt(a^2 - ab -ac + b^2 - bc + b^2) / 6 For double roots a^2 - ab -ac + b^2 - bc + c^2 = 0 a^2 + b^2 + c^2 - (ab + bc + ac) = 0 This only works if ab = a^2, bc = b^2 and ac = c^2... which means a = b = c
```
7 months ago
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• 101 Video Lectures
• Revision Notes
• Test paper with Video Solution
• Mind Map
• Study Planner
• NCERT Solutions
• Discussion Forum
• Previous Year Exam Questions