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for what value of non negative integer a will the e system x²-y²=0,(x-a)²+y²=1 have exactly three real solutions

```
3 years ago

Arun
25768 Points
```							given =  x²-y²=0               (x-k)²+y²=1  => x²- 2kx + k²+y² =1 by adding this two equations we will get,2x² - 2kx + k²= 0 and now we all know that if the discriminant is zero i.e., b² + 4ac=0 then the quadratic equation has one solution. Then,            4k² - 8k = 0           4k²=8k           4k = 8             k = 8/4 = 2  so the correct answer is a) 2
```
7 months ago
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• 731 Video Lectures
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• Test paper with Video Solution
• Mind Map
• Study Planner
• NCERT Solutions
• Discussion Forum
• Previous Year Exam Questions