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Posted by Shivam Yadav 6 years, 2 months ago
- 4 answers
Yogita Ingle 6 years, 2 months ago
LHS: (2x–1)(x–3) =2x2–6x–x+3 = 2x2–7x+3
RHS: (x+5)(x–1) = x2–x+5x–5 = x2 + 4x – 5
Now; 2x2–7x+3 = x2+4x–5
Or, 2x2–7x+3–x2−4x+5 = 0
Or, x2–11x+8=0
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Shivam Yadav 6 years, 2 months ago
0Thank You