Find the zeros of y=x^2-8x-3 by completing the square
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There are no real roots for the equation since the determinant is negative.
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Answer: The answer is 4+√19 and 4-√19.
Step-by-step explanation:
Given that, y = x² - 8x - 3
We find zeroes, so we put y = 0
Then, x² - 8x - 3 = 0
=> (x)² - (2× x × 4) + (4)² - (4)² - 3 = 0
=> (x -4)² - 16 -3 = 0
=> (x-4)² = 19
=> x = 4 +-√19
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