Find zeroes of the polynomial : 1) 7y^2-11/3y-2/3 Answer it only if you know
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→ 7y² - 11/3y - 2/3
→ Comparing with ax² + bx + c
→ we get,
→ a=7 , b=-11/3 , c=-2/3
→ Now using quadratic equation
→ x= (-b±√b²-4ac)/2a
→ x = [-(-11/3) ± √(-11/3)² - 4(7)(-2/3)]/14
→ x = [11/3 ± √121/9 + 56/3]/14
→ x = [11/3 ± √121/9 + 168/9]/14
→ x = [11/3 ± √289/9]/14
→ x = [11/3 + √289/9]/14 (First zero)
→ x = [11/3 - √289/9]/14 (second zero)
→ So the two zeros are these written above.
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