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The zeros of the polynomial 18x^2-27x+7 are ___________(a) \(\frac {7}{6}, \frac {1}{3}\)(b) \(\frac {-7}{6}, \frac {1}{3}\)(c) \(\frac {7}{6}, \frac {-1}{3}\)(d) \(\frac {7}{3}, \frac {1}{3}\)I got this question in class test.This is a very interesting question from Zeros and Coefficients of Polynomial topic in chapter Polynomials of Mathematics – Class 10

Answer»

The CORRECT answer is (a) \(\frac {7}{6}, \frac {1}{3}\)

The best EXPLANATION: 18x^2-27x+7=0

18x^2-21x-6x+7=0

3x(6x-7)-1(6x-7)=0

(6x-7)(3x-1)=0

x=\(\frac {7}{6}, \frac {1}{3}\)

The ZEROS are \(\frac {7}{6}\) and \(\frac {1}{3}\).



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