Question

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 1

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 \(\frac {1}{3}\).