Finding the Zero of Polynomial 9x^2 - 5

To find the zero of the polynomial 9x^2 - 5, we need to set it equal to zero and solve for x.

9x^2 - 5 = 0
9x^2 = 5
x^2 = 5/9
x = ±sqrt(5/9)

Therefore, the zeroes of the polynomial 9x^2 - 5 are x = sqrt(5/9) and x = -sqrt(5/9).

Relation between Zeroes and Coefficients

The relation between the zeroes and coefficients of the polynomial 9x^2 - 5 can be explained using Vieta's formulas.

Vieta's formulas state that for a quadratic polynomial ax^2 + bx + c, the sum of the zeroes is -b/a and the product of the zeroes is c/a.

Applying Vieta's formulas to 9x^2 - 5, we get:

Sum of zeroes = -b/a = 0/9 = 0
Product of zeroes = c/a = -5/9

Therefore, the sum of the zeroes of the polynomial 9x^2 - 5 is zero, and the product of the zeroes is -5/9.

This means that the polynomial can be written as:

9x^2 - 5 = 9(x - sqrt(5/9))(x + sqrt(5/9))

Where (x - sqrt(5/9)) and (x + sqrt(5/9)) are the factors of the polynomial.

Conclusion

In conclusion, we found that the zeroes of the polynomial 9x^2 - 5 are x = sqrt(5/9) and x = -sqrt(5/9), and we explained the relation between the zeroes and coefficients using Vieta's formulas. Understanding this relationship can help in factoring and solving quadratic equations.