Question

Find the zeroes of the quadratic polynomial and verify the relationship between the zeroes and coefficients 6x^2-3-7x

Answer 1

see explanation

Explanation:

Consider the factors of the product of the x² term and the constant term which sum to give the coefficient of the x- term.

Express in standard form, that is 6x² - 7x - 3

product = 6 × - 3 = - 18, sum = - 7

The factors are - 9 and + 2

Use these factors to split the middle term

6x² - 9x + 2x - 3 ( factor the first/second and third/fourth terms )

= 3x(2x - 3) + 1(2x - 3) ← factor out (2x - 3)

= (2x - 3)(3x + 1)

To obtain zeros equate to zero

(2x - 3)(3x + 1) = 0

Equate each factor to zero and solve for x

2x - 3 = 0 ⇒ x =
`(3)/(2)`

3x + 1 = 0 ⇒ x = -
`(1)/(3)`

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The relationship between the zeros and the coefficients is

sum of zeros = -
`(b)/(a)` and

product of zeros =
`(c)/(a)`

-
`(b)/(a)` = -
`(7)/(6)` and

`(3)/(2)` -
`(1)/(3)` =
`(7)/(6)`

`(3)/(2)` × -
`(1)/(3)` = -
`(1)/(2)` and

`(c)/(a)` =
`(-3)/(6)` = -
`(1)/(2)`

Verifying both relationships