Question

Find the zeroes of the following quadratic equation polynomial and verify the relationship between the zer
oes and the cofficients
1)6x2-7x-3

Answer 1

Answer:

Step-by-step explanation:

Let one zero of the polynomial be "a" and another zero be "b"

Let p(x) 6x² - 7x - 3 = 0

Now, 6x² - 7x - 3 = 0

= 6x² - 9x + 2x - 3 = 0

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

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

= 3x + 1 = 0 & 2x - 3 = 0

= x = -1/3 & x = 3/2

Now, sum of zeroes = 3/2 - 1/3 = 9-2/6 = 7/6

- coefficient of x/coefficient of x²= -(-7)/6 = 7/6

Product of zeroes = (3/2)(-1/3) = -3/6 = -1/2

Constant term/coefficient of x² = -3/6 = -1/2

Hence, sum of zeroes = -coefficient of x/coefficient of x²

Product of zeroes = constant term/coefficient of x²

Verified

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