Question

find the zeroes of the polynomial x^2 + 1/6x - 2​

Answer 1

Answer:

Polynomial :

${x}^{2} + \frac{1}{6} x - 2$

Comparing with standard equation, we get

a = 1, b = 16 , c = -2

To find zero, equate it with 0

${x}^{2} + \frac{1}{6} x - 2 = 0$

Multiply by 6

$6 {x}^{2} + x - 12 = 0$

$6 {x}^{2} + 9x - 8x - 12 = 0$

=> 3x(2x+3) - 4(2x+3) = 0

=> (2x+3)(3x-4) = 0

2x+3 = 0 Or 3x-4 = 0

=> x = −3/2or 4/3

We know

Sum of zeroes = −b/a

=> −3/2+4e3=−1e6

=> −9+8/6=−16

=> −1/6=−1/6

Also

Product of zeroes = ca

=> −3/2×4/3=−2/1

=> -2 = -2

hence verified !!