Question

the zeros of a quadratic polynomial x²- 2x-8 are​

Answer 1

Answer:

To find zeroes:-

x²-2x - 8= 0

x² - 4x+2x -8 = 0

X(x-4) +2(x-4) = 0

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

x-4 = 0

x = 4

and x+ 2 = 0

x = -2

hence, 4 and -2 are zeroes of the polynomial

Answer 2

Step-by-step explanation:

${x}^{2} - 2x - 8 = 0$

By Shridhracharya formula,

$x = \frac{ - b \pm \sqrt{ {b}^{2} - 4ac} }{2a}$

$x = \frac{ - ( -2) \pm \sqrt{ {( - 2)}^{2} - 4 \times 1 \times ( - 8)} }{2 \times 1}$

$x = \frac{2 \pm \sqrt{4 + 32} }{2}$

$x = \frac{2 \pm \sqrt{36} }{2}$

$x = \frac{2 \pm6}{2} \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: (\sqrt{36} = 6)$

$x = \frac{2 + 6}{2} , \frac{2 - 6}{2}$

$x = \frac{8}{2} , \frac{ - 4}{2}$

$\boxed{x = 4, - 2}$

Hence the roots are (4,-2)