Question

What are the zeroes of quadratic polynomial x^2-x-4 ?

Answer 1

x²-x-4

a= 1

b=(-1)

c=(-4)

By completing square method .☺️

$d = {b}^{2} - 4ac \\ \\ d = { (- 1)}^{2} -4 \times 1 \times ( - 4) \\ \\ d = 1 + 16 \\ \\ d = 17$

$since \: d \: is \: positive \\ \\ x = \frac{ - b + - \sqrt{d} }{2a} \\ \\ x = \frac{ - ( - 1) + \sqrt{17} }{2 \times 1} \: \: \: \: \: \: \: \: x = \frac{ - ( - 1) - \sqrt{17} }{2 \times 1} \\ \\ x = \frac{1 + \sqrt{17} }{2} \: \: \: \: \: \ \: x = \frac{1 - \sqrt{17} }{2}$

zeroes are

$\frac{1 + \sqrt{17} }{2} \: and \: \frac{1 - \sqrt{17} }{2}$