Question

HW
find its roots using methods ​

Answer 1

Answer:

roots are 3+(or)-√5

hence it showed

Answer 2

Question:-

Find the roots of

${x}^{2} - 6x + 4 = 0$

Solution:-

We know that

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

It is quadratic formula

Here,

a = 1

b = -6

c = 4

By using Quadratic formula

$x = \frac{ - ( - 6) \frac{ + }{} \sqrt{ {( - 6)}^{2} - 4 \times 1 \times 4 } }{2 \times 1}$

$x = \frac{6 \frac{ + }{} \sqrt{36 - 16} }{2}$

$x = \frac{6 \frac{ + }{} \sqrt{20} }{2}$

$x = \frac{6 \frac{ + }{}2 \sqrt{5} }{2}$

Take 2 in common

$x = \frac{2(3 ± \sqrt{5}) }{2}$

Finally,

$x = 3 \frac{ + }{} \sqrt{5}$

i.e.,

$x = 3 + \sqrt{5} \: \: or \: \:x = 3 - \sqrt{5}$