Question

write the nature of the roots of
${x}^{2} - 4 \sqrt{3} x + 12 = 0$
​

Answer 1

Answer:

it is real

Step-by-step explanation:

$a = 1 \\ b = - 4 \sqrt{3} \\ c = 12 \\ d = {b}^{2} - 4ac \\ d = { (- 4 \sqrt{3 }) }^{2} - 4(1)(12) \\ 0 = 16 \times 3 - 48 \\ 0 = 48 - 48 \\ 0 = 0 \\ it \: has \: real \: roots$