Question

The quadratic equation 3x2 – 4 √3 x + 4 = 0 has​

Answer 1

Answer:

roots real and equal

Step-by-step explanation:

D=$b^{2}$-4ac

b=4$\sqrt{3\\}$

a=3

c=4

D=$(4\sqrt{3})^{2}$-4*3*4

=16*3-48

48-48

0

so roots real and equal

ROOTS- (-b±√D)/2

(-4√3)/2=2√3

Answer 2

$\sf \: 3 {x}^{2} - 4 \sqrt{3} x + 4 = 0$

Comparing with ax²+ bx+ c=0 ,

a = 3 , b= -4√3 , c= 4

$\sf \: {b}^{2} - 4ac = {( - 4 \sqrt{3}) }^{2} - 4 \times 3 \times 4$

$\begin{aligned} \sf \: = 16 \times 3 - 4 \times 3 \times 4 \\ \sf \: = 16 \times 3 - 48 \\ \sf \: = 48 - 48 \\ \sf \therefore \green{{b}^{2} - 4ac = 0} \end{aligned}$

By quadratic formula,

$\sf \: x = \dfrac{ - b \pm \: \sqrt{ {b}^{2} - 4ac} }{2a}$

$\sf \: x = \dfrac{ - ( - 4) \times \sqrt{3} \pm \: \sqrt{0} }{2 \times 3}$

$\sf \: x = \dfrac{4 \sqrt{3} \pm \: \sqrt{0} }{6}$

$\sf \: x = \dfrac{4 \sqrt{3} \pm \: 0 }{ 6}$

$\sf \: x = \dfrac{4 \sqrt{3} }{6}$

$\sf \: x = \dfrac{ \cancel4 \sqrt{3} }{ \cancel6}$

$\green{ \boxed{\sf \: x = \dfrac{2 \sqrt{3} }{3}}}$

Additional information :

To find equation -

$\sf \: {x}^{2} - ( \alpha + \beta )x + ( \alpha \beta ) = 0$

$\sf \alpha + \beta = - \dfrac{b}{a}$

$\sf \alpha\beta = \dfrac{c}{a}$

$\sf {x}^{2} - (sum \: of \: roots)x + (product \: of \: roots)$

$\sf \: Quadratic \: formula - \\ \sf \: x = \dfrac{ - b \pm \sqrt{ {b}^{2} - 4ac} }{2a}$