Question

Show that2x²-6x+3=0 has real roots also find the roots

Answer 1

Answer:

since the value of discriminant is greater than zero ,it has distinct real roots

Answer 2

AnswEr

The roots of the equation are :

(3 + √3)/2 and (3 - √3)/2

GivEn

The quadratic equation is

• 2x² - 6x + 3 = 0

Task

• To show that the given equation has real roots
• To find the roots of the equation

SoluTion

Given ,

2x² - 6x + 3 = 0

We know that , a quadratic equation has real roots if its discriminant is greater than or equal to zero i.e.

➜ b² - 4ac ≥ 0

Here in the equation ,

a = 2 , b= -6 and c = 3

Thus ,

$\sf \dashrightarrow {( - 6)}^{2} + 4 \times 3 \times 2 \\ \\ \sf \dashrightarrow 36 - 24 \\ \\ \sf \dashrightarrow 12$

which is greater than zero

Thus , discriminant , b² - 4ac > 0 so the given equation has real roots

Now , by quadratic formula we have

$\sf ↦ x = \dfrac{ - b \pm \sqrt{b {}^{2} - 4ac} }{2a} \\ \\ \implies \sf x = \dfrac{ - ( - 6) \pm \sqrt{12} }{2 \times 2} \\ \\ \sf \implies{x} = \dfrac{6 \pm2 \sqrt{3} }{4} \\ \\ \sf\implies{x} = \dfrac{3 \pm \sqrt{3} }{2}$

Thus , the roots are ,

$\sf \longrightarrow{x} = \dfrac{3 + \sqrt{3} }{2} \: \: and \: \longrightarrow x = \dfrac{3 - \sqrt{3} }{2}$