Question

find nature of roots ,it root real then find 2x sq-6x+3=0​

Answer 1

$\sf{\underline{\boxed{\huge{\pink{Question}}}}}$

• find the nature of the roots $2x^2 - 6x + 3 = 0$

$\sf{\underline{\boxed{\huge{\pink{Solution}}}}}$

$\sf\implies 2x^2 - 6x + 3 = 0$

⠀⠀⠀⠀⠀⠀⠀

• compare the eq with

$\sf{\underline{\bold{ax^2 + bx + c = 0 }}}$

⠀⠀⠀⠀⠀⠀⠀

☯ a = 2

☯ b = -6

☯ c = 3

$\sf{\underline{\boxed{\blue{\large{\bold{ D = b^2 - 4ac}}}}}}$

⠀⠀⠀⠀⠀⠀⠀

• finding value of D.

⠀⠀⠀⠀⠀⠀⠀

$\sf\implies D = b^2 - 4ac$

$\sf\implies D = (-6)^2 - 4 \times 2 \times 3$

$\sf\implies D = 36 - 24$

$\sf\implies D = 6$

$\sf{\underline{\boxed{\blue{\large{\bold{ D = 6}}}}}}$

• compare the value of D with the criteria

$\sf \implies D > 0$

$\sf\implies b^2- 4ac > 0$

$\sf\implies 6 > 0$

• Therefore , the equation have real and unequal roots .

$\sf{\underline{\boxed{\blue{\large{\bold{ Real \: and \: Unequal \: roots}}}}}}$