Question

find the nature of the root of the following quadratic equation . if the real root exist , find them​

Answer 1

Answer:

( 3 ± √3 ) / 2

Step-by-step explanation:

Discriminant of standard eq. is given by b^2 - 4ac.

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

 ⇒ Discriminant

 ⇒ ( - 6 )^2 - 4( 2 * 3 )

 ⇒ 36 - 4( 6 )

 ⇒ 36 - 24

 ⇒ 12

Hence, nature of roots is  real and unequal.

 Using Quadratic formula:

⇒ x = ( - b ± √D ) / 2a

      =   ( 6 ± √12 ) / 2( 2 )

      = ( 6 ± 2√3 ) / 4

      = ( 3 ± √3 ) / 2

Answer 2

Answer:

⋆ Given Polynomial : 2x² – 6x + 3 = 0

where ; a = 2⠀⠀b = – 6⠀⠀c = 3

$\rule{100}{0.8}$

☯ $\underline{\textsf{Using Discriminat Formula to find roots :}}$

$\dashrightarrow\sf\:\:2x^2-6x+3=0\\\\\\\dashrightarrow\sf\:\:x=\dfrac{-\:b\pm\sqrt{b^2-4ac}}{2a}\\\\\\\dashrightarrow\sf\:\:x=\dfrac{-\:( - 6)\pm\sqrt{( - 6)^2-(4 \times 2 \times 3)}}{2 \times 2}\\\\\\\dashrightarrow\sf\:\:x=\dfrac{6\pm\sqrt{36-24}}{4}\\\\\\\dashrightarrow\sf\:\:x=\dfrac{6\pm\sqrt{12}}{4}\\\\\\\dashrightarrow\sf\:\:x = \dfrac{6 \pm2\sqrt{3}}{4}\\\\\\\dashrightarrow\sf\:\:x = \dfrac{2(3 \pm\sqrt{3})}{4}\\\\\\\dashrightarrow\:\:\underline{\boxed{\sf x = \dfrac{3 \pm \sqrt{3}}{2}}}$

⠀

$\therefore\:\textsf{Roots of the Quadratic Equation are \textbf{ \dfrac{\text3\pm\sqrt{\text3}}{\text2} }}$