Question

solve the equation by shreedharacharya 2x² + 5 root 3x + 6 = 0​

Answer 1

${\underline{\underline{\mathfrak{\green{\huge{Answer}}}}}}$

Hey Mate!!

p(x) = 2x² + 5√3x +6

→ 2x² + 4√3x + √3x + 6

→ 2x(x + 2√3) + √3(x + 2√3)

→ (2x + 2√3)(x + 2√3)

→ 2(x + √3)(x + 2√3)

⇒ x = -√3    OR     -2√3.

Answer 2

Given:-

→p(x)=0

→2x²+5√3x+6=0

$\rule{200}{1}$

To Find:-

→The Value of x, Using Shreedharacharya's Formula or Discriminant Formula?

$\rule{200}{1}$

AnsWer:-

→2x²+5√3x+6=0

•Using Discriminant Formula•

→d=b²-4ac

๛a=2,b=5√3,c=6

♦Putting the Values♦

→d=(5√3)²-4×2×6

→d=75-48

→d=27

→√d=±√27

→√d=3√3

$\rule{200}{1}$

•Taking x(+)=$\frac{-b+ \sqrt{d}}{2a}$

→x=$\frac{-(5 \sqrt{3}+3 \sqrt{3})}{2×2}$

→x=$\frac{-2 \sqrt{3}}{4}$

→x=$\frac{\cancel{-2}(1 \sqrt{3} )}{\cancel{4}}$

→x=$\frac{- \sqrt{3}}{2}$

$\rule{200}{1}$

•Taking x(-)=$\frac{-b- \sqrt{d}}{2a}$

→x=$\frac{-(5 \sqrt{3}-3 \sqrt{3} )}{2×2}$

→x=$\frac{-8 \sqrt{3}}{4}$

→x=$\frac{\cancel{-8}( 1 \sqrt{3})}{\cancel{4}}$

→x=$\frac{-2 \sqrt{3}}{1}$

→x=-2√3

$\rule{200}{2}$