Question

Find the root of the following quadratis
equation (if they exist) by the
Method of ouaduatic Formula.
2x^2 + 5root3x + 6 = 0
​​

Answer 1

Given:-

P(x)=0

2x²+5√3x+6=0

$\rule{200}{1}$

To Find:-

The Zeros of p(x)?[p(x)=Equation]

$\rule{200}{1}$

Answer:-

→2x²+5√3x+6=0

๛Using Discriminant formula๛

•a=2,b=5√3,c=6

→d=b²-4ac ๛[Discriminant Formula]

→d=(5√3)²-4×2×6

→d=75-48

→d=27

→√d=√b²-4ac

→√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=$\frac{-5 \sqrt{3}+3 \sqrt{3}}{4}$

→x=$\frac{-2 \sqrt{3}}{4}$

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

→x=$\frac{-1 \sqrt{3}}{2}$

$\rule{200}{1}$

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

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

→x=$\frac{-5 \sqrt{3}-3 \sqrt{3}}{4}$

→x=$\frac{-8 \sqrt{3}}{4}$

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

→x=$\frac{-2 \sqrt{3}}{1}$

$\rule{200}{2}$

Answer 2

Step-by-step explanation:

Given -

polynomial is 2x² + 5√3x + 6 = 0

To Find -

Roots of the polynomial by Quadratic formula.

Now,

2x² + 5√3x + 6 = 0

here,

a = 2

b = 5√3

c = 6

Quadratic formula = - b ± √b² - 4ac/2a

= -(5√3) ± √(5√3)² - 4 × 2 × 6/2(2)

= -5√3 ± √75 - 48/4

= -5√3 ± √27/4

= -5√3 ± 3√3/4

Hence,

The roots of the polynomial 2x² + 5√3x + 6 = 0 is

x = -5√3 + 3√3/4

= -2√3/4

= -√3/2

and

x = -5√3 - 3√3/4

= -8√3/4

= -2√3

Formula Used -

x = -b ± √b² - 4ac/2a