Question

show that the quadratic 2 X square + 5 root 3 X + 6 has real root and always solve it​

Answer 1

Answer:

Step-by-step explanation:

2x^2 + 5√3x+6 = 0

Here a=2, b = 5√3< c=6

Fir a quadratic equation to have real roots

b^2- 4ac > or = 0

(5√3)^2 - 4 ×2 ×6

= 25 × 3 -48

=75 - 48

=27 > 0

Hence real roots exist

Now finding the roots

Splitting the middle term

2x^2 + 4√3x + √3 x + 6 =0

2x( x +2√3) + √3 (x + 2√3) = 0

(2x+√3)(x+2√3)=0

So roots are

x = -√3/2, x= -2√3

Answer 2

step by step explanation;

2x^2 + 5√3x+6 = 0

Here a=2, b = 5√3< c=6

Fir a quadratic equation to have real roots

b^2- 4ac > or = 0

(5√3)^2 - 4 ×2 ×6

= 25 × 3 -48

=75 - 48

=27 > 0

Hence real roots exist

Now finding the roots

Splitting the middle term

2x^2 + 4√3x + √3 x + 6 =0

2x( x +2√3) + √3 (x + 2√3) = 0

(2x+√3)(x+2√3)=0

So roots are

x = -√3/2, x= -2√3