Question

(1) Determine the nature of the roots for the quadratic equation
√3x² + 2√2x – 2√3=0​

Answer 1

Answer:

The given quadratic equation is

\sqrt{3}x^2+\sqrt{2}x-2\sqrt{3}=0

3

x

2

+

2

x−2

3

=0

A quadratic equation is ax^2+bx+c=0ax

2

+bx+c=0 .

If b^2-4ac < 0b

2

−4ac<0 , then the equation have two complex roots.

If b^2-4ac=0b

2

−4ac=0 , then the equation have equal real roots.

If b^2-4ac > 0b

2

−4ac>0 , then the equation have two distinct real roots.

In the given equation,

a=\sqrt{3},b=\sqrt{2},c=-2\sqrt{3}a=

3

,b=

2

,c=−2

3

b^2-4ac=(\sqrt{2})^2-4(\sqrt{3})(-2\sqrt{3})=2+24=26b

2

−4ac=(

2

)

2

−4(

3

)(−2

3

)=2+24=26

Since b^2-4ac > 0b

2

−4ac>0 , therefore, the given quadratic equation have two distinct real root.

Answer 2

The given equation is

$(\sqrt{3})x^{2} + (2\sqrt{2})x +(-2\sqrt{3}) =0\\where\:a=\sqrt{3},\: b=2\sqrt{2}\: and \:c = -2\sqrt{3}\\Determinant\:of\:the\:equation\:(\Delta) = b^{2}-4ac\\=(2\sqrt{2})^{2}-4\times\sqrt{3}\times(-2\sqrt{3})\\=8+8\times3\\=32$

Since Δ > 0, and Δ is not a perfect square

∴The two roots are unequal to each other and are irrational