Question

√3x²+ √2x - 2√3 =0 determine the nature of the root for the quadratic equation​

Answer 1

EXPLANATION.

Quadratic Equation,

⇒ F(x) = √3x² + √2x - 2√3 = 0.

As we know that,

⇒ D = b² - 4ac.

⇒ D = (√2)² - 4(√3)(-2√3).

⇒ D = (2) + 24.

⇒ D = 26.

⇒ D > 0.  Or b² - 4ac > 0,

Roots are real and unequal.

                                                                                         

MORE INFORMATION.

Conditions of common Roots,

Let quadratic equation are a₁x² + b₁x + c₁ = 0  and  a₂x² + b₂x + c₂ = 0.

(1) = If only one Root is Common.

x = b₁c₂ - b₂c₁/a₁b₂ - a₂b₁.

y = c₁a₂ - c₂a₁/a₁b₂ - a₂b₁.

(2) = If both roots are common

a₁/a₂ = b₁/b₂ = c₁/c₂.

Nature of the factors of the quadratic expression.

(1) = Roots are different, if b² - 4ac > 0.

(2) = Rational and different, if b² - 4ac is a perfect square.

(3) = Real and equal, if b² - 4ac = 0.

(4) = If D < 0 Roots are imaginary and unequal or complex conjugate.

Answer 2

$\sf\huge { \underline {Given} }\: \large { is \: a \: quadratic \: equation} \\ \\ \sf⇢ \sqrt{3} {x}^{2} + \sqrt{2} x - 2 \sqrt{3} = 0$

$\huge \sf\underline{To \: find }$

The nature of the roots;

$\sf➞F(x)= \sqrt{3} {x}^{2} + \sqrt{2} x - 2 \sqrt{3} = 0$

We know that:

$\sf➣ \: D = {b}^{2} - 4ac$

$\sf\rightarrow \: D = ( \sqrt{2} {)}^{2} - 4( \sqrt{3} )( - 2 \sqrt{3} )$

$\rightarrow \sf \: D = 2 + 24$

$\implies \sf \: D = 26$

$\therefore \sf D > 0 \: or \: {b}^{2} - 4ac > 0$

So we have found the nature of the given roots i.e. real and unequal.