Question

Determine the nature of roots for each of the quadratic equation.
(1) $3x^2-5x+7=0$
(2) $\sqrt{3}{x}^2+\sqrt{2}x-2\sqrt{3}=0$
(3) $m^2-4x-3=0$

Answer 1

(1) $3x^2-5x+7=0$
In order to find the nature of the roots we need to find the discriminant.
$$\begin{gathered} D=b^2-4ac \\ a=3,b=-5,c=7 \\ D={\left(-5\right)}^2-4\times 3\times 7=25-84=-59 \end{gathered}$$
Since, the D < 0 so, there is no real root.

(2) $\sqrt{3}{x}^2+\sqrt{2}x-2\sqrt{3}=0$
In order to find the nature of the roots we need to find the discriminant.
$$\begin{gathered} D=b^2-4ac \\ a=\sqrt{3},b=\sqrt{2},c=-2\sqrt{3} \\ D={\left(\sqrt{2}\right)}^2-4\times \sqrt{3}\times \left(-2\sqrt{3}\right)=2+24=26>0 \end{gathered}$$
Since, the D > 0 so, real and unequal roots.

(3) $m^2-4x-3=0$
In order to find the nature of the roots we need to find the discriminant.
$$\begin{gathered} D=b^2-4ac \\ a=1,b=-4,c=-3 \\ D={\left(-4\right)}^2-4\times \left(1\right)\times \left(-3\right)=16+12=28 \end{gathered}$$
Since, the D > 0 so, real and unequal roots.