#### Question: Find the nature of the roots of the following quadratic equations. If real roots exist, find them:

i. 2x^{2} – 3x + 5 = 0

ii. 3x^{2} - 4\\sqrt3)x + 4 = 0

iiii. 2x^{2} – 6x + 3 = 0

#### Explanatory Answer

##### i. 2x^{2} – 3x + 5 = 0

In this equation, a = 2, b = -3 and c = 5

∴ Discriminant b^{2} – 4ac = (-3)^{2} – 4 × 2 × 5

= 9 – 40 = -36 < 0

The discriminant is negative.

So, the equation has __NO real roots__.

##### ii. 3x^{2} - 4 √3 x + 4 = 0

In this equation, a = 3, b = -4 √3 and c = 4

∴ discriminant b^{2} – 4ac = (-4 √3)^{2} – 4 × 3 × 4 = 48 – 48 = 0

Because the discriminant is zero, the equation has two equal real roots.

x = \\frac{-(-4\sqrt3)}{2 × 3}) = \\frac{4\sqrt3}{6}) = \\frac{2\sqrt3}{3}) = \\frac{2}{\sqrt{3}})

The roots are \\frac{2}{\sqrt{3}}) and \\frac{2}{\sqrt{3}})

##### iii. 2x^{2} – 6x + 3 = 0

In this equation a = 2, b = -6, c = 3

∴ discriminant b^{2} - 4ac = (-6)^{2} – 4 × 2 × 3

= 36 – 24 = 12 > 0

The discriminant is positive.

So, the equation has two distinct real roots.

The roots are \\frac {-b\pm \sqrt {{b}^{2}-4ac}} {2a})

x = \\frac {-(-6)\pm \sqrt {12}} {2 \times 2}) = \\frac {6\pm \sqrt {12}} {4}) = \\frac {6\pm 2\sqrt {3}} {4})

x = \\frac{3 + \sqrt{3}}{2}) and x = \\frac{3 - \sqrt{3}}{2})