Question: Find the nature of the roots of the following quadratic equations. If real roots exist, find them:
i. 2x2 – 3x + 5 = 0
ii. 3x2 - 4\\sqrt3)x + 4 = 0
iiii. 2x2 – 6x + 3 = 0
Explanatory Answer
i. 2x2 – 3x + 5 = 0
In this equation, a = 2, b = -3 and c = 5
∴ Discriminant b2 – 4ac = (-3)2 – 4 × 2 × 5
= 9 – 40 = -36 < 0
The discriminant is negative.
So, the equation has NO real roots.
ii. 3x2 - 4 √3 x + 4 = 0
In this equation, a = 3, b = -4 √3 and c = 4
∴ discriminant b2 – 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. 2x2 – 6x + 3 = 0
In this equation a = 2, b = -6, c = 3
∴ discriminant b2 - 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})