#### Question: Find the roots of the quadratic equations given in Q.1 above by applying the quadratic formula:

i. 2x^{2} - 7x + 3 = 0

ii. 2x^{2} + x – 4 = 0

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

iv. 2x^{2} + x + 4 = 0

#### Video Explanation

#### Explanatory Answer

##### i. 2x^{2} - 7x + 3 = 0

The roots of ax^{2} + bx + c = 0 are \\frac {-b \pm \sqrt {{b}^{2}-4ac}} {2a})

In this question a = 2, b = -7, c = 3

So, b^{2} – 4ac = (-7)^{2} – 4 * 2 * 3

= 49 - 24 = 25 > 0

∴ x = \\frac{-(-7){\pm }{\sqrt{25}}}{2 * 2}) = \\frac{7 {\pm } 5}{4})

x = \\frac{7 + 5}{4}) = 3 and x = \\frac{7 - 5}{4}) = \\frac{1}{2})

##### ii. 2x^{2} + x – 4 = 0

The roots of ax^{2} + bx + c = 0 are \\frac {-b\pm \sqrt {{b}^{2}-4ac}} {2a})

In this question a = 2, b = 1 and c = -4

b^{2} – 4ac = (1)^{2} - 4 * 2 * (-4) = 1 + 32 = 33 > 0

∴ x = \\frac{-1{\pm }{\sqrt{33}}}{2*2}) = \\frac{-1{\pm }{\sqrt{33}}}{4})

x = \\frac{-1+{\sqrt33}}{4}) and \\frac{-1-{\sqrt33}}{4})

##### iii. 4x^{2} + 4√3 x + 3 = 0

The roots of ax^{2} + bx + c = 0 are \\frac {-b\pm \sqrt {{b}^{2}-4ac}} {2a})

In this question a = 4, b = 4 √3 and c = 3

b^{2} – 4ac = (4 √3)^{2} - 4 * 4 * 3

= 48 - 48 = 0, which is not negative

∴ x = \\frac{-4{\sqrt3}{\pm } 0}{2 * 4}) = -\\frac{\sqrt3}{2}) and -\\frac{\sqrt3}{2})

##### iv. 2x^{2} + x + 4 = 0

The roots of ax^{2} + bx + c = 0 are \\frac {-b\pm \sqrt {{b}^{2}-4ac}} {2a})

In this question a = 2, b = 1 and c = 4

b^{2} – 4ac = (1)^{2} - 4 * 2 * 4 = 1 - 32 = - 31 < 0

Square root of a negative number is not real.

Hence, this equation does not have real roots.