Question: Find the roots of the quadratic equations given in Q.1 above by applying the quadratic formula:
i. 2x2 - 7x + 3 = 0
ii. 2x2 + x – 4 = 0
iii. 4x2+ 4\\sqrt3)x + 3 = 0
iv. 2x2 + x + 4 = 0
Video Explanation
Explanatory Answer
i. 2x2 - 7x + 3 = 0
The roots of ax2 + bx + c = 0 are \\frac {-b \pm \sqrt {{b}^{2}-4ac}} {2a})
In this question a = 2, b = -7, c = 3
So, b2 – 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. 2x2 + x – 4 = 0
The roots of ax2 + bx + c = 0 are \\frac {-b\pm \sqrt {{b}^{2}-4ac}} {2a})
In this question a = 2, b = 1 and c = -4
b2 – 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. 4x2 + 4√3 x + 3 = 0
The roots of ax2 + bx + c = 0 are \\frac {-b\pm \sqrt {{b}^{2}-4ac}} {2a})
In this question a = 4, b = 4 √3 and c = 3
b2 – 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. 2x2 + x + 4 = 0
The roots of ax2 + bx + c = 0 are \\frac {-b\pm \sqrt {{b}^{2}-4ac}} {2a})
In this question a = 2, b = 1 and c = 4
b2 – 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.