Question

Brahmagupta's solution to a quadratic equation of the form ax2 + bx= c involved only one solution. Which solution

would he have found for the equation 3x² + 4x=6?
Ox= √√4(3)(6) +4²-4 _ 2√22-4
2(3)
○ x= √4(3)(4) +6² − 6 = 2√21-6 = √21-3
2(3)
6
Ox= √4(4)(6) +3²-3
2(4)
=
√22-2
3
105-3
8
Ox= √√4(3)(6) 4² +4 _ 2√14+4 _ √14+2
2(3)
6

Answer 1

Explanation:

Brahmagupta's solution to a quadratic equation of the form ax² + bx = c is given by x = (-b ± √(b² - 4ac)) / 2a.

For the equation 3x² + 4x = 6, we have a = 3, b = 4, and c = 6. Plugging these values into the formula, we get:

x = (-4 ± √(4² - 4(3)(6))) / 2(3)

x = (-4 ± √(16 - 72)) / 6

x = (-4 ± √(-56)) / 6

Since the square root of a negative number is not a real number, the quadratic equation has no real solutions. Therefore, Brahmagupta would not have found a solution for this particular equation using his method.