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?
O x= 4(3)(6) + 42 - 4
2 (3)
222-4 22-2
3
6
O x = 4(3)(4) + 62 - 6 _ 2/21-6
21-3
3
2 (3)
6
O x= V4(4)(6) + 32 - 3
V105 - 3
8
2(4)
O x = 4(3)(6) - 42 + 4
214 + 4
6
14+2
3
2 (3)

set each factor equal to 0

4x-3 = 0

2x-1 = 0

solve for x.

4x-3=0

4x=3

x= 3/4

solve for x. 

2x-1= 0

2x= 1

x= 1/2

eventually after a few more steps, your answer should be 1/2 is less than or equal to x and 3/4 is greater than or equal to x

ιт ιѕ 6 6/23

step-by-step explanation:

нσρє ι ¢συℓ∂ нєℓρ уσυ συт

the answer is b.6

step-by-step explanation:

area of rectangle a is 2 times that of rectangle b that means 2x+x =   36cm^2

3x = 36cm ^2

x = 36/3

x= 12cm^2

and 2x = 24cm^2

now side of square = are divided by sides = 36/4 = 6cm

that means length of rectangle a and b is 6cm too

now for rectangle a,

l*b = area

= 6*y=24

y=24/6 = 4

therefore dimensions of rectangle a are- length = 6cm, breadth = 4cm

and for rectangle b, length = 6cm and breadth = 6-4 = 2cm.

hope this .