how would you solve this quadratic equation 12abx-(a-b)x+ab=0?

Answer 1

That is not a quadratic equation... a quadratic equation has a tem of some multiple of x², a term of some multiple of x, and a constant term. That has three variables (unless you meant a and b were constants), and no term of x².

If you meant that the equation is:
12abx² - (a-b)x + ab = 0
where a and b are constants, you would find the answer using the quadratic formula...
x = [a - b +/- sqrt(a² - 2ab + b² - 48a²b²)]/[24ab]

//edit: puggy has got the wrong sign on his (a-b) term, the function you gave has -(a-b) and the negative signs cancel each other in the formula.

Answer 2

For firsts, the discriminant is b^2-4*a*c. Now let's seem on the three circumstances. efficient Discriminant: y = x^2 + 5x + 6 The discriminant is 5^2 - 4*a million*6 = a million. this suggests there are 2 different roots of the equation. In graphing words, the graph of the equation hits the x-axis two times. also, because the discriminant is a acceptable sq., the roots are rational numbers. 0 Discriminant: y = x^2 - 6x + 9 The discriminat is (-6)^2 - 4*a million*9 = 0. this suggests there is in problem-free words one root of the equation. at the same time as graphed, the equation hits the x-axis in problem-free words once, on the vertex of the parabola. unfavorable Discriminant: y = 2x^2 + x + 7 The discriminant is a million^2 - 4*2*7 = -fifty 5. this suggests there are not any actual roots of the equation. The graph of this equation does no longer hit the x-axis in any respect.

Answer 3

Since you say it's quadratic, I'll assume you mean:

12abx² - (a-b)x + ab = 0

And you'd solve it the same way you'd solve any quadratic equation: complete the square or use the quadratic equation.

I pick the quadratic.

x = ((a - b) ± √((a - b)² - 4(12ab)(ab)))/2(12ab)
x = (a - b ± √(a² -2ab + b² - 48a²b²))/24ab

That's pretty ugly, but I think that's as far as you can go with it, pretty much.

Answer 4

12abx² - (a-b)x + ab=0
So x² - (a-b)/12ab x + 1/12 = 0

ie x² -1/12(1/b - 1/a) x = -1/12

ie x² - 2*(1/24(1/b - 1/a)) x + (1/24(1/b - 1/a)² = (1/24(1/b - 1/a)² -1/12 * (2*24)/(2*24)

(x - (1/24)(1/b - 1/a))² = (1/24)²(1/b - 1/a))² - 48/24²

x - (1/24)(1/b - 1/a) = ±√[(1/b - 1/a))² - 48]/ 24

x = (1/24)[(1/b - 1/a)±√[(1/b - 1/a))² - 48] ]

Now clearly this will only have real solutions if (1/b - 1/a))² ≥48

ie (a - b)² ≥ 48a²b²

ie a - b ≥ ie a - 4√3 ab ≥ b ie a ≥ b/(1 - 4b√3)

OR a - b ≤ - 4√3 ab ie a ≤ b/(1 + 4b√3)

Answer 5

12abx² - (a-b)x + ab=0
So x² - (a-b)/12ab x + 1/12 = 0

ie x² -1/12(1/b - 1/a) x = -1/12

ie x² - 2*(1/24(1/b - 1/a)) x + (1/24(1/b - 1/a)² = (1/24(1/b - 1/a)² -1/12 * (2*24)/(2*24)

(x - (1/24)(1/b - 1/a))² = (1/24)²(1/b - 1/a))² - 48/24²

x - (1/24)(1/b - 1/a) = ±√[(1/b - 1/a))² - 48]/ 24

x = (1/24)[(1/b - 1/a)±√[(1/b - 1/a))² - 48] ]

Now clearly this will only have real solutions if (1/b - 1/a))² ≥48

ie (a - b)² ≥ 48a²b²

ie a - b ≥ ie a - 4√3 ab ≥ b ie a ≥ b/(1 - 4b√3)

OR a - b ≤ - 4√3 ab ie a ≤ b/(1 + 4b√3)

Answer 6

For one thing, I think you meant to type "12abx^2". All you have to do is plug in the appropriate values for the coefficients.

x = { -(a - b) +/- sqrt[ (a-b)^2 - 4(12ab)(ab) ] } / 2(12ab)

x = { (a - b) +/- sqrt [ (a^2 - 2ab + b^2) - 48a^2b^2 ] } / [24ab]

From here, it's a matter of reduction.

//edit: Steve is absolutely correct. It should be (a - b) instead of (b - a) because I have to take -(-(a-b))). Corrected.

Answer 7

That is not a quadratic equation at all ..... it is linear eqn. Solving foy x gives x = ab / (12ab - a + b)

Answer 8

12abx-ax-bx+ab=0
12abx=ax+bx
12=(ax+bx)/abx
12=x(a+b)/abx
12=a+b/ab
so a can 12 and b can be 0 or
b can be 12 abd a cab be 0

Answer 9

its called copying off somebody elses paper.