Question

I need someone good at math Please tell me how to factor to solve: (2x-3)(3x+2)≤ 0. The answer to this equation is (-2/3) ≤ x ≤ (3/2)

Answer 1

help please

Answer 2

You have two brackets on the left hand side, but when multiplied you want the answer to be less than or equal to zero.

There are two ways you can get a negative answer. The first is that bracket one is negative and bracket two is positive \((-)\times (+)\). Solve this and see if it is possible. The other option is bracket one is positive and bracket two is negative \((+)\times (-)\). Again solve and see if it gives any possible solutions.

Option 1.
\((2x-3)\leq 0 \textrm{ AND } (3x+2)\geq 0\)

Option 2.
\((2x-3)\geq 0 \textrm{ AND } (3x+2)\leq 0\)

Answer 3

what? i provided the answer and original equation i just need to know how to factor it correctly or sole it correctly

Answer 4

the original equation is what it is, its not changable.

Answer 5

according to your options 1 and 2 i can use neither of them because they are not what the original question is. Im positive that this is what the original question is.

Answer 6

That is how you solve it. I have not changed the equation at all. The equation is two things multiplied together. for the product to be negative, one bracket must be positive and the other negative. Think of (-2) time (3), this gives you -6.

If you look at option 1, were we take the first bracket as negative and the second as positive, then that gives you \(2x-3\leq 0\) and \(3x+2\geq0\) which to be true means that \(x\leq (3/2)\) and \(x\geq (-2/3)\) which written together gives you your answer \((-2/3)\leq x \leq (3/2)\). Option 2, does not give any feasible answers in this case