Solve for x: 2x^2 = 5x + 12
The first step to solving any equation is to work out what kind of equation it is. As this particular equation has only x's and numbers, and it has a x2 term, we know that it is quadratic.Now that we know that it is quadratic we know that they can only be solved when the equation is equal to 0. In this case that requires some rearrangement. 2x2 - 5x - 12 = 0 (We could have -2x2 + 5x + 12 = 0, but it is generally easier to have the x2 term be positive). There are 3 main methods for solving quadratic equations; fatorising, completing the square and the quadratic formula. Factorising is by far the easiest, so it is normally a good idea to start here, however if you cannot come up with a solution in 10-20 seconds you should try a different method, especially in an exam as a lot of the time the factors are not immeaditly obvious, or difficult to spot. (Square roots, fractions ect.) The factorising method involves finding 2 numbers that add to make the coefficient of the x term and multiply to make the number term, however in cases such as this were the coefficient of the x2 term isn't 1, we need to multiply number term by that. We need 2 numbers that add to make -5, and multiply to make -24. As our number term is negitive, we know we ae looking for 1 positive and one negitive number. Now we just think of pairs of factors of 24 and see if the difference between them is 5. 1 and 24 doesn't work, niether does 2 and 12, but 3 and 8 does work. ( 3 x 8 = 24 and 3-8 = -5)
Now we can break our -5x into +3x and -8x in our equation. 2x2 - 8x + 3x -24 = 0. Next we find a common factor between our x2 term and one of the x terms, and one between the number term and the second x term. In this case we can have x(2x +3) and -4(2x + 3) The factor taken out should be the same, if not we have either taken out the wrong factor, or paired up the wrong x terms to the wrong 'other' term. Now this 2x + 3 is a factor of the whole equation leaving us with (2x + 3)(x - 4) = 0. Now this can only occur when either 2x + 3 = 0, or when x - 4 = 0. This means the solution to our original expression, is either when x = 4 (From the x-4 term) or when x = -3/2 (from the 2x + 3 term)
