Question

What is the easiest way to solve a quadratic equation?​

Answer 1

Answer: This is an opinion, look below! :)

Step-by-step explanation:

Factor the expression. To factor the expression, you have to use the factors of the {\displaystyle x^{2}}x^{2} term (3), and the factors of the constant term (-4), to make them multiply and then add up to the middle term, (-11). Here's how you do it:

Since {\displaystyle 3x^{2}}3x^{2} only has one set of possible factors, {\displaystyle 3x}3x and {\displaystyle x}x, you can write those in the parenthesis: {\displaystyle (3x\pm ?)(x\pm ?)=0}(3x\pm ?)(x\pm ?)=0.

Then, use process of elimination to plug in the factors of 4 to find a combination that produces -11x when multiplied. You can either use a combination of 4 and 1, or 2 and 2, since both of those numbers multiply to get 4. Just remember that one of the terms should be negative, since the term is -4.[3]

By trial and error, try out this combination of factors {\displaystyle (3x+1)(x-4)}(3x+1)(x-4). When you multiply them out, you get {\displaystyle 3x^{2}-12x+x-4}3x^{2}-12x+x-4. If you combine the terms {\displaystyle -12x}-12x and {\displaystyle x}x, you get {\displaystyle -11x}-11x, which is the middle term you were aiming for. You have just factored the quadratic equation.

As an example of trial and error, let's try checking a factoring combination for {\displaystyle 3x^{2}-11x-4=0}3x^{2}-11x-4=0 that is an error (does not work): {\displaystyle (3x-2)(x+2)}(3x-2)(x+2) = {\displaystyle 3x^{2}+6x-2x-4}3x^{2}+6x-2x-4. If you combine those terms, you get {\displaystyle 3x^{2}-4x-4}3x^{2}-4x-4. Though the factors -2 and 2 do multiply to make -4, the middle term does not work, because you needed to get {\displaystyle -11x}-11x, not {\displaystyle -4x}-4x.