|
>> how do you solve it without know the quadratic formula or some such? This is actually possible, if you've been taught to solve problems instead of being taught to apply magic formulas. Of course, one is rarely taught math this way now. Let me try. Suppose you have 3x^2+5+2x=7. Now, you have no idea how to solve for x. But what you could solve? If you had something like (ax+b)^2 = c, then probably this would be easy for you to solve, right (of course, provided c is not negative, won't get into that now)? We just take square root of both parts, and then it's a simple linear equation. Now, we can notice we can also get rid of a, for simplicity, by dividing both parts by a and get something like (x+b)^2 = c. Which in our example should represent 3x^2+5+2x=7 or alternatively x^2 + 2/3x - 2/3 = 0. But how we find proper b and c? So, is there a way to make that equation look like that with suitable values? Let's look at (x+b)^2 = x^2 + 2bx + b^2. We have 2bx = 2/3x which means b = 1/3. So we have (x+1/3)^2 on one side, and on the other side we have to get 2/3 and also additional 1/9 which is the b^2, meaning c would be 7/9. So, x + 1/3 is sqrt(7/9) with + or - sign (since square on non-zero root always has 2 options), and x would be that minus 1/3. As you can see, we used no special magic formula - only thing we needed is how to write (x+b)^2, which is easy to see just by multiplying (x+b)*(x+b) by hand. You can do cubic equations this way too, but it is considerably more laborious. Look up Cardano and Vieta methods. Doable but I wouldn't enjoy doing it by hand too much. |
Remember, the goal of an exam is to measure if you have the knowledge. Whether you got the knowledge by studying hard in a traditional classroom setting as scheduled by a syllabus is irrelevant.