Question

I’ve noticed here are around 4 ways to solve a Quadratic Equation. Factoring, Completing the square, The quadratic Formula, and graphing. With some equations, it seems that all 4 can be used for solving. But for a lot of equations, sometimes factoring doesn’t work, and you’ll have to use the Quadratic formula. Or maybe the Quadratic Formula is too slow and you need to factor instead etc. Is there an easy or quick way to tell, just by looking at a quadratic equation, which method is best for solving?

Answer 1

Answer:

In my preferred order, ...

• use a graphing calculator
• factor
• use the quadratic formula

Step-by-step explanation:

My personal favorite is to use a graphing calculator. If it shows the zeros are whole numbers, your solution is at hand. If it shows the zeros are recognizable decimal fractions*, the solution is also at hand. If it shows the zeros are unrecognizable as fractions, they are likely irrational. Then the vertex of the graph gets you most of the way to an exact solution.

If the quadratic's coefficients are not integers, then the quadratic formula is least likely to give difficulty.

Example

The attached is a graph of the quadratic ...

  y = 2x^2 -3x -4

We can see from the x-intercepts that the roots are irrational (none are fractions with a denominator that is a factor of 2). The vertex of the graph is (3/4, -5 1/8), so the vertex form of the equation for this graph is ...

  y = 2(x -3/4)^2 -(5 1/8)

Solving this for y=0, we find ...

  0 = 2(x -3/4)^2 -(5 1/8)

  5 1/8 = 2(x -3/4)^2 . . . . . .add the constant (vertex y-value)

  (5 1/8)/2 = (x -3/4)^2 . . . . divide by 2 (the leading coefficient)

  2 9/16 = (x -3/4)^2

  x = 3/4 ± √(41/16) . . . . . . . add the horizontal offset (vertex x-value)

  x = (3 ± √41)/4 . . . . . . . . . . simplify

If you see what we've done here, you can write down the answer from the vertex and the leading coefficient:

  x = (3/4) ± √((5 1/8)/2) . . . . . vertex (3/4, 5 1/8), leading coefficient 2

__

With a little practice, factoring can be a relatively simple way to find the solution. As a preliminary step, the equation should be divided by any common factor of the coefficients, so they are all relatively prime. It helps immensely to know your times tables thoroughly. I like this method, because it can get a solution with fewer steps than using the quadratic formula.

The product of two binomials is ...

  (ax +b)(cx +d) = acx^2 +(ad+bc)x +bd

The product of the leading coefficient (ac) and the constant (bd) is abcd. It has factors of ad and bc, whose sum is the coefficient of the linear term. So, if you can factor the product of the leading and trailing coefficients and find a sum of factors that is the middle coefficient, you're home free. Sometimes the factors jump out at you, and sometimes you need to list a bunch of them before you see what you need.

In our example, we would be looking for factors of 2(-4) = -8 that have a sum of -3.

  -8 = (-8)(1) = -4(2) = (-2)(4) = (-1)(8)

These pairs have sums of -7, -2, 2, 7. None of the sums is -3, so factoring by integers is not possible. You can see that it is unnecessary to list the pairs with positive sums in this case.

Once you have the values ad and bc, you can see that 'a' is the greatest common divisor of ac and ad. Then ad/a = d, and ac/a = c. Now you know 3 of the 4 numbers in the binomial factors.

_____

* The Rational Root Theorem can tell you what reasonable fractions might be for your quadratic. You should be familiar with the decimal equivalents of fractions with small integer denominators.

Answer 2

Answer:

See explanation

Step-by-step explanation:

Usually, I just find the quadratic formula the easiest to go, but here are some tips to tell if which you should use.

1. First see if you can get integer values for the roots to multiply to c and add to b.

This is the easiest way to solve a quadratic equation, and I always check for whether I can easily find the roots by multiply two numbers for c and adding them for b.

For instance, the Quadratic $x^2 + 5x + 6$

I can easily see that 2 and 3 are the roots here - as $2\cdot3=6$ and $2+3=5$. If you can do this and receive integer values, then great! I'd definitely stick with this.

2. See if you can complete the square

If you have a quadratic which is a perfect square, you'll be able to tell if c is a perfect square, and b is twice that number. This will be easiest. Any quadratic equation can be solved by completing the square, but I personally stick to this method only if I can tell it's a perfect square quadratic.

3. Stick to the Quadratic Formula

If all else fails, go with the Quadratic formula. It's not that hard and it gets the job done. I would highly discourage going with graphing, as it takes too much time and it's quite a hassle unless you have a graphing calculator.

I really hope this helped!