What is the difference between solving a quadratic by factoring and using the quadratic formula?

Question

Factoring is efficient when the quadratic is "nice," however, using the Quadratic formula works every time (unless there isn't a solution...)

Answer 1

quadratic formula works every time

while factoring doesn"t

Answer 2

When I am presented with a quadratic in standard form--Ax²+By+C--to solve, I have a list of 4 methods:

The first 3 work when A=1. If A≠1 you an either factor A out, or go directly to method 4

1. Is it a perfect square--can it be written in the form y=(x-a)²
2. Quick check for this is to ask: Is (B/2)²=C
3. If no then go to the next method 2.
4. If yes, then y=(x-√c)²
5. Can it be easily factored?
6. Ask the question: are there 2 numbers (a and b) which multiply to get C and add to get B.
7. If no go to method C
8. If yes then y=(x-a)(x-b)
9. Complete the Square
10. Rewrite the equation y=[ x²+Bx+_____] +C. Since A=1 we do not have to write it.
11. you are just rewriting the equation leaving a space to complete the square.
12. In the space you add (B/2)² and then subtract (B/2)² at the end.
13. y= [x-(B/2)]² + [C-(B/2) this gives you the Vertex Form, which allows you to easily graph and find the roots.
14. Quadratic Formula which is a packaged up version of completing the square. this gives you the roots.

This is not a complete lesson on solving Quadratics as there are other methods and techniques that can be used. Generally using the Quadratic Formula is always my last resort.

Answer 3

Factoring is basically a shortcut -- if you have a sufficiently "nice" quadratic expression, you can find the solution(s) much more quickly and easily than with the quadratic formula. But, if you can't figure out a way to factor the expression, you can always use the quadratic formula to get the answer instead (it's always guaranteed to give you the right answer as long as you plug everything in correctly).

Take the equation x² + 5x + 6 = 0 for example. To factor the expression on the left, you want to try to think of two numbers that add up to 5 (the coefficient for the x term) and multiply to 6 (the constant term). The two numbers you're looking for turn out to be 2 and 3 (2 + 3 = 5, and 2*3 = 6). So, the equation becomes (x + 2)(x + 3) = 0, meaning the solution are x = -2 and x = -3. Way simpler than using the quadratic formula.

But now let's say you're asked to solve x² - x - 1 = 0. You need to think of two numbers that add up to -1 (the coefficient for the x term) and also multiply to -1 (the constant term). That's actually pretty hard. In this situation, when you can't come up with a way to factor the quadratic expression, you have to resort to either the quadratic formula or completing the square. If you plug everything into the quadratic formula, you get x = x = 1/2 ± √5/2 as the solution.

Answer 4

The question "what is the difference between solving a quadratic by factoring and using the quadratic formula?" is similar to asking about the difference between getting into your car by opening the car door and stepping in, versus climbing through the car's window to get in. Both ways would get in to the car, one is more athletic than the other.

If you could easily spot a factoring solution, then you need not climb through the window.

If the coefficients of x's are helping, the quadratic formula would do the trick of finding the imaginary terms.

Example: 7 x² + 2 x - 5 = 0.

Here, you could easily spot the factors: ( 7 x -5 ) and (x + 1), that give the roots x = -1 and 5 / 7.

If you opt to use the quadratic formula, then you could burn more calories, do more work, as follows:

x = ( - b ± sqrt( b² - 4 a c ) ) / 2 a

With

a = 7,

b = 2, and

c = -5

We get

x = ( - b ± sqrt( b2 - 4 a c ) ) / 2 a = ( - 2 ± sqrt( 2² - 4 (7)(-5 ) ) / (2 *7)

= ( - 2 ± sqrt( 4 + 140 ) ) / 14

= ( - 2 ± 12 ) / 14

Which gives, x = 1 and -5/7

Answer 5

Both factoring and the quadratic formula are just two different ways to solve a quadratic equation, and each has its own strengths. Factoring is usually quicker if the numbers work out nicely, like when the equation breaks down cleanly into two binomials. For example, if you have something like x² + 5x + 6 = 0, you can factor it easily because you're looking for two numbers that multiply to 6 and add to 5. In that case, 2 and 3 work perfectly, so it factors into (x+2)(x+3) = 0.

But now take something like x² - 2x - 7 = 0. To factor this, you'd need two numbers that multiply to -7 and add to -2. When you try the factor pairs of -7, which are -7 and 1 or 7 and -1, neither of those add up to -2. That means it doesn’t factor nicely using integers. So instead of wasting time trying to factor something that doesn’t work, you go straight to the quadratic formula, which always works no matter what the equation looks like.

Answer 6

Hello, thank you for taking the time to post your question!

It just depends on the type of quadratic that you are dealing with! The quadratic formula works for all functions that begin with x^2, so it’s most useful when you have roots that are not nice and neat. Factoring is great when you expect the answer to have simple integer values and you see a way to neatly factor it out.

So if you think the equation will have nice roots, you can try to factor. If you think the factoring could get messy or are unsure how to approach factoring, the quadratic formula is a great choice.

I hope that helps get you moving in the right direction! Feel free to reach out if you still have questions beyond that :)

Answer 7

The quadratic formula is derived from a completion of the square on the most general trinomial equation, ax² + bx + c = 0.

Solving by factoring is based on a knowledge of how two binomials are multiplied and the resulting product, i.e., the FOIL method, and a strong knowledge of the multiplication tables.

Answer 8

Factoring is more like guessing the answers based on properties you know the two solutions to have. It is really relying on a general property known as Vieta's formulas, which say that for a monic polynomial, it must factor (over the complex numbers at least) as (x-r1)(x-r2) where r1 and r2 are the roots. Therefore, multiplying this out, we get x²-(r1+r2)x + r1r2. Therefore, if we know that the linear term will be negative the sum of the roots and the constant term will be the product of the two roots. If we find two numbers r1 and r2 that satisfy this, then they must be the roots of the polynomial.

Therefore, if we have a polynomial with "easy" roots, we can often see the solutions to this system more easily than plugging into the quadratic formula. If, for example, we have the polynomial x²-11x + 18 or something like that, we know that the two roots add up to 11 and multiply to 18, so we can just try numbers until we get 2 and 9. However, if we instead had x²-10x + 18, we aren't going to guess that the roots are 5+sqrt(7) and 5-sqrt(7) which still add to 10 and multiply to 18.

Answer 9

Let's say we have the following binomial:

x²−5x+6

This can be easily factored into the following to find solutions:

(x−2)(x−3)

The solutions to this binomial are x = 2 and x = 3.

By using the quadratic formula, we can also find our solutions using our binomials coefficients and constants as inputs.

ax² + bx + c

Quadratic formula: x = -b +/- √(b² - 4ac)

2a

When we plug in our values using the +/-, we get x = 2 and x = 3.

Use factoring when it’s simple and the quadratic is easily factorable. Use the quadratic formula when:

1. The equation doesn't factor nicely
2. You want to be sure you didn’t miss a solution

Both methods are valid—just pick based on the problem's difficulty.