Quadratic equations appear all the time in math, and you need to know how to solve them. The quadratic formula is one method that helps you to solve these equations.

So, when do you use the quadratic formula? **You can use the quadratic formula to find the zeros (x-intercepts) of a parabola. You can also use the quadratic formula to find the roots of a quadratic equation if factoring is difficult. In addition, the quadratic formula is useful in physics to deal with gravity and falling objects.**

Of course, there are certain situations when factoring is easier than using the quadratic formula.

In this article, we’ll talk about when to use the quadratic formula. We’ll also take a look at some examples to illustrate the idea.

Let’s begin.

## When To Use The Quadratic Formula

The quadratic formula is used in several different scenarios in math and physics, including:

**Finding zeros of a parabola**(finding the x-intercepts on the graph of a quadratic).**Finding roots of a quadratic equation**(when it is difficult to factor).**Problems that involve gravity**(tracking the position of falling objects).

The quadratic formula gives the solution for the standard form of the quadratic equation

**ax**(where a is not zero).^{2}+ bx + c = 0

The solutions are given by:

The quadratic formula is a shortcut for the method of completing the square. In fact, we complete the square for the standard form of the quadratic equation in order to derive the quadratic formula.

## Can You Always Use The Quadratic Formula?

You can always use the quadratic formula to solve a quadratic equation. There is no restriction on the values of the coefficients a, b, and c, except that *a cannot be zero* (otherwise, we would not have a quadratic in the first place – we would have a line).

No matter what coefficients you have in the quadratic equation, the quadratic formula will always find a solution. Of course, there are some situations when the solutions will be complex or imaginary.

When you use the quadratic formula, there are three basic cases:

**1.) Two Real Solutions (Two Distinct Real Roots) [Positive Discriminant]****2.) One Real Solution (One Repeated Real Root) [Zero Discriminant]****3.) Two Complex Solutions (Two Distinct Complex Conjugate Roots) [Negative Discriminant]**

Remember that the discriminant is the expression b^{2} – 4ac (which is found under the radical in the numerator of the quadratic formula).

However, the quadratic formula is one tool in your mathematical toolkit. Just because you *can* always use the quadratic formula doesn’t mean you *should* always use it.

Let’s take a look at some scenarios where you might use methods other than the quadratic formula to solve a quadratic equation.

## When To Use The Quadratic Formula vs Factoring

If a quadratic is not easily factorable, then go ahead and use the quadratic formula. If the quadratic is easy to factor, then you might want to skip the quadratic formula and just factor.

The reason is that the quadratic formula involves both a radical and a fraction. This can make your calculations more error-prone, and it can take more time than an easy factor.

Still, this raises the question of when a quadratic is easy to factor. It takes some practice to recognize those cases, but there are ways to tell if a quadratic is factorable.

### How To Tell If A Quadratic Is Factorable

There are lots of tips and tricks to tell when a quadratic is factorable, but here are some of the common ones:

**Difference of Squares****Perfect Square Trinomial****Greatest Common Factor****Sum & Product of Roots**

Let’s take a look at each of these in turn.

#### Difference Of Squares

A difference of squares has the form A^{2} – B^{2}. It factors as (A + B)(A – B), which you can verify with FOIL:

(A + B)(A – B) = A^{2} – AB + AB – B^{2} = A^{2} – B^{2} (The middle terms from FOIL cancel each other out.)

When looking for a difference of squares, the key things to watch out for are:

**A Difference Of Terms**– a difference of squares has exactly two terms: one positive and the other negative.**Perfect Square Coefficients**– look for numbers like 1, 4, 9, 16, 25, 36, etc. as coefficients of the terms.**Even Exponents**– look for numbers like 2, 4, 6, 8, etc. as the exponents of any variables in the terms.

Here are some examples to help drive the point home.

##### Example 1: Difference Of Squares

Given 4x^{2} – y^{2}, we know that we have a difference of squares, because:

**The terms have opposite signs (one positive, one negative)****The coefficient of 1 and 4 are perfect squares (1**^{2}= 1 and 2^{2}= 4)**The exponents on x and y are both even (both are 2).**

Rewriting in terms of squares, we get 4x^{2} – y^{2} = (2x)^{2} – (y)^{2}. This gives us A = 2x and B = y (using the notation from before).

Then our factors are (A + B)(A – B), or (2x + y)(2x – y).

##### Example 2: Difference Of Squares

Given 9x^{2} – 16y^{4}, we know that we have a difference of squares, because:

**The terms have opposite signs (one positive, one negative)****The coefficient of 9 and 16 are perfect squares (3**^{2}= 9 and 4^{2}= 16)**The exponents on x and y are both even (2 for x and 4 for y).**

Rewriting in terms of squares, we get 9x^{2} – 16y^{4}= (3x)^{2} – (4y^{2})^{2}. This gives us A = 3x and B = 4y^{2} (using the notation from before).

Then our factors are (A + B)(A – B), or (3x + 4y^{2})(3x – 4y^{2}).

##### Example 3: Difference Of Squares

Given 25c^{6}x^{2} – d^{10}y^{4}, we know that we have a difference of squares, because:

**The terms have opposite signs (one positive, one negative)****The coefficient of 25 and 1 are perfect squares (5**^{2}= 25 and 1^{2}= 1)**The exponents on c, d, x and y are all even (6 for c, 10 for d, 2 for x, and 4 for y).**

Rewriting in terms of squares, we get 25c^{6}x^{2} – d^{10}y^{4}= (5c^{3}x)^{2} – (d^{5}y^{2})^{2}. This gives us A = 5c^{3}x and B = d^{5}y^{2} (using the notation from before).

Then our factors are (A + B)(A – B), or (5c^{3}x + d^{5}y^{2})( 5c^{3}x – d^{5}y^{2}).

#### Perfect Square Trinomial

A perfect square trinomial has the form A^{2} + 2AB + B^{2}. It factors as (A + B)^{2}, which you can verify with FOIL:

(A + B)^{2} = (A + B)(A + B) = A^{2} + AB + AB + B^{2} = A^{2} + 2AB + B^{2} (The middle terms from FOIL are combined, since they are like terms.)

A slight variation has the form A^{2} – 2AB + B^{2}. It factors as (A – B)^{2}, which you can verify with FOIL:

(A – B)^{2} = (A – B)(A – B) = A^{2} – AB – AB + B^{2} = A^{2} – 2AB + B^{2} (The middle terms from FOIL are combined, since they are like terms.)

When looking for a perfect square trinomial, the key things to watch out for are:

**Three Terms**– a perfect square trinomial has exactly three terms.**Two Perfect Square Terms**– for the two “end” terms, look for perfect squares (like 1, 4, 9, 16, 25, 36, etc.) as coefficients and even numbers (such as 2, 4, 6, 8, etc.) as exponents.**Even Coefficient**– the “middle” term will have an even coefficient.

Here are some examples to help drive the point home.

##### Example 1: Perfect Square Trinomial

Given 4x^{2} + 4xy + y^{2}, we know that we may have a difference of squares, because:

**There are three terms.****The coefficients of 4 and 1 on the “end terms” are perfect squares (2**^{2}= 4 and 1^{2}= 1), and the exponents on the variables are both even (2 for x and 2 for y).**The “middle” term has an even coefficient (4).**

Taking the square root of the “end” terms, we get A = 2x and B = y (using the notation from before).

Then we can factor as (A + B)^{2}, or (2x + y)^{2}. As a final step, we can use FOIL to verify that (2x + y)^{2} is the same as 4x^{2} + 4xy + y^{2}.

##### Example 2: Perfect Square Trinomial

Given 9x^{4} + 30x^{2}y + 25y^{2}, we know that we may a difference of squares, because:

**There are three terms.****The coefficients of 9 and 25 on the “end terms” are perfect squares (3**^{2}= 9 and 5^{2}= 25), and the exponents on the variables are both even (4 for x and 2 for y).**The “middle” term has an even coefficient (30).**

Taking the square root of the “end” terms, we get A = 3x^{2} and B = 5y (using the notation from before).

Then we can factor as (A + B)^{2}, or (3x^{2} + 5y)^{2}. As a final step, we can use FOIL to verify that (3x^{2}+ 5y)^{2} is the same as 9x^{4} + 30x^{2}y + 25y^{2}.

#### Greatest Common Factor

Sometimes, you may need to find a greatest common factor to “reveal” one of the methods mentioned earlier (such as difference of squares or perfect square trinomial).

Remember that if all of the terms in an expression have a common factor, then we can factor it out. For example, we can factor 15x^{2} – 5y as 5(3x^{2} – y).

Let’s see an example to show how it works.

##### Example 1: Greatest Common Factor

Given 27x^{2} – 48y^{4}, we know that we can factor out a 3, since both 27 and 48 are divisible by 3. This gives us:

27x^{2} – 48y^{4} = 3(9x^{2} – 16y^{2})

The part in parentheses should look familiar, since it is the difference of squares that we factored earlier in this post!

The complete factorization is:

3(3x + 4y^{2})(3x – 4y^{2}).

#### Sum & Product Of Roots

Sometimes, we may have to do a little more work to recognize a quadratic that is easy to factor. It all comes down to the sum and product of the roots.

Let’s say we want to factor the quadratic x^{2} – 5x + 6 = 0. There are two things we need from the solutions R and S:

**The roots R and S must add up to -5**(-5 is the coefficient of x in the quadratic equation)**.****The roots R and S must have a product of 6 when we multiply them together**(6 is the constant term in the quadratic equation).

The numbers -2 and -3 come to mind, since -2 + -3 = -5 and (-2)*(-3) = 6. This means that we can factor as:

x^{2} – 5x + 6 = (x – 2)(x – 3)

which you can verify by using FOIL and combining like terms.

This means that the solutions to this quadratic equation are x = 2 and x = 3.

### When To Use The Quadratic Formula vs Completing The Square

As mentioned earlier, the quadratic formula is a shortcut for completing the square. However, there is a certain case when you might need to know how to complete the square.

Remember: to complete the square, take the coefficient of the x term, divide it by 2, square the result, and then add the result to both sides of the equation.

Let’s say you have the equation x^{2} + y^{2} – 2x + 4y = 11 and you want to graph it. Even if I told you that it is a circle, it might not be easy to find out the center and radius.

However, completing the square (twice!) may be able to help us. Rearranging the terms so that variables are grouped together, we get:

x^{2} – 2x + y^{2} + 4y = 11

Completing the square for x (divide -2 by 2 to get -1, then square -1 to get 1):

(x^{2} – 2x + 1) + y^{2} + 4y = 11 + 1 [add 1 to both sides of the equation]

Completing the square for y (divide 4 by 2 to get 2, then square 2 to get 4):

(x^{2} – 2x + 1) + (y^{2} + 4y + 4) = 11 + 1 + 4 [add 4 to both sides of the equation]

Simplifying by using perfect square trinomials on the left and adding like terms on the right:

(x – 1)^{2} + (y + 2)^{2} = 16

Remember that the equation for a circle is given as

(x – a)^{2} + (y – b)^{2} = R^{2}

Where (a, b) is the center of the circle and R is its radius.

Rewriting our equation in this form gives us:

(x – 1)^{2} + (y – (-2))^{2} = 4^{2}

So, this circle has a center at (1, -2) and a radius of 4.

### When To Use The Quadratic Formula In Physics

The quadratic formula is used in physics when dealing with gravity and falling objects. Remember that the acceleration due to gravity is -9.8 meters per second (or -32 feet per second).

If we know the starting speed and position of the object, we can use some calculus to come up with a quadratic equation that would model the object’s position at time t.

For example, if you drop a ball from a height of 800 feet in a vacuum, its initial position is 800 (the starting height of the ball) and its initial velocity is zero (since the ball was dropped from rest, not thrown downwards).

This gives us a position (height) equation of y = -16t^{2} + 800.

If we were to solve this quadratic equation with the quadratic formula, we would get a time of about 7.071 seconds as the (repeated real) root of the equation.

After t = 1 second, the ball is at a height of 784 feet.

After t = 2 seconds, the ball is at a height of 736 feet.

After t = 3 seconds, the ball is at a height of 656 feet.

After t = 4 seconds, the ball is at a height of 544 feet.

After t = 5 seconds, the ball is at a height of 400 feet.

After t = 6 seconds, the ball is at a height of 224 feet.

After t = 7 seconds, the ball is at a height of 16 feet.

After t = 7.071 seconds, the ball is at a height of 0 feet (it hits the ground).

## Conclusion

Now you know when to use the quadratic formula – and when not to!

You might want to read my article on whether parabolas are one-to-one functions or my article on quadratic functions.

You might also be interested to read my article about how to factor a quadratic binomial (you can use a special case of the quadratic formula).

This article explains what the solutions of a quadratic formula represent.

I hope you found this article helpful. If so, please share it with someone who can use the information.

Don’t forget to subscribe to my YouTube channel & get updates on new math videos!

~Jonathon