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## Algebra (all content)

### Unit 9: Lesson 7

The quadratic formula- The quadratic formula
- Understanding the quadratic formula
- Using the quadratic formula
- Worked example: quadratic formula
- Worked example: quadratic formula (example 2)
- Worked example: quadratic formula (negative coefficients)
- Quadratic formula
- Using the quadratic formula: number of solutions
- Number of solutions of quadratic equations
- Proof of the quadratic formula
- Quadratic formula review
- Discriminant review
- Quadratic formula proof review

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# Quadratic formula review

CCSS.Math: ,

The quadratic formula allows us to solve any quadratic equation that's in the form ax^2 + bx + c = 0. This article reviews how to apply the formula.

## What is the quadratic formula?

The quadratic formula says that

for any quadratic equation like:

### Example

We're given an equation and asked to solve for q:

This equation is already in the form a, x, squared, plus, b, x, plus, c, equals, 0, so we can apply the quadratic formula where a, equals, minus, 7, comma, b, equals, 2, comma, c, equals, 9:

Let's check both solutions to be sure it worked:

q, equals, minus, 1 | q, equals, start fraction, 9, divided by, 7, end fraction |
---|---|

$\begin{aligned}0&=-7q^2+2q+9\\\\0&=-7(-1)^2+2(-1)+9 \\\\0&=-7(1)-2+9 \\\\0&=-7-2+9\\\\0&=0\end{aligned}$ | $\begin{aligned}0&=-7q^2+2q+9\\\\0&=-7\left(\dfrac{9}{7}\right)^2+2\left (\dfrac{9}{7}\right)+9 \\\\0&=-7\left(\dfrac{81}{49}\right)+\left (\dfrac{18}{7}\right)+9 \\\\0&=-\left(\dfrac{81}{7}\right)+\left (\dfrac{18}{7}\right)+9 \\\\0&=-\left(\dfrac{63}{7}\right) +9 \\\\0&=-9 +9 \\\\0&=0\end{aligned}$ |

Yep, both solutions check out.

*Want to learn more about the quadratic formula? Check out this video.*

*Want more practice? Check out this exercise*.

## Want to join the conversation?

- what if the equation doesn't equal zero(0 votes)
- then just subtract the non-zero number from the RHS to the LHS and make the RHS equal to zero.(10 votes)

- can you reccomend other math websites for algebra 1 and 2(3 votes)
- can you reccomend other math websites for algebra 1 and 2(1 vote)
- IXL is a good one, but without a login given to you by an administrator, you have to pay a membership fee.(2 votes)

- How do you get from (-4 +-2 sqrt of 34)/-6

to (-2+- sqrt of 34)/-3 ?

I get that it’s divided by 2 but it’s strange that the square root of 34 doesn’t change.(1 vote)- You can only remove the factor of 2 once from each term. There are two terms: -4 and +/- 2 sqrt(34)

-4/2 = -2

+/- 2 sqrt(34)/2 = +/- sqrt(34)

Also, the 2 contained in the sqrt(34) is sqrt(2) which can't be divided by 2. They are not equal values.

Hope this helps.(2 votes)

- A garden in the shape of a rectangle is surrounded by a walkway of uniform width. The dimensions of the garden only are 35 by 24. The area of the garden and the walkway together is 1,530 square feet. What is the width of the walkway in feet? a. 4 ft b. 5 ft c. 34.5 ft(1 vote)
- Assume the length of the walkway to be
**x**. If you illustrate the data as a drawing on a piece of paper, you will see that the sides of the garden + walkway are (35+2x) and (24 + 2x). Since AREA = LENGTH x WIDTH, we get to the equation:`(35 + 2x) (24 + 2x) = 1530`

**By expanding the L.H.S, we get**`840 + 70x + 48x + 4x^2 = 1530`

**Now after rearranging**`4x^2 + 118x + 840 - 1530 = 0 -----> 4x^2 + 118x - 690 = 0`

**After applying quadratic formula (a = 4, b = 118, c = -690) we get****x = 5 ft**OR**x = -34.5 ft**

Since the length can NEVER be negative, we can exclude x = -34.5 ft. *SO THE ANSWER IS*b.5ft**(1 vote)

- I learned so many different methods to solve for x but can I use the quadratic formula at all times? or does it not work with every situation(0 votes)
- It works for all quadratics (Ax^2 + Bx + C = 0). It won't work for linear equations, or 3rd degree equations, etc. However, I suggest you have this conversation with your teacher. Chances are very good that you need to know the other methods for solving quadratics as well.

Hope this helps.(3 votes)

- are there any shortcuts or patterns we can use to make calculation quicker?(1 vote)
- I do not understand this. Can someone please explain(1 vote)
- This is a formula, so if you can get the right numbers, you plug them into the formula and calculate the answer(s). We always have to start with a quadratic in standard form: ax^2+bx+c=0. Making one up, 3x^2+2x-5=0, we see a=3, b=2, c=-5. I teach my students to start with the discriminant, b^2-4ac. Also, especially in the beginning, put the b value in parentheses so that you square a negative number if b is negative. In our example, this gives (2)^2-4(3)(-5) = 4+60=64. If I take √64 = 8. Filling out the formula, we get x=(-2±8)/(2(3)) or breaking it into the two parts x=(-2+8)/6=1 and (-2-8)/6=-10/6=-5/2. Where is the confusion? It is always hard to answer when we cannot figure out what you do understand and where you are confused.(1 vote)

- What if there was a negative number under the square root? What would you do then?(1 vote)
- If there is a negative number inside the radical, then the roots will be complex numbers. See this link: https://www.khanacademy.org/math/algebra2/polynomial-functions/quadratic-equations-with-complex-numbers/v/complex-roots-from-the-quadratic-formula

If you haven't learned about "i", the imaginary number, then you should start at this link: https://www.khanacademy.org/math/algebra2/introduction-to-complex-numbers-algebra-2/the-imaginary-numbers-algebra-2/v/introduction-to-i-and-imaginary-numbers(1 vote)

- How do you know what should be under the square root?(1 vote)