Main content
Algebra 1
Unit 14: Lesson 6
The quadratic formula- The quadratic formula
- Understanding the 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
- Quadratic formula review
- Discriminant 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 |
---|---|
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?
- Sal, How does the quadratic formula relate to business and economics?(7 votes)
- It helps in lots of ways. It can possibly predict the future path of certain things, especially if your graph is exponential.(16 votes)
- what if the equation doesn't equal zero(6 votes)
- then just subtract the non-zero number from the RHS to the LHS and make the RHS equal to zero.(11 votes)
- Can u ask Sal to watch an education rap, it's called:
Bring It Back
by Balistik (ZT)
There's a cool little bar about education= the key to success(8 votes) - can you reccomend other math websites for algebra 1 and 2(4 votes)
- What happens when the discriminant is a negative number? If it is negative would your answer be imaginary?(2 votes)
- Yes... If the discriminant is negative, then there are 2 roots, but they are complex numbers.(4 votes)
- If a quadratic equation ax^2+bx+c=0 has more than two roots, then it becomes an identity a=b=c=0.
Can someone please prove this above statement to me?
- How did the identity part come about? How can a=b=c=0?(1 vote)- First, in order for an equation to be a quadratic, the "a" (in ax^2+bx+c = 0) can't = 0
As soon as "a = 0", you no longer have a 2nd degree equation. And, your graph will not be a parabola.
If a=b=c=0, then the left side of the equation becomes 0. Do the substitution and you'll see.
Your equation becomes 0=0. This is an equation that is always true, and is called an identity. But, again, it is no longer a quadratic. "a" will never = 0 in a quadratic equation. Quadratics only have 0, 1, or 2 real roots. If they have 0 real roots, then they will have 2 complex roots.(4 votes)
- What happens when a discriminant is a negative number? If it is negative would your answer be imaginary?(2 votes)
- Yes. If the disciminant is negative, then the quadratic solutions are 2 complex numbers.(2 votes)
- if the quadratic equation that we are given can be factored, do we factor it down first before plugging it into the quadratic formula?(1 vote)
- If you can factor it, you do not need to use the quadratic formula. If you do both, they had better match or you made a mistake.(3 votes)
- I don't understand how to do this out of solving by graphing, solving by factoring, solving by completing the square, and simplifying square roots, this should be the second easiest behind graphing! Yet somehow, no matter how many times I watch the three videos and read the two articles I never get any of the questions in the Quadratic Formula practice correct, 0/4,0/4,0/4,0/4,0/4, I know I have a bad work ethic but I don't know why I should continue trying to learn this! It's just too hard!(1 vote)
- There are several advantages of the quadratic formula including knowing how many roots it has as well as finding answers with irrational numbers (cannot completely simplify the square root). While completing the square also allows you to do this, graphing and factoring cannot accomplish this task. If you want to present one of your attempts step by step, maybe someone can find what you are doing wrong. Several regular issues come up if b is negative. The beginning of the formula, -b, would change the formula to -(-b) or positive b, and if you are squaring this, you have to make sure you are doing (b)^2 inside parentheses rather than b^2 which also messes with apporpriate signs.
If we have 2x^2 - 4x +5, a=2, b=-4, and c=5. (-b+/- sqrt(b^2-4ac)/(2a) is best done with inside the square root first, so (-4)^2-4(2)(5) = 16-40 which tells us that there are no real solutions. Change c = -5, we end up with (-4)^2-4(2)(-5)=16+40=56. We then have to see if we can simplify sqrt(56). If you try -4^2-4(2)(-5) you get a wrong answer.(2 votes)
- How do you find the vertex using the quadratic formula?(2 votes)