Sal, How does the quadratic formula relate to business and economics?

It helps in lots of ways. It can possibly predict the future path of certain things, especially if your graph is exponential.

what if the equation doesn't equal zero

then just subtract the non-zero number from the RHS to the LHS and make the RHS equal to zero.

What happens when the discriminant is a negative number? If it is negative would your answer be imaginary?

Yes... If the discriminant is negative, then there are 2 roots, but they are complex numbers.

What happens when a discriminant is a negative number? If it is negative would your answer be imaginary?

Yes. If the disciminant is negative, then the quadratic solutions are 2 complex numbers.

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?

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.

if the quadratic equation that we are given can be factored, do we factor it down first before plugging it into the quadratic formula?

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.

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!

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.

Main content

Algebra 1

Unit 14: Lesson 6

The quadratic formula

Quadratic formula review

CCSS.Math:

HSA.REI.B.4

HSA.REI.B.4b

Google Classroom

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

x, equals, start fraction, minus, start color #e07d10, b, end color #e07d10, plus minus, square root of, start color #e07d10, b, end color #e07d10, squared, minus, 4, start color #7854ab, a, end color #7854ab, start color #e84d39, c, end color #e84d39, end square root, divided by, 2, start color #7854ab, a, end color #7854ab, end fraction

for any quadratic equation like:

start color #7854ab, a, end color #7854ab, x, squared, plus, start color #e07d10, b, end color #e07d10, x, plus, start color #e84d39, c, end color #e84d39, equals, 0

Example

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

0, equals, minus, 7, q, squared, plus, 2, q, plus, 9

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:

\begin{aligned} q &= \dfrac{-b \pm \sqrt{b^2 - 4ac}}{2a} \\\\ q &= \dfrac{-2 \pm \sqrt{2^{2} - 4 (-7) (9)}}{2(-7)} \\\\ q &= \dfrac{-2 \pm \sqrt{4 +252}}{-14} \\\\ q &= \dfrac{-2 \pm \sqrt{256}}{-14} \\\\ q &= \dfrac{-2 \pm 16}{-14} \\\\ q &= \dfrac{-2 + 16}{-14} ~~,~~ q = \dfrac{-2 - 16}{-14} \\\\ q &= -1 ~~~~~~~~~~~~,~~ q = \dfrac{9}{7} \end{aligned}

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.

Practice

Solve for x.

minus, 4, plus, x, plus, 7, x, squared, equals, 0

Choose 1 answer:

(Choice A)
x, equals, start fraction, minus, 7, plus minus, square root of, 65, end square root, divided by, 8, end fraction
(Choice B)
x, equals, start fraction, minus, 1, plus minus, square root of, 5, end square root, divided by, 2, end fraction
(Choice C)
x, equals, start fraction, minus, 1, plus minus, square root of, 113, end square root, divided by, 14, end fraction
(Choice D)
x, equals, start fraction, 3, plus minus, square root of, 21, end square root, divided by, 6, end fraction

Want more practice? Check out this exercise.

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Charlie Williams
Posted 3 years ago. Direct link to Charlie Williams's post “Sal, How does the quadrat...”
Sal, How does the quadratic formula relate to business and economics?
(7 votes)
Answer
- Victor
  Posted 3 years ago. Direct link to Victor's post “It helps in lots of ways....”
  It helps in lots of ways. It can possibly predict the future path of certain things, especially if your graph is exponential.
  (16 votes)
kaycoach03
Posted 3 years ago. Direct link to kaycoach03's post “what if the equation does...”
what if the equation doesn't equal zero
(6 votes)
Answer
- amy
  Posted 3 years ago. Direct link to amy's post “then just subtract the no...”
  then just subtract the non-zero number from the RHS to the LHS and make the RHS equal to zero.
  (11 votes)
joeygold2016
Posted 3 years ago. Direct link to joeygold2016's post “Can u ask Sal to watch an...”
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)
Answer
Tiggertast
Posted 3 years ago. Direct link to Tiggertast's post “can you reccomend other m...”
can you reccomend other math websites for algebra 1 and 2
(4 votes)
Answer
Drake
Posted 4 years ago. Direct link to Drake's post “What happens when the dis...”
What happens when the discriminant is a negative number? If it is negative would your answer be imaginary?
(2 votes)
Answer
- Kim Seidel
  Posted 4 years ago. Direct link to Kim Seidel's post “Yes... If the discriminan...”
  Yes... If the discriminant is negative, then there are 2 roots, but they are complex numbers.
  (4 votes)
Palomi Mangesh Kurade
Posted 6 years ago. Direct link to Palomi Mangesh Kurade's post “If a quadratic equation a...”
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)
Answer
- Kim Seidel
  Posted 6 years ago. Direct link to Kim Seidel's post “First, in order for an eq...”
  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)
yaneiri-reyes-gonz
Posted 8 months ago. Direct link to yaneiri-reyes-gonz's post “What happens when a discr...”
What happens when a discriminant is a negative number? If it is negative would your answer be imaginary?
(2 votes)
Answer
- Kim Seidel
  Posted 8 months ago. Direct link to Kim Seidel's post “Yes. If the disciminant ...”
  Yes. If the disciminant is negative, then the quadratic solutions are 2 complex numbers.
  (2 votes)
sheilla Muligande
Posted 3 years ago. Direct link to sheilla Muligande's post “if the quadratic equation...”
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)
Answer
- David Severin
  Posted 3 years ago. Direct link to David Severin's post “If you can factor it, you...”
  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)
Mark
Posted 4 months ago. Direct link to Mark's post “I don't understand how to...”
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)
Answer
- David Severin
  Posted 4 months ago. Direct link to David Severin's post “There are several advanta...”
  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)
SerafinaFifi
Posted 3 years ago. Direct link to SerafinaFifi's post “How do you find the verte...”
How do you find the vertex using the quadratic formula?
(2 votes)
Answer