Algebra 1 (Eureka Math/EngageNY)
- 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
Sal solves the equation -7q^2+2q+9=0 by using the quadratic formula. Created by Sal Khan and Monterey Institute for Technology and Education.
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- Where do the constants come from and what do they do?(37 votes)
- the constants come from the "standard form" of a quadratic equation.
which is ax^2 + bx + c = 0
example: 2x^2 + 4x + 6 = 0 (so a = 2, b = 4, c = 6)
so the constants are A B and C. and if you look at a quadratic equation presented to you in standard form, you can see what A B and C are.
if youre still not sure, look at my equations again and look at a few quadratic equations and it should start to make sense
ax^2 + bx + c = 0
2x^2 + 4x + 6 = 0 (a=2,b=4,c=6)
3x^2 + 6x + 9 = 0 (a=3,b=6,c=9)
and remember if it just looks like x^2 + 4x + 8 = 0 (meaning there is nothing before the x^2 so it looks like A doesnt exist, it is there in the form of 1)
x^2 is the same as (1)(x^2) so in that case, A = 1
hope that helps!(37 votes)
- what if a is an irrational number? My teacher gave us a problem that like that and when i asked of that was possible she said yes. Is it?(13 votes)
- It sure is. Irrational numbers are just numbers, and they can be algebraically manipulated just like any others. Let's say we have the equation √5 x^2 + √3 x - π = 0. Is that a valid quadratic equation? Yes, it is. We have a = √5, b = √3, and c = -π. Plugging those into the quadratic formula gives x = [-√3 ± √((√3)^2 + 4π√5)] / 2√5. That doesn't actually simplify very much (although (√3)^2 simplifies to 3, of course), but if you use your calculator you'll see that the roots are approximately x = 0.8597 and x = -1.6343.(19 votes)
- Why does it have to be plus or minus?(9 votes)
- We used the quadratic equation because we are solving a quadratic expression. What is that? It is an expression that has a variable raised to the power 2 AND that the power 2 is the largest power in the expression: eg x²+2x+1 is a quadratic expression, but 2x+1 is not (no x²), nor is x³+x²+2x+1 (the power 3 in x³ is too big.)
OK with me so far?
Here is a simple example:
y=x² is a quadratic expression since it has a x² term and there is no other term with a greater exponent.
Now suppose I wrote 9=x². What values of x satisfy the equation?
I am sure you know that 3x3=9, so x=3 is a solution, right?
But what about -3? (-3)(-3) also equals 9, (-3)(-3)=9, so x=-3 as well. So this quadratic equation has TWO solutions.
We can write these two solutions as x=±2, which means x=2 and x=-2.
So you can see we have to use plus or minus, because when you solve a quadratic equation, very often there are two solutions, those being some value, lets call it 'a', and its negative, that is '-a'.
Keep Studying and Keep Asking Questions!(15 votes)
- when doing the quadractic formula would you always need a -b(7 votes)
- if there is nothing in front of the x (coefficient) then you take that to mean 1.
for example, x^2-x+44 is exactly the same as 1x^2-1x+44.(9 votes)
- When you have a quadratic equation, (i.e., 2x^2-12x-14) how would you simplify with the two negatives? I know you'd have to chose two numbers that multiply to -14 but add to -12 but how does that work? You could only get -14 and 12 instead of -12. Help!(5 votes)
For 2x^2-12x-14 your first factor out a 2
2(x^2 - 6x-7) Then you need numbers that add to -6 and multiply to -7
I hope that helps make it click for you.(6 votes)
- What if the radical number isn't prime? Like 16 went perfectly into 256 but what if there's a problem in which its not like that(5 votes)
- Good question, the answer to a problem like this is usually written as the square root of the number. For example, if we had an equation like x^2 = 57, where the square root of 57 would be the value for x, then it's normally written √57. If your teacher wants you to use a number and not use a calculator to find the value, then Sal made a video about how to do this here:
- 2x^2 do you times it or leave it(5 votes)
- Since the 2x are left out of the parenthesis, it's just 2(x^2).(4 votes)
- This topic is generally featured in an algebra 1 course, correct?(4 votes)
- I would think both. I used them in Algebra 1 and 2. It should be introduced, however, during Algebra 1 and used through your math career. Even Calculus.(2 votes)
- Isn't -2-16 equal -18 not 18?(1 vote)
- Yes, -2-16 = -18
But notice, Sal has
(-2-16)/(-14) = 18/14because he cancelled out the minus in the -18 with the minus in the -14.
Hope this helps.(4 votes)
Use the quadratic formula to solve the equation, 0 is equal to negative 7q squared plus 2q plus 9. Now, the quadratic formula, it applies to any quadratic equation of the form-- we could put the 0 on the left hand side. 0 is equal to ax squared plus bx plus c. And we generally deal with x's, in this problem we're dealing with q's. But the quadratic formula says, look, if you have a quadratic equation of this form, that the solutions of this equation are going to be x is going to be equal to negative b plus or minus the square root of b squared minus 4ac-- all of that over 2a. And this is actually two solutions here, because there's one solution where you take the positive square root and there's another solution where you take the negative root. So it gives you both roots of this. So if we look at the quadratic equation that we need to solve here, we can just pattern match. We're dealing with q's, not x's, but this is the same general idea. It could be x's if you like. And if we look at it, negative 7 corresponds to a. That is our a. It's the coefficient on the second degree term. 2 corresponds to b. It is the coefficient on the first degree term. And then 9 corresponds to c. It's the constant. So, let's just apply the quadratic formula. The quadratic formula will tell us that the solutions-- the q's that satisfy this equation-- q will be equal to negative b. b is 2. Plus or minus the square root of b squared, of 2 squared, minus 4 times a times negative 7 times c, which is 9. And all of that over 2a. All of that over 2 times a, which is once again negative 7. And then we just have to evaluate this. So this is going to be equal to negative 2 plus or minus the square root of-- let's see, 2 squared is 4-- and then if we just take this part right here, if we just take the negative 4 times negative 7 times 9, this negative and that negative is going to cancel out. So it's just going to become a positive number. And 4 times 7 times 9. 4 times 9 is 36. 36 times 7. Let's do it up here. 36 times 7. 7 times 6 is 42. 7 times 3, or 3 times 7 is 21. Plus 4 is 25. 252. So this becomes 4 plus 252. Remember, you have a negative 7 and you have a minus out front. Those cancel out, that's why we have a positive 252 for that part right there. And then our denominator, 2 times negative 7 is negative 14. Now what does this equal? Well, we have this is equal to negative 2 plus or minus the square root of-- what's 4 plus 252? It's just 256. All of that over negative 14. And what's 256? What's the square root of 256? It's 16. You can try it out for yourself. This is 16 times 16. So the square root of 256 is 16. So we can rewrite this whole thing as being equal to negative 2 plus 16 over negative 14. Or negative 2 minus-- right? This is plus 16 over negative 14. Or minus 16 over negative 14. If you think of it as plus or minus, that plus is that plus right there. And if you have that minus, that minus is that minus right there. Now we just have to evaluate these two numbers. Negative 2 plus 16 is 14 divided by negative 14 is negative 1. So q could be equal to negative 1. Or negative 2 minus 16 is negative 18 divided by negative 14 is equal to 18 over 14. The negatives cancel out, which is equal to 9 over 7. So q could be equal to negative 1, or it could be equal to 9 over 7. And you could try these out, substitute these q's back into this original equation, and verify for yourself that they satisfy it. We could even do it with the first one. So if you take q is equal to negative 1. Negative 7 times negative 1 squared-- negative 1 squared is just 1-- so this would be negative 7 times 1, right? That's negative 1 squared. Negative 1 times 2 is minus 2 plus 9. So it's negative 7 minus 2, which is negative 9, plus 9, does indeed equal 0. So this checks out. And I'll leave it up to you to verify that 9 over 7 also works out.