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Quadratic Equation Solver

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Quadratic Equation Solver

We can help you solve an equation of the form “ax2 + bx + c = 0” Just enter the
values of a, b and c below:

Is it Quadratic?

Only if it can be put in the form ax2 + bx + c = 0, and a is not zero.

The name comes from “quad” meaning square, as the variable is squared (in other words x2).

These are all quadratic equations in disguise:

In disguise
In standard form
a, b and c
x2 = 3x -1
x2 – 3x + 1 = 0
a=1, b=-3, c=1
2(x2 – 2x) = 5
2×2 – 4x – 5 = 0
a=2, b=-4, c=-5
x(x-1) = 3
x2 – x – 3 = 0
a=1, b=-1, c=-3
5 + 1/x – 1/x2 = 0
5×2 + x – 1 = 0
a=5, b=1, c=-1

How Does this Work?

The solution(s) to a quadratic equation can be calculated using the Quadratic Formula:

The “±” means we need to do a plus AND a minus, so there are normally TWO solutions !

The blue part (b2 – 4ac) is called the “discriminant”, because it can “discriminate” between the possible types of answer:

when it is positive, we get two real solutions,
when it is zero we get just ONE solution,
when it is negative we get complex solutions.

Learn more at Quadratic Equations

Note: you can still access the old version here.

Quadratic Equations with No Real Solution Tutorial | Sophia Learning

Quadratic Equations with No Real Solution Tutorial | Sophia Learning – Determine if a quadratic equation has real or non-real solutions by finding the value of the discriminant. So I hope that these key points and examples helped you understand a little bit more about solving quadratic equations with no real solution.Solution: D = 32 – 4 ⋅ 4 = -7 < 0 So the equation has 0 real roots. Problem 2. What is the value of the greater root of the equation [tex]x^2-5x+4=0 Solve the quadratic equation [tex]x^2+14x+45=0[/tex] In the answer box, write the roots separated by a comma. Solution: The discriminant is [tex]D=14^2-4…Algebra Quadratic Equations and Functions Solutions Using the Discriminant. The discriminant is the portion #(b^2-4ac)# if it is negative then the roots include a term which is the square root of a negative number; since there are no Real numbers whose square root is negative, when the…

Quadratic Equations: Problems with Solutions – Reviews how to solve quadratics by using the Quadratic Formula. But the Quadratic Formula will always spit out an answer, whether or not the quadratic expression was factorable. Let's try that first problem from the previous page again, but this time we'll use the Quadratic Formula instead of the…Applications of Quadratic Equations. Related Topics: More Lessons for Basic Algebra Math Worksheets. Scroll down the page for more examples and solutions for quadratic equations in vertex form. The Vertex and Axis of Symmetry In a parabola, the vertex is the highest or lowest point…Solve quadratic equations by factorising, using formulae and completing the square. Each method also provides information about the corresponding quadratic graph. If the product of two numbers is zero then one or both of the numbers must also be equal to zero.

Quadratic Equations: Problems with Solutions

How do you use the discriminant to find the number of real solutions… – Quadratic Equation Solver. We can help you solve an equation of the form "ax2 + bx + c = 0" Just enter the values of a, b and c below How Does this Work? The solution(s) to a quadratic equation can be calculated using the Quadratic Formula: The "±" means we need to do a plus AND a minus…Solution: To check whether the equation has a solution or not, quadratic formula for discriminant is used. Check whether one root of the Quadratic Equation is twice of other or not.To solve first move the 98 to the other side by subtracting it from both sides. Then divide by – 1 on both sides to that the x^2 is no longer a negative. You are left with: Now square root both sides to get. The two factors that make up 98 are 49 and 2. 49 can be square rooted.

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Quadratic Applications – Vertical Motion (Algebraically) – .

How to find the foci, center and vertices, and asymptotes of a hyperbola – So ladies and gentlemen,
we have a problem, or we have an equation.
We want to write it
into are standard form. And then, we want to
plot the asymptotes and make a little
sketch of the graph. How does that sound? That's awesome. Terrible. Yay. OK, so the first
thing is again, if you know our definition
of our parabolas, we need to first
before we can even determine if it's going to
be horizontal or vertical, we need to at least be able
to see what the graph is going to look like. So the first thing
I'm going to do is I'm going to
combine my 4x squared plus 8x minus 3y squared
equals a negative 16. So I put the constant on
the other side and I arrange my x's and my y's. And I do that because
now I want to make sure that I can combine
them, because actually, I forgot to go over this. Let's go over the
formulas right that we're trying to figure out. This one is y minus k. OK, so we have our two formulas. What we need to do
is to make sure we can write this in a formula. So the first I
need to do is now, I'm going to have to
complete the square. Before I can complete
the square of the x, I need to make sure
I factor out the 4. I can't complete this
square with my negative 3. But that's fine. That just means that my k
is going to be zero, right? Now, I complete the
square over here. So I have 8. I've got a background
of 4, so that's a 2. So I have 2 divided by 2
squared, which equals 1. So I have 4 times x squared
plus 2x plus 1 minus 3y squared equals negative 16. However ladies and gentlemen,
notice that I'm not adding a 1. I'm adding a 1 that's
being multiplied by 4. So therefore, I'm going to have
to add a 4 over here as well. So therefore now, I
simplify this back down to the binomial square. So it's x plus 1 squared minus
3y squared equals negative 4. Now again, remember, both
of our formulas equal 1. Right? Both those formulas equal 1. So let's divide by
negative 12 and we'll do this by both terms. Therefore, by simplifying this,
I now have a negative 3 times x plus 1 squared plus
y squared over 4. I'm sorry, that's a 3
on the bottom right? Whoa, so you made the top
negative, not the bottom? It doesn't matter. Negative 4 divided
by 2 is the same as– Why are you adding? I thought it was multiplying. Because– The negative divided
by another negative makes it positive, right? True. So now, you're right though. This needs to be a
subtraction problem, right? So rather than saying a
negative number plus a positive, can rewrite this is as
y squared over 4 minus x plus 1 squared over 3 equals 1. So you're right. Right? You want to write in that
format because now you have it as a subtraction, you know
now that a squared equals 4 and b squared equals 3. Right? How do you though? Because it's always
a squared minus b squared for a hyperbola,
a squared minus b squared. Huh? a is always the one. No, not necessarily. Oh god. It's just always in the formula
a squared minus b squared. That's the formula
that we're always going to have in this form. So in hyperbolas it's on no
matter whether it's big or not? You're not always going to
have a larger or smaller. But I thought you
said that asymptote– On your ellipses, that's
the way that you're always going to have it. OK? So how do you know
what [INAUDIBLE] You just write it
add or subtraction, one minus the other. Then whatever that is,
whatever you subtracted from, that's your a minus your b. So it could be negative? It could be 4 minus 5. Yeah. That was our last
example, right? So now, let's going and take
a look at our h and our k. So if we know our h now
represents our– so let's do our center. So now I can say the center
is going to be negative 1,0. Yes? Yes, no, maybe so? OK. So ladies and gentlemen,
if we have our 4 and our y, if our a squared is
under our y, now that means we're going to have
a vertical one, right? Vertical. So we know our center
is at negative 1,0. The vertices are going
to be plus or minus 4. So therefore, if a
squared equals 4, we know that a equals 2. So to find the vertices,
ladies and gentlemen, are your vertices– Please pardon the
interruption, please turn your TV sets to channel
6 for the Mustang news. Since this has a
vertical transverse axis, our vertices are gong to be on
the transverse axis as well. So our vertices are going to
be negative 1, plus or minus 2, which will be negative 1,
negative 1 and negative 1, 2. Wait, did we ever
get formulas to try to figure out the vertices? Or how do you
figure out vertices? It's the same distance. It's a. So it's the same as an ellipse. It's just a definition of a. The only thing you
need to know is it's on the same as
your transverse axis. Remember, ellipses? They were all on the
same major access. Well, we're on the same what
we call a transverse axis. So the h is not
changing in this case because you have a
vertical transverse axis. The same thing now
goes with our foci. Now, we need to determine
what the foci is. And remember, the relationship
we have for our hyperbolas– obviously, I think that was a
bad reason for you to come back in class at this point in time– Because remember the definition
for the foci that we had was c squared equals a
squared plus b squared. Well, we have a squared is
4, plus b squared which is 3. Right? b is 3c squared
though not c, right? I'm sorry? Do you only need
c squared plus c? No, we actually need to
figure out what c is. Because c is the difference,
c is the distance from your center to your foci. So c is the square root of 7. So our foci are going to be
negative 1, plus or minus the square root of 7,
which is negative 1, 0. Sorry, it's 0 plus or minus 2, 0
plus or minus square root of 7. So Does that matter? Yeah, why does it? Yeah, of course it matters. That's a huge mistake. Guys, you're taking
your center and adding the vertices to your k, to the
k coordinate of your vertices. This one just happened to
be c, so it didn't matter. Yeah, but why do you
have to put 0 plus? No, you don't have to. And that's why I lazy
and I didn't write it. But you need to
understand you're adding into the k form of your center. Yeah. You just need to make
sure you understand that. In this case, yeah, you're going
to have a square root of 7. And in this case,
you have negative 1 minus the square root of 7. Well, guys you can write
those in your decimal form. The square root of 7
is roughly 2.6457513. My mom is about negative
square root of 7. Oo! Raise your hand. Sorry. I'm sorry, say that again. Are you allowed to put the
negative square root of 7? Yeah, as long as
you're not taking the square root of negative 7. But yeah, leave it
in that exact answer. So your foci are going to be
somewhere right around here. Now, last thing
ladies and gentlemen, remember they're in between. They're in between the center
and the vertices or your foci. Now, that's an
approximate value. The last thing, ladies and
gentlemen, we are trying to do is find our asymptotes. So remember, for a
vertical asymptote, we have y equal k plus or minus
a divided by b times h minus h. So all we're going to do
to find our asymptotes is plug in these values. So we have our k which is
0 plus or minus a over b. So our a is 2 and our b
is the square root of 3. Times x plus 1. Times x minus h,
which is negative 1. You can just use
it as negative 1? Oh. So then when I
distribute that through, you're going to have that line. Oh, I'm sorry. That's plus or minus, sorry. Did we need to do the 3? We will learn how to graph
them, but we'll practice that next class period. This is so you guys understand
how to find the asymptote, OK? So that's all you guys
are really simply doing, is plugging them in today. .

Ch 6: Teaching Textbooks Algebra 1 (v2.0) Chapter Test Bank Answers Explained – .