Equations

[JellyFish]

Hi, I was wondering if anyone could help me with this problem:
"Given that a = u + v and b = uv, solve for u and v in terms of a and b".

The next part says: "Use your solutions to help solve the quadratic equation x^2 + ax + b = 0".

I would really appreciate any assistance!
Thanks

 

[stapel]

"Given that a = u + v and b = uv, solve for u and v in terms of a and b".

Try solving each of the above equations for, say, "v=". Set the results equal, multiply through the clear the denominator, and then solve the resulting quadratic for u in terms of a and b. Then back-solve to find v in terms of a and b.

"Use your solutions to help solve the quadratic equation x^2 + ax + b = 0".

From multiplying polynomials, you know that x[sup:3l7ywxlm]2[/sup:3l7ywxlm] + ax + b came from some product (x + u)(x + v) = x[sup:3l7ywxlm]2[/sup:3l7ywxlm] + (u + v)x + uv. Note how this relates to the previous part.

If you get stuck, please reply showing how for you have gotten, and what your thoughts are. Thank you!

Eliz.

 

[mmm4444bot]

This exercise requires you to have a good understanding of factoring quadratic polynomials, of using the quadratic formula, and of knowing which symbols represent constants versus variables.

It also helps if you've learned how to solve a system of two equations using substitution.

You did not tell me what you know about these things, so I'm not sure how and where to begin helping.

a = u + v

b = u * v

To solve this system of two equations for u and v, we use substitution and the quadratic formula.

If we divide both sides of the second equation by v, then we get u = b/v.

We substitute this expression for u into the first equation.

a = b/v + v

Solve for v in the usual way. Clear the fraction by multiplying both sides by v. You'll get a quadratic equation in which v is the variable. Now use the quadratic formula.

You're probably familiar with the following representation of a quadratic equation:

ax^2 + bx + c = 0

Be careful not to confuse the symbols a and b in this equation with the symbols a and b in your exercise. They're not the same.

You've factored quadratic polynomials, right?

For example, if you're asked to factor x^2 + 9x + 20, then you're looking for the following form:

(x + m) * (x + n)

where m + n = 9 and m * n = 20.

(I hope this looks familiar, to you.)

Now look at the quadratic polynomial in your exercise:

x^2 + ax + b

The symbols a and b represent constants, just like the 9 and 20 in my example above.

If we wanted to factor x^2 + ax + b, then we could go like this:

(x - u) * (x - v)

Just like above, we want to find two numbers u and v such that

u + v = a

u * v = b

Do you see the relationship to your exercise?

Please show as much work as you can doing the first part. Let us know if you have any questions.

We'll continue from there.

 

[JellyFish]

I appologize for not posting what I had so far I may not have read the "Read this before posting thread" before I posted, oops.

This is how we did it....

u = a-v, sub this into the other one,

(a-v)v = b
av - v^2 = b

Then we did tried it two different ways the first we just did:
v(a-v) = b
so then v = b OR a-v = b
-v = b - a
v = a - b

We subbed this back into to u+v = a and got solutions but when we checked them it didnt work out for uv = b

The other way:
-v^2 + av - b = o
Then used the quadratic formula but the solutions we got are:
-a + sqrt a^2 +4b/2 and -a - sqrt a^2 +4b/2

Are those correct and if so how the heck do you put them back into one of the equations?

Thanks for all of your help!

 

[mmm4444bot]

… we did [try] it two different ways …

v(a-v) = b

so then v = b OR a-v = b <<<< This reasoning works only if b = 0.

(Look up "Zero-Product Property", if you don't understand.)

We don't know what number b stands for, so it might not be zero.



Then [we] used the quadratic formula but the solutions we got are:

v = [-a + sqrt(a^2 +4b)]/2 or v = [-a - sqrt(a^2 +4b)]/2

The quadratic formula is the way to go.

Note my edits in red; we use grouping symbols to clearly show what part's the numerator and what part's under the square-root sign.

However, check your algebra. I think you have a sign error on each of those.

You could check by redoing the quadratic formula, but multiply both sides of the equation by -1, first:

v^2 - ax + b = 0

After you're sure that you've got the correct results for v, do not put them into the other equation to find u. Start over from the very beginning, instead.

This time, solve one of the two equations for v, and substitute the result for v in the other equation. This will give you a quadratic in the variable u, rather than v.

Tell me what you notice about this new polynomial when you compare it to the one you just did with v.

If you're on the ball, you'll be able to immediately write down expressions for u (just like you did above for v) without having to do the quadratic-formula steps.

 

[JellyFish]

Thank you so much.

I think I corrected my error and then got:

v = [a + sqrt(a^2 +4b)]/2 or v = [a - sqrt(a^2 +4b)]/2


When I solved for u I got the same equation as I had with v but with the variable being u, i.e. u^2 - au +b = 0 so then the two solutions for u are the exact same as the ones for v?


With regard to the other question: "Use your solutions to help solve the quadratic equation x^2 + ax + b = 0". I read the other post so I would have something like
x^2 + (u + v)x + uv = 0 which I understand but there must be more to it since I haven't used my solutions yet?

 

[mmm4444bot]

Good job.

Your new expressions for v are correct, and you've realized that solving for u gives the same expressions.

We get the same quadratic polynomials, with only the symbol representing the variable being different (v in one and u in the other). Since the two expressions for u are identical to the two expressions for v, one of them must equal u and the other is then v.

In other words, we arbitrarily assign one to u and the other to v.

These solutions for u and v are the answers for the first part of your exercise.

Now they bring in a new equation: x^2 + ax + b = 0

The second part of the exercise is to solve for x.

Previously, I showed you how this quadratic polynomial could be factored.

x^2 + ax + b = (x - u) * (x - v)

with u + v = a and u * v = b.

This fact associates the expressions that you just found for u and v to this quadratic equation.

Do you see why?

 

[JellyFish]

Because when you solve for x you get x = u or x = v but if we use the values for u and v from the other problem, they are actually the "normal soultions" so the quadratic equation?

 

[mmm4444bot]

You got it; u and v are the solutions to x^2 + ax + b = 0.

Here's a final thought. Consider this exercise in terms of actual values for u, v, a, and b; maybe you'll gain deeper insight into something.

If you were to try factor x^2 - (2/3)x - 1/9, you would be looking for two numbers whose product is negative 1/9 and whose sum is negative 2/3, right?

And, you could also find this polynomial's zeros by using the Quadratic Formula.

Now, you are given the following system of equations.

\(\displaystyle -\frac{2}{3} \; = \; \frac{1 + \sqrt{2}}{3} \; + \; \frac{1 - \sqrt{2}}{3}\)

\(\displaystyle -\frac{1}{9} \; = \; \begin( \frac{1 + \sqrt{2}}{3} \end) \cdot \begin( \frac{1 - \sqrt{2}}{3} \end)\)

Can you see u, v, a, and b?

What are the zeros of x^2 - (2/3)x - 1/9?

MY EDITS: spelling and unclear language corrections

 

[JellyFish]

Thank you so much mmm4444bot, I really appreciate you talking me through it! Much easier then I was making it out to be!

 

[mmm4444bot]

… The other way:

-v^2 + av - b = o

Then used the quadratic formula but the solutions we got are:

-a + sqrt a^2 +4b/2 and -a - sqrt a^2 +4b/2 …

Oh, I just realized something.

When you said that you "got" [-a +/- sqrt(a^2 + 4b)]/2, were you telling me that this was the final result for u?

In other words, when I first read that post, I saw your expressions for v as containing a sign error. However, if you had actually taken correct expressions for v and substituted them back into one of the equations in the system to find u, then what you typed above is correct for what you would have obtained for u.

Doing that algebra is what Elizabeth suggested (backsolving), so maybe that's what you did.

(Even with the minus sign in front of a, it's actually okay, because we could factor out a negative one from each expression. I did not recognize this fact until I did the algebra by hand.)

I also want to note that Elizabeth's approach is just another way to arrive at the same results. (I skipped it because I was being lazy. Maybe, I should not have given you the short-cut. Perhaps, you should try backsolving for u, sometime -- good practice for using a conjugate expression to rationalize a denominator.)