Can anyone teach me how to find the solutions to these equations?

Question

I went over the examples they provided us with in the book, but it skipped a few steps and I just can't get it right. Can you please teach me how to find the solutions?

Problems:
1.)x^4+5x^2-36=0

2.)6(s/s+1)^2+5(s/s+1)-6=0

3.)3x^1/3+2x^2/3=5

The whole assignment revolves around these equations. Once I figure out (or understand) how to do them, I can try the other ones on my own. Please help!

Answer 1

These problems are quadratic equations in disguise. To solve, use a substitution, we'll use the letter z.

For #1, let z = x^2. We then have

(x^2)^2 + 5x^2 - 36 = 0
z^2 + 5z - 36 = 0
(z + 9)(z - 4) = 0
z = 4, -9
So then
x^2 = 4, -9
and x = +/-2

For number 2, we make a similar substitution. Let z = s/(s+1).. We then have

6(s/(s + 1))^2 + 5(s/(s + 1)) - 6 = 0
6z^2 + 5z - 6 = 0
(3z - 2)(2z + 3) = 0
z = 2/3, -3/2
s/(s+1) = 2/3, -3/2
3s = 2s + 2 OR 2s = -3s - 3
s = 2 OR 5s = -3
s = 2,-3/5

Try doing #3 yourself, using z = x^(1/3).

Good luck!

Answer 2

All of those equations are quadratic, so all of them are subject to the standard methods of solving quadratics: factoring, completing the square, and the Quadratic Formula. For simplicity's sake, I'm going to assume you know how to solve quadratics.

Take your first one:

x^4 + 5x^2 - 36 = 0

Note that the powers are 4, 2 and 0 - the sure sign of a quadratic. It might help if we make a substitution: let y = x^2. The problem then becomes

y^2 + 5y - 36 = 0

If we factor that, we get

(y + 9)(y - 4) = 0

So y = -9 or y = 4

Resubstitute:

x^2 = -9 or x^2 = 4

x = +/- 3i or x = +/-2

The equation is fourth order, and we have four solutions - just as it should be.

For the second one, make y = s/s+1. So:

6y^2 + 5y -6 = 0

Solve that, then resubstitute as we did above.

For the last one, make y = x^1/3:

3y + 2y^2 = 5

and go.

Once you get used to this, you can drop the substitution thing and go directly to the quadratic solution, but you may find it easier to begin with. Just remember: look at the powers first. If they go as 2 times something, the something, and a constant, then you're looking at a disguised quadratic.

Answer 3

The three equations collectively suggest using a substitution of variables approach. To use this approach you must be able to recognize a familiar problem-type within the given problem.

In this case all three problems are basically quadratic equations;
e.g. 2y^2 - 5y + 7 = 0.

Such problems can always be solved using the quadratic formula.
You're job is to first massage the given problem to the quadratic form by making a variable substituiton. Choosing the correct substitution is the real trick.

For the second one try this: u = s/s+1

When you replace s/s+1 by u the equation becomes"

6u^2 +5u - 6 = 0 which is a simple quadratic. Solve this to get answers for u and then use the answers for u to get the answers for s.

The other two are similar.

Answer 4

Simple math.

As an easier example:
#2

It looks complicated but make (s/s+1) = x
Then re-write your question as:

6x^2 + 5x - 6 = 0

Then solve for x. Which I got to be x= +2/3 x=-3/2

then substitute for both circumstances:

S/S+1 = 2/3, S/S+1 = -3/2

In each mini equation, solve for s, which is relatively simple. That would be your answer.

For #1 and 3 it is the exact same.

#1 define y = x^2 then you'll get:

y^2 + 5y - 36

solve for y. Then using that solution substitute and solve for x, like in #2.

for #3, it's a little trickier, but substitute again, u = x^1/3 you'll get

3u + 2u^2 = 5 Then solve for u. Subsitutie what u is defined as with your solutions and vola!



These problems are trying to teach you about substituting variables to make problems easier.

Answer 5

These are all quadratic equations. I HATED them in high school. ew.
look here, I hope it helps.
http://en.wikipedia.org/wiki/Quadratic_equation

1. substitute x^ 2 for z, you'll have a normally looking quadratic equation:
z^2 + 5z - 36 = 0

2. same appleis here: sub ( s/s+1) with z; you've got:
6z^2 + 5z - 6 =0

3. Sub x^ 1/3 for z:
3z + 2z^2 -5 = 0
2z^2 + 3z - 5=0

Now, you can start drawing the parabola: we used to do it during maths classes in Russia, I wonder if you do it here?

Answer 6

1 put 36 on the other side and divide out

Answer 7

I don't want to tell you how to do them step by step, but I can tell you that they are all quadratic equations. This means that you can factor them like you would factor regular quadratic equations or use the quadratic formula.

Answer 8

I can probably solve it if you can tell me what ^ is.