If your solution to stated problem does not match the given solution, you should self-critique per instructions at http://vhcc2.vhcc.edu/dsmith/geninfo/labrynth_created_fall_05/levl1_22/levl2_81/file3_259.htm.

 

Your solution, attempt at solution:

 

If you are unable to attempt a solution, give a phrase-by-phrase interpretation of the problem along with a statement of what you do or do not understand about it. This response should be given, based on the work you did in completing the assignment, before you look at the given solution.

 

 

 

 

 

 

012. `*   12

 

 

*   1.4.12 (was 1.4.6). Explain how you found the real solutions of the equation (1-2x)^(1/3) - 1 = 0

 

Your solution:

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Confidence Assessment:

Given Solution:

* *  Starting with

 

(1-2x)^(1/3)-1=0

add 1 to both sides to get

(1-2x)^(1/3)=1

then raise both sides to the power 3 to get

[(1-2x)^(1/3)]^3 = 1^3.

 

Since [(1-2x)^(1/3)]^3 = (1 - 2x) ^( 1/3 * 3) = (1-2x)^1 = 1 - 2x we have

 

1-2x=1.

 

Adding -1 to both sides we get

-2x=0

so that

x=0.

 

Self-critique (if necessary):

 

 

 

 

 

 

 

 

 

 

Self-critique Rating:

Question: *   1.4.28 (was 1.4.18). Explain how you found the real solutions of the equation sqrt(3x+7) + sqrt(x+2) = 1.

 

Your solution:

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Confidence Assessment:

Given Solution:

* *  Starting with

 

sqrt(3x+7)+sqrt(x+2)=1

 

we could just square both sides, recalling that (a+b)^2 = a^2 + 2 a b + b^2.

 

This would be valid but instead we will add -sqrt(x+2) to both sides to get a form with a square root on both sides. This choice is arbitrary; it could be done either way. We get

 

sqrt(3x+7)= -sqrt(x+2) + 1 .

Now we square both sides to get

sqrt(3x+7)^2 =[ -sqrt(x+2) +1]^2.

Expanding the right-hand side using (a+b)^2 = a^2 + 2 a b + b^2 with a = -sqrt(x+2) and b = 1:

3x+7= x+2 - 2sqrt(x+2) +1.

Note that whatever we do we can't avoid that term -2 sqrt(x+2).

 

Simplifying

3x+7= x+ 3 - 2sqrt(x+2)

then adding -(x+3) we have

3x+7-x-3 = -2sqrt(x+2).

Squaring both sides we get

(2x+4)^2 = (-2sqrt(x+2))^2.

 

Note that when you do this step you square away the - sign.  This can result in extraneous solutions.

 

We get

 

4x^2+16x+16= 4(x+2).

Applying the distributive law we have

4x^2+16x+16=4x+8.

Adding -4x - 8 to both sides we obtain

4x^2+12x+8=0.

Factoring 4 we get

4*((x+1)(x+2)=0

and dividing both sides by 4 we have

(x+1)(x+2)=0

Applying the zero principle we end up with

(x+1)(x+2)=0

so that our potential solution set is

x= {-1, -2}.

 

Both of these solutions need to be checked in the original equation sqrt(3x+7)+sqrt(x+2)=1

 

As it turns out:

 

the solution -1 gives us sqrt(1) + sqrt(1) = 1 or 2 = 1, which isn't true,

while

the solution -2 gives us sqrr(1) + sqrt(0) = 1 or 1 + 0 = 1, which is true.

 

ANOTHER SOLUTION

 


<h3>@&
When you square both sides, you square the entire side on each.

This would give you

(sqrt(3x+7) + sqrt(x+2) )^2 = 1^2,

not

Sqrt(3x+7)^2 + sqrt(x+2)= 1^2

and not even

Sqrt(3x+7)^2 + sqrt(x+2)^2 = 1^2

The left-hand side is the square of a binomial:

(sqrt(3x+7) + sqrt(x+2) )^2 = (sqrt(3x+7) + sqrt(x+2) ) * (sqrt(3x+7) + sqrt(x+2) )

which is equal to

sqrt(3x+7) * (sqrt(3x+7) + sqrt(x+2) ) + sqrt(x+2) * (sqrt(3x+7) + sqrt(x+2) ) .

This is equal to

sqrt(3x+7) * sqrt(3x+7) + sqrt(3x+7) * sqrt(x+2) + sqrt(x+2) sqrt(3x+7) + sqrt(x+2) sqrt(3x+7)

which in turn is equal to

(3x + 7) + 2 * sqrt(3x+7) sqrt(x+2) + (x+2).

So the equation becomes

(3x + 7) + 2 * sqrt(3x+7) sqrt(x+2) + (x+2) = 1

*@</h3>

<h3>@&
As you see there are still square roots in the equation. However they are in only one term as opposed to two temrs, so we're making progress.

All we need to do is isolate that one term on one side of the equation. We do this by subtracting the 3x+7 and the x+2 from both sides, getting

2 sqrt(3x+7) * sqrt(x+2) = 1 - (3x+7) - (x+2)

We first simplify the right-hand side to get

2 sqrt(3x+7) * sqrt(x+2) = -4x - 8

Then we square both sides to get

4 ( 3x+7) (x+2) = 16 x^2 +64 x + 64.

There is still work to do. We have to expand the left-hand side and move all terms to one side of the equation to get a quadratic equation, which we can solve by factoring.
*@</h3>



<h3>@&
When we do this we get the equation

4 x^2 + 12 x + 8 = 0

We can divide both sides by 4 to get

x^2 + 3·x + 2 = 0

then factor to get

(x+2)(x+1) = 0

which gives us solutions x = -2 and x = -1.


*@</h3>



<h3>@&
When we do this we get the equation

4 x^2 + 12 x + 8 = 0

We can divide both sides by 4 to get

x^2 + 3·x + 2 = 0

then factor to get

(x+2)(x+1) = 0

which gives us solutions x = -2 and x = -1.


*@</h3>



<h3>@&
It might seem that we're done, but we still have to check our solutions in the original equation.

Having squared both sides it is possible that we gained extraneous solutions that won't work in the original equation.

Plugging our solutions x = -2 and x = -1 into the original equation

sqrt(3x+7) + sqrt(x+2) = 1

we find that only the x = -2 solution works. The x = -1 solution results in the equation 2 = 1, and has to be rejected

Our solution is therefore just

x = -2.

*@</h3>

 

STUDENT QUESTION

 

I got to the third step but I got confused on what to eliminate or substitute in, looking at the solution, im still a little confused on how it all worked out. U got any suggestions on how to look at it in a better way???

INSTRUCTOR RESPONSE


You're pretty much stuck with this technique and this way of looking at the problem.

It should be pretty clear to you that

(sqrt(x+3))^2 is just x + 3.

Squaring the expression [ -sqrt(x+2) +1] is a little more challenging.

We could use the distributive law:

[ -sqrt(x+2) +1]^2
= [-sqrt(x + 2) + 1 ] * [-sqrt(x + 2) + 1 ]
= -sqrt(x+2) * [-sqrt(x + 2) + 1 ] + 1 * [-sqrt(x + 2) + 1 ]
= -sqrt(x+2) * (-sqrt(x + 2) ) + (-sqrt(x + 2) * 1 + 1 * (-sqrt(x + 2) ) + 1 * 1
= (x + 2) - sqrt(x + 2) - sqrt(x + 2) + 1
= x+2 - 2sqrt(x+2) +1.

Once we get the equation

3x+7= x+2 - 2sqrt(x+2) +1

we see that we still need to 'get to' that x within the square root. To do that we rearrange the equation so that the square root is on one side, all by itself, so we can square it without dragging a lot of other stuff along.

So we do a couple of steps and we get

3x+7-x-3 = -2sqrt(x+2).

If we square both sides of this equation, we get rid of all the square roots and we get x out where we can deal with it.

The details are in the given solution, but we get the equation

4x^2+16x+16= 4(x+2).

This equation now has x^2 and x terms, so we know it's a quadratic, and we rearrange and solve it as such. The details are in the given solution.

COMMON ERROR

 

sqrt(3x+7)^2= (-sqrt(x+2)+1)^2
(3x+7)=(x+2)+1

EXPLANATION


(a + b)^2 = (a + b) * (a + b) = a ( a + b) + b ( a + b) = a * a + a * b + b * a + b * b.

So (a + b)^2 is not equal to a^2 + b^2.

 

Nor is it so that

 

(-sqrt(x+2)+1)^2 is equal to (x+2) + 1.

When you square the right-hand side you can't write it

-sqrt(x+2)+1 ^2

because written this way only the 1 gets squared.

You have to write it

(-sqrt(x+2)+1) ^2,

which means

(-sqrt(x+2) + 1) * (-sqrt(x+2) + 1).

When multiplied out using the distributive law, using the same sequence of steps used above to find (a + b) ( a + b), you should get

 

(x + 2) - 2 sqrt( x + 2) + 1.
 

Self-critique (if necessary):

 

 

 

 

 

 

 

 

 

 

Self-critique Rating:

Question: *   1.4.40 (was 1.4.30). Explain how you found the real solutions of the equation x^(3/4) - 9 x^(1/4) = 0.

 

Your solution:

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Confidence Assessment:

Given Solution:

* *  Here we can factor x^(1/4) from both sides:

 

Starting with

 

x^(3/4) - 9 x^(1/4) = 0

we factor as indicated to get

x^(1/4) ( x^(1/2) - 9) = 0.

Applying the zero principle we get

x^(1/4) = 0 or x^(1/2) - 9 = 0

which gives us

x = 0 or x^(1/2) = 9.

 

Squaring both sides of x^(1/2) = 9 we get x = 81.

 

 

 

 

Self-critique (if necessary):

 

 

 

 

 

 

 

 

 

 

Self-critique Rating:

Question: *   1.4.46 (was 1.4.36). Explain how you found the real solutions of the equation x^6 - 7 x^3 - 8 =0

 

Your solution:

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Confidence Assessment:

Given Solution:

* *  Let a = x^3.

 

Then a^2 = x^6 and the equation x^6 - 7x^3 - 8=0 becomes

 

a^2 - 7 a - 8 = 0.

This factors into

(a-8)(a+1) = 0,

with solutions

a = 8, a = -1.

 

Since a = x^3 the solutions are

 

 

We solve these equations to get

 

and

STUDENT QUESTION

 

I am confused as to why you substituted the a. I know how to do this on the calculator by using y = and 2nd graph to get the solution (-1, 2)

INSTRUCTOR RESPONSE


If you substitute a for x^3, then you end up with a quadratic equation that can be easily factored.

If a = x^3, then x^6 = a^2 so the equation becomes

a^2 - 7 a - 8 = 0.

We factor this and find that a can be either 8 or -1.

So x^3 can be either 8 or -1.

Thus x can be either 2 or -1.
 

 

Self-critique (if necessary):

 

 

 

 

 

 

 

 

 

 

Self-critique Rating:

Question: *   1.4.64 (was 1.4.54). Explain how you found the real solutions of the equation x^2 - 3 x - sqrt(x^2 - 3x) = 2.

 

Your solution:

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Confidence Assessment:

Given Solution:

* *  Let u = sqrt(x^2 - 3x).

 

Then u^2 = x^2 - 3x, and the equation is

 

u^2 - u = 2.

 

Rearrange to get

 

u^2 - u - 2 = 0.

 

Factor to get

 

(u-2)(u+1) = 0.

 

 

Substituting x^2 - 3x back in for u we get

 

sqrt(x^2 - 3 x) = 2

and

sqrt(x^2 - 3 x) = -1.

 

The second is impossible since sqrt can't be negative.

 

The first gives us

 

sqrt(x^2 - 3x) = 2

so

x^2 - 3x = 4.

Rearranging we have

x^2 - 3x - 4 = 0

so that

(x-4)(x+1) = 0

and

x = 4 or x = -1.

 

STUDENT QUESTION

 


I got stuck on this part u=(-sqrt2+-sqrt10)/2, but after I looked at the solution, it made a little more sense to me, but im not real confident. Got any suggestions on how to approach it in a different way???

INSTRUCTOR RESPONSE


Plugging into the quadratic formula we get

u=(-sqrt2+-sqrt10)/2,

meaning u can take one of the two values

u=(-sqrt2+sqrt10)/2

or

u=(-sqrt2-sqrt10)/2.

These quantities are just plain old numbers, which you could evaluate (up to some roundoff) on your calculator.

The first possible value of u is about equal to about .874.

The second possible value of u is negative.

Now u stands for x^2, so we ignore the negative value of u (this since x^2 can't be negative).

So we're left with

x^2 = u = .874.

So x = +- sqrt(.874), giving us the values of x in the given solution.

STUDENT QUESTION

 

I still do not understand using u. I can do it from the 2nd step. Problem: u^2 - u - 2 = 0. Factor: (u-2)(u + 1) = 0. You get u = 2, -1. You will solve x^2 - 3x -4. Factor will be (x - 4)(x + 1) = 0. Solutions are 4, -1.

INSTRUCTOR RESPONSE


The left-hand side consists of

x^2 - 3x

and

the square root of x^2 - 3x.

So instead of

x^2 - 3 x - sqrt(x^2 - 3x)

we write the left-hand side as

u - sqrt(u),

which is easier to deal with.

We solve for u, then come back and figure out what value(s) of x give us our values of u.
 

 

 

Self-critique (if necessary):

 

 

 

 

 

 

 

 

 

 

Self-critique Rating:

Question: *   1.4.92 \ 90 (was 1.4.66). Explain how you found the real solutions of the equation x^4 + sqrt(2) x^2 - 2 = 0.

 

Your solution:

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Confidence Assessment:

Given Solution:

* *  Starting with

 

x^4+ sqrt(2)x^2-2=0

 

we let u=x^2 so that u^2 = x^4 giving us the equation

 

u^2 + sqrt(2)u-2=0

 

Using the quadratic formula

 

u=(-sqrt2 +- sqrt(2-(-8))/2

so

u=(-sqrt2+-sqrt10)/2

 

Note that u = (-sqrt(2) - sqrt(10) ) / 2 is negative, and u = ( -sqrt(2) + sqrt(10) ) / 2 is positive.

 

u = x^2, so u can only be positive. Thus the only solutions are the solutions to the equation come from

 

x^2 = ( -sqrt(2) + sqrt(10) ) / 2.

The solutions are

x = sqrt( ( -sqrt(2) + sqrt(10) ) / 2 )

and

x = -sqrt( ( -sqrt(2) + sqrt(10) ) / 2 ).

 

Approximations to three significant figures are

 

and

 

Self-critique (if necessary):

 

 

 

 

 

 

 

 

 

 

Self-critique Rating:

 

Add comments on any surprises or insights you experienced as a result of this assignment.