Assignment 12 14

course MTH 158

????????????\??assignment #012

012. `query 12

College Algebra

10-25-2008

......!!!!!!!!...................................

20:26:16

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

......!!!!!!!!...................................

RESPONSE -->

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

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

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

1 - 2x = 1

-2x = 1 - 1

x = 0/-2

x = 0

Check

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

The solution set is {0}.

.................................................

......!!!!!!!!...................................

20:27:06

10-25-2008 20:27:06

** 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. **

......!!!!!!!!...................................

NOTES -------> ok

.......................................................!!!!!!!!...................................

20:33:08

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

......!!!!!!!!...................................

RESPONSE -->

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

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

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

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

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

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

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

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

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

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

(4x^2 + 16x + 16)/4 = (4x + 8)/4

x^2 + 4x + 4 = x + 2

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

x^2 + 3x + 2 = 0

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

x + 1 = 0 x + 2 = 0

x = -1 x = -2

Check:

x = -1: sqrt(3x + 7) + sqrt(x + 2) = sqrt(3(-1) + 7) + sqrt(-1 + 2) = sqrt(-3 + 7) + sqrt(1) = sqrt(4) + sqrt(1) = 2 + 1 = 3

Not a solution because 3 does not equal 1

x = -2: sqrt(3x + 7) + sqrt(x + 2) = sqrt(3(-2) + 7) + sqrt(-2 + 2) = sqrt(-6 + 7) + sqrt(0) = sqrt(1) = 1 = 1

The solution set is {-2}.

.................................................

......!!!!!!!!...................................

20:33:19

10-25-2008 20:33:19

** 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, which 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

It turns out that the -1 gives us sqrt(4) + sqrt(1) = 1 or 2 + 1 = 1, which isn't true, while -2 gives us sqrr(1) + sqrt(0) = 1 or 1 + 0 = 1, which is true.

x = -1 is the extraneous solution that was introduced in our squaring step.

Thus our only solution is x = -2. **

......!!!!!!!!...................................

NOTES -------> ok

.......................................................!!!!!!!!...................................

20:42:35

**** query 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.

......!!!!!!!!...................................

RESPONSE -->

I didn't think about it until just now, but on the last question was the reason that they divide by 4, rather than setting it equal to 0, because it would be set to 0 and not a variable?

4*(x+1)(x+2)=0 if 4 = 0 or if x+1 = 0 or if x+2=0. Of course 4 can't equal 0; it's therefore convenient to divide both sides by 4, since 4 doesn't contribute to the solution.

Back to problem 40:

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

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

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

x^3 = 9x

You need to take the 4th power of 9 as well as the fourth power of x^(1/4).

x^3 - 9x = 0

x(x^2 - 9) = 0

x(x + 3)(x - 3) = 0

x = 0 x + 3 = 0 x - 3 = 0

x = -3 x = 3

Check:

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

x = -3: x^(3/4) - 9x^(1/4) = (-3)^(3/4) - 9(-3)^(1/4): Not a real solution

x = 3: x^(3/4) - 9x^(1/4) = (3)^(3/4) - 9(3)^(1/4) = ((3)^(3/4))^4 - (9(3)^(1/4))^4 = 3^3 - 9(3) = 27 - 27 = 0

The solution set is {0, 3}.

.................................................

......!!!!!!!!...................................

21:09:31

10-25-2008 21:09:31

** 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.

So our solution set is {0, 81). **

......!!!!!!!!...................................

NOTES -------> I understand how and why factoring works, in fact, I can't believe I missed that. I have re-worked the problem by factoring and adjusted my assignment folder.

My concern is why solving it as if it were a radical equation doesn't work. It works on paper but produces a different solution set.

I went back and reworked the problem, the same way as I did before, expecting it not to work, but it does. The only place that I can see where I may be MAKING THE STEPS WORK is in the Check section for x = 3.

If x = 3: x^(3/4) - 9x^(1/4) = (3)^(3/4) - 9(3)^(1/4)

(RIGHT HERE) = ((3)^(3/4))^4 - (9(3)^(1/4))^4 = 3^3 - 9(3) = 27 - 27 = 0

When I plugged 3^(3/4) - 9(3)^(1/4) into my calculator I got -9.565..., so that makes me think that it an extraneous solution.

I still don't understand why one method for solving an equation would work while another wouldn't? It seems to me that all equations would produce the same solutions no matter what method was used to solve.

Your method of solving the equation would work here, but you failed to take the 4th power of 9 (see my note).

When you take an even power of both sides of the equation, you can introduce extraneous solutions. If you have the equation x = -1, for example, and square both sides you get x^2 = 1. This equation has two solutions, x = 1 and x = -1. However x = 1 is not a solution of the original equation.

Squaring both sides of an equation therefore does not necessarily give you an identical equation. Any solution of the original equation will be a solution of the squared equation, but not all solutions of the squared equation are necessarily solutions of the original.

.......................................................!!!!!!!!...................................

21:15:44

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

......!!!!!!!!...................................

RESPONSE -->

x^6 - 7x^3 - 8 = 0

Let y = x^3, thus y^2 = x^6. The original equation becomes:

y^2 - 7y - 8 = 0

(y + 1)(y - 8) = 0

y + 1 = 0 y - 8 = 0

y = -1 y = 8

Solve for x:

x^3 = y

x^3 = -1

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

x = -1

&

x^3 = y

x^3 = 8

(x^3)^(1/3) = 8^(1/3)

x = 2

Check:

x = -1: x^6 - 7x^3 = 8 = (-1)^6 - 7(-1)^3 - 8 = 1 - 7 * -1 - 8 = 1 + 7 - 8 = 8 - 8 = 0 = 0

&

x = 2: x^6 - 7x^3 - 8 = (2)^6 - 7(2)^3 - 8 = 64 - 56 - 8 = 64 - 64 = 0 = 0

The solution set is {-1, 2}.

.................................................

......!!!!!!!!...................................

21:15:57

10-25-2008 21:15:57

** 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 x^3 = 8 and x^3 = -1.

We solve these equations to get

x = 8^(1/3) = 2 and x = (-1)^(1/3) = -1. **

......!!!!!!!!...................................

NOTES -------> ok

.......................................................!!!!!!!!...................................

21:23:51

**** query 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.

......!!!!!!!!...................................

RESPONSE -->

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

Let y = sqrt(x^2 - 3x), thus y^2 = x^2 - 3x.

The original equation becomes:

y^2 - y = 2

y^2 - y - 2 = 0

(y + 1)(y - 2) = 0

y + 1 = 0 y - 2 = 0

y = -1 y = 2

Solve for x:

y = sqrt(x^2 - 3x)

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

The negative solution makes this a non-real solution because the principle square root is nonnegative.

y = sqrt(x^2 - 3x)

2 = sqrt(x^2 - 3x)

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

4 = x^2 - 3x

0 = x^2 - 3x - 4

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

x + 1 = 0 x - 4 = 0

x = -1 x = 4

Check:

x = -1: x^2 - 3x - sqrt(x^2 - 3x) = (-1)^2 - 3(-1) - sqrt((-1)^2 - 3(-1)) = 1 + 3 - sqrt(1 + 3) = 4 - sqrt(4) = 4 - 2 = 2 = 2

x = 4: x^2 - 3x - sqrt(x^2 - 3x) = (4)^2 - 3(4) - sqrt(4^2 - 3(4)) = 16 - 12 - sqrt(16 - 12) = 4 - sqrt(4) = 4 - 2 = 2 = 2

The solution set is {-1, 4}.

.................................................

......!!!!!!!!...................................

21:24:47

10-25-2008 21:24:47

** 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.

Solutions are u = 2, u = -1.

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.

DER **

......!!!!!!!!...................................

NOTES -------> Shouldn't your solutions have been x = 4 and x = -1? Because x - 4 = 4 and x + 1 = -1.

You are right. Careless typing on my part.

.......................................................!!!!!!!!...................................

21:36:37

**** query 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.

......!!!!!!!!...................................

RESPONSE -->

I am actually glad you asked this one because I had trouble with it. This is what I got:

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

Let y = x^2, thus y^2 = x^4. The original equation becomes y^2 + sqrt(2)y - 2 = 0. Using the quadratic equation:

y = (-b +/- sqrt(b^2 - 4ac))/(2a)

y = (-(sqrt(2) +/- sqrt((sqrt(2)^2) - 4(1)(-2))/(2 * 1)

y = (-sqrt(2) +/- sqrt(2) + 8))/2

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

Solve for x:

x^2 = y

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

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

Check:

x = sqrt((-sqrt(2) + sqrt(10)/2): x^4 + sqrt(2)x^2 - 2 = ((-sqrt(2) + sqrt(10))/2)^2 + sqrt(2)((-sqrt(2) + sqrt(10))/2) - 2 = 12/4 - (2/2) + (sqrt(2 * 10))/2 - 2 = sqrt(20/2)

x = sqrt((-sqrt(2) - sqrt(10))/2): x^4 + 2sqrt(2)x^2 - 2 = ((-sqrt(2) - sqrt(10))/2)^2 + sqrt(2)((-sqrt(2) - sqrt(10))/2)^2 - 2 = (-8/4) - (2/2) - (sqrt(20)/2) - 2 = -2 - 1 - (sqrt(20/2)) - 2 = -5 - (sqrt(20)/2)

And I'M LOST!!!!

.................................................

......!!!!!!!!...................................

22:04:33

10-25-2008 22:04:33

** Starting with

x^4+ sqrt(2)x^2-2=0 we let u=x^2 so that u^2 = x^4:

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

using 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

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 are x = .935 and x = -.935. **

......!!!!!!!!...................................

NOTES -------> So all that extra work I was doing and stressing over was really just a matter of plugging it into the calculator.

But, I was still not working because somehow I couldn't make my checks = 0. I always ended up with a weird square root, I've erased it now, but I think it was sqrt(20)/2.

But I have adjusted my work to reflect this answer.

The solution you give includes the correct exact solution, sqrt( -sqrt(2) +- sqrt(10) ) / 2.

.......................................................!!!!!!!!...................................

22:06:05

10-25-2008 22:06:05

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

......!!!!!!!!...................................

NOTES -------> On page 121 of the text, Example 7:

Why didn't they factor x^2 - 4 out to (x - 2)(x + 2)? It would have produced the same solutions. Is this a necessary step or should we just take the square root of the result? Or does it matter which method you use?

.................................................

&#Your work looks very good. Let me know if you have any questions. &#