Choose a suitable substitution to solve this equation

[LostInLunacy]

Hi. I have been doing these 'quadratic equations in disguise' questions, but I am finding them quite challenging.

Exact wording:
Choose a suitable substitution and solve the equation:

1) [MATH]x + \sqrt{x} - \frac{3}{4} = 0[/MATH]
2) x⁶ - 3x³ + 2 = 0

I have attached my working for the first question, however I am not sure if it's correct, and I cannot do the second question as I get stuck half way through.

My understanding is that you invent a variable (I used y) and assign it a value that enables you to write the equation as a quadratic. However that does not seem possible on the second question as you are left with a cubed term, so I cannot get past that point.

I think an understanding of how someone would approach the question would help me. Any help is appreciated!

 

[Dr.Peterson]

Good work on the first, but check for extraneous solutions. That can occur because of an unstated restriction on y, namely that as a square root it can't be negative.

For the second, what substitution did you make. If you write it out, you may see for yourself either that it's the wrong substitution, or that it doesn't leave a cube! I suspect you made a silly mistake.

 

[JeffM]

This is really about pattern recognition. You get to something that looks like this

[MATH]ax^n + bx^{n/2} + c [/MATH]
Oh, you say, "That looks sort of like a quadratic." You do the following substitution

[MATH]\text {Let } u = x^{n/2} \implies u^2 = (x^{n/2})^2 = x^n \implies\\ ax^n + bx^{n/2} + c \text { means } au^2 + bu + c.[/MATH]Now you are working with a quadratic.

 

[Steven G]

#1) You skipped a step that caused you to see that one of your answers is not correct.

You wrote x= (-3/2)^2.

Since y = sqrt(x) and y = -3/2 you should have written that sqrt(x) = -3/2. Since sqrt( no value) = -3/2 you know that there is no solution.

#2) You want to get y^2 - 3y^1 + 2 = 0 Not that the powers of y are 2 and 1.

In x^6 - 3x^3 + 2 = 0 the powers of x are 6 and 3. What can you multiply/divide 6 and 3 by to get 2 and 1. That is the secret to get the correct substitution.

Please post back with your work.

 

[JeffM]

#1) You skipped a step that caused you to see that one of your answers is not correct.

You wrote x= (-3/2)^2.

Since y = sqrt(x) and y = -3/2 you should have written that sqrt(x) = -3/2. Since sqrt( no value) = -3/2 you know that there is no solution.

#2) You want to get y^2 - 3y^1 + 2 = 0 Not that the powers of y are 2 and 1.

In x^6 - 3x^3 + 2 = 0 the powers of x are 6 and 3. What can you multiply/divide 6 and 3 by to get 2 and 1. That is the secret to get the correct substitution.

Please post back with your work.

Jomo

Your response is of course correct. But I think it is simpler to just test whether the smaller exponent is exactly half the larger. If so, you have found your substitution.

 

[Steven G]

Jomo

Your response is of course correct. But I think it is simpler to just test whether the smaller exponent is exactly half the larger. If so, you have found your substitution.

I thought that was basically what I said. You are the expert so I will rethink how to say it better. Thanks!

 

[LostInLunacy]

Thank you all for your responses.
They were all helpful and putting them together I think I understand a bit more now. My confusion was in part due to some false knowledge I had in my head about exponents.

I corrected question 1 to show just one valid solution and attempted question 2.

 

[Dr.Peterson]

You wrote (u-1) + (u-2) = 0 where you meant, of course, (u-1)(u-2) = 0.
But the significant error is in not writing out what you were thinking behind the long arrow. Write those steps, and ponder the signs ...

Or, check your answers!! These are not extraneous, just erroneous.

 

[Steven G]

I thought that was basically what I said. You are the expert so I will rethink how to say it better. Thanks!

I really need to get back in the classroom because I am forgetting too much. The term with the lower exponent (usually) is what you call y and the other term best be y^2, that is the power must be twice the other power.

 

[Steven G]

Thank you all for your responses.
They were all helpful and putting them together I think I understand a bit more now. My confusion was in part due to some false knowledge I had in my head about exponents.

I corrected question 1 to show just one valid solution and attempted question 2.

There you go again not writing x^3 = 1 and x^3 =2

Also if x^3 = 1, x is NOT +/- 1. Note that (-1)^3 is NOT 1!

 

[LostInLunacy]

There you go again not writing x^3 = 1 and x^3 =2

Also if x^3 = 1, x is NOT +/- 1. Note that (-1)^3 is NOT 1!

Didn't realise I needed that extra working but I'll bear it in mind

On the second point: oh right yeah, I feel silly now...
Yeah it should just be x = 1 and x = the positive cube root of 2, right? I did actually put that initially but then changed my mind and did +/-

 

[LostInLunacy]

You wrote (u-1) + (u-2) = 0 where you meant, of course, (u-1)(u-2) = 0.
But the significant error is in not writing out what you were thinking behind the long arrow. Write those steps, and ponder the signs ...

Or, check your answers!! These are not extraneous, just erroneous.

Ah yes, the sign was just a mistake in writing it out! Whoops.