Make x the subject of √(x - a) + b = 1/c

[Phavonic]

Hi, I got this question wrong on a test.

Make x the subject of [h=1]√(x - a) + b = 1/c[/h]It looks like it should be x = (1/c - b)[FONT=Nimbus Roman No9 L, Times New Roman, Times, serif]^2 + a [/FONT]

But this is not right and is a bit more complicated. I'd appreciate it if someone could show me the correct workings out. Thanks.

 

[mmm4444bot]

How did you begin, when you tried it?

Did you subtract b from each side, followed by squaring each side?

 

[HallsofIvy]

Hi, I got this question wrong on a test.

Make x the subject of [h=1]√(x - a) + b = 1/c[/h]It looks like it should be x = (1/c - b)[FONT=Nimbus Roman No9 L, Times New Roman, Times, serif]^2 + a [/FONT]

But this is not right and is a bit more complicated. I'd appreciate it if someone could show me the correct workings out. Thanks.

Subtract b from both sides to get \(\displaystyle \sqrt{x- a}= 1/c- b= \frac{1- bc}{c}\). Now square both sides: \(\displaystyle x- a= \left(\frac{1- bc}{c}\right)^2\). Finally, add a to both sides: \(\displaystyle x= \left(\frac{1- bc}{c}\right)^2+ a\). Why do you say that is not right? Were you clear, on your test paper, that you meant "\(\displaystyle (1/c- b)^2\)" and not "\(\displaystyle 1/(c- b)^2\)".

 

[Denis]

Make x the subject of √(x - a) + b = 1/c

Sometimes these go faster with temporary substitutions; like u = x - a
Then:
√u + b = 1/c
√u = 1/c - b
u = (1/c - b)^2
....
substitute back in ... get my drift?

 

[Phavonic]

faulty answers section atback of book....!

Subtract b from both sides to get \(\displaystyle \sqrt{x- a}= 1/c- b= \frac{1- bc}{c}\). Now square both sides: \(\displaystyle x- a= \left(\frac{1- bc}{c}\right)^2\). Finally, add a to both sides: \(\displaystyle x= \left(\frac{1- bc}{c}\right)^2+ a\). Why do you say that is not right? Were you clear, on your test paper, that you meant "\(\displaystyle (1/c- b)^2\)" and not "\(\displaystyle 1/(c- b)^2\)".

[FONT=&quot]Because according to my textbook the answer should be

x = (b^2c^2 - 2bc +1 + ac^2)/c^2

This is not the first time it's given me an overly complicated answer. For instance, it asks me to find the value of u using the quadratic formula

u + 1 + 3/(u + 1) = 8

but it says u = (-2.58) or 0.58, whereas I always get the answer 6.61 and (-0.61) !!....similarly, I was told when asking others that I was correct.

The book is mathematics for today (second edition!) by G. D. Buckwell and A. D. Ball. But thanks anway for clearing that [/FONT]

 

[mmm4444bot]

[FONT=&quot]

… according to my textbook the answer should be

x = (b^2c^2 - 2bc +1 + ac^2)/c^2

It would have been helpful had you stated this reason, when you started the thread. We could have immediately told you that their answer is simply a different form of what you obtained. They expanded your result, and then combined terms into single ratio.

Unless they specifically instructed you to report your answer as a single ratio, your result is fine.

This is not the first time it's given me an overly complicated answer. For instance, it asks me to find the value of u using the quadratic formula

u + 1 + 3/(u + 1) = 8

but it says u = (-2.58) or 0.58, whereas I always get the answer 6.61 and (-0.61) !! …

[/FONT]Their answer is not complicated; it's just written in a different form. (Let us know, if you can't see how to go back and forth between the different forms.)

Their rounded approximations for u are definitely wrong. (Yours are correct.) Math texts contain some error; we all have to deal with this and move on. :cool:

 

[Dr.Peterson]

Because according to my textbook the answer should be

x = (b^2c^2 - 2bc +1 + ac^2)/c^2

This is not the first time it's given me an overly complicated answer. For instance, it asks me to find the value of u using the quadratic formula

u + 1 + 3/(u + 1) = 8

but it says u = (-2.58) or 0.58, whereas I always get the answer 6.61 and (-0.61) !!....similarly, I was told when asking others that I was correct.

The book is mathematics for today (second edition!) by G. D. Buckwell and A. D. Ball. But thanks anway for clearing that

Do you see that your answer and theirs are equivalent? They just expanded the square and combined to make one fraction. There is a sense in which this is simpler. But there is no reason to suppose that they would say you are wrong, or for you to say that they are wrong. They are just two forms of the same answer, perhaps reflecting different senses of what is simple, or of what is an appropriate final answer.

The other clearly is wrong (unless you misread the problem).

 

[Denis]

u + 1 + 3/(u + 1) = 8

but it says u = (-2.58) or 0.58, whereas I always get the answer 6.61 and (-0.61)

Your answer is KEERECT!!

 

[Phavonic]

Further workings out...

Hi, I got this question wrong on a test.

Make x the subject of √(x - a) + b = 1/c

It looks like it should be x = (1/c - b)^2 + a

But this is not right and is a bit more complicated. I'd appreciate it if someone could show me the correct workings out. Thanks.

OK, thanks. So it only leaves me to ask someone to help me with the rest of the workings out. I originally got stuck at (1/c - b)^2 + a, before expanding, so I'm not so good at squaring brackets with a fraction in them. Anyone....?

Thanks.

 

[Dr.Peterson]

OK, thanks. So it only leaves me to ask someone to help me with the rest of the workings out. I originally got stuck at (1/c - b)^2 + a, before expanding, so I'm not so good at squaring brackets with a fraction in them. Anyone....?

Thanks.

You're asking for help expanding (1/c - b)^2 + a? It would be more helpful if you had shown your attempt, so we could see HOW you are stuck.

There are two main ways to start. One is to realize that (1/c - b)^2 = (1/c - b)(1/c - b), and just distribute as you would (a+b)(c+d). Some call this "FOIL". Can you do that?

The more efficient way is to use the formula (a + b)^2 = a^2 + 2ab + b^2, which is obtained by FOIL, and then memorized because it makes the work easier. So you would square 1/c, double the product of 1/c and -b, and square -b.

Let's see what you can do.

 

[JeffM]

OK, thanks. So it only leaves me to ask someone to help me with the rest of the workings out. I originally got stuck at (1/c - b)^2 + a, before expanding, so I'm not so good at squaring brackets with a fraction in them. Anyone....?

Thanks.

\(\displaystyle \sqrt{x - a} + b = \dfrac{1}{c} \implies \sqrt{x - a} = \implies \dfrac{1}{c} - b \implies\)

\(\displaystyle x - a = \left ( \dfrac{1}{c} - b \right )^2 \implies x =\left ( \dfrac{1}{c} - b \right )^2 + a.\)

Perfectly good answer.

You square a binomial with a fraction just like any other binomial.

\(\displaystyle (u \pm v)^2 = u^2 \pm 2uv + v^2.\) So

\(\displaystyle x =\left ( \dfrac{1}{c} - b \right )^2 + a = \dfrac{1^2}{c^2} - 2b * \dfrac{1}{c} + b^2 + a = \dfrac{1}{c^2} - \dfrac{2b}{c} + b^2 + a.\)

Another perfectly good answer. I'd likely to do it the way the book did.

\(\displaystyle \sqrt{x - a} + b = \dfrac{1}{c} \implies \sqrt{x - a} = \implies \dfrac{1}{c} - b = \dfrac{1 - bc}{c} \implies\)

\(\displaystyle x - a = \left ( \dfrac{1 - bc}{c} \right )^2 = \dfrac{(1 - bc)^2}{c^2} = \dfrac{1 - 2bc + b^2c^2}{c^2} \implies\)

\(\displaystyle x = \dfrac{1 - 2bc + b^2c^2}{c^2} + a = \dfrac{1 - 2bc + b^2c^2 + ac^2}{c^2}.\)

Another good answer.