Solving for x

[Sabi]

Hi guys,

Looking for a bit of clarification on a couple questions I have done.

After doing these questions. My teacher started talking about alternative methods, so now I'm questioning what what I've done. He made it sound like he's after a single value.

Input on answers/method would be great. Sorry it's a bit messy

Many thanks

 

[Deleted member 4993]

Hi guys,

Looking for a bit of clarification on a couple questions I have done.

After doing these questions. My teacher started talking about alternative methods, so now I'm questioning what what I've done. He made it sound like he's after a single value.

Input on answers/method would be great. Sorry it's a bit messy

Many thanks

View attachment 22940

First of all, your work is exceptionally legible - not messy at all.

However, you did not tell us:

Where (starting which line) do you need clarification?​

 

[Sabi]

First of all, your work is exceptionally legible - not messy at all.

However, you did not tell us:

Where (starting which line) do you need clarification?​

Thanks ha,

The clarification is if I have used the appropriate method to solve these questions. My teacher gave the impression that we would would come to a single value for a final result so I'm thinking perhaps I should of used another method. Unless I misunderstood him.

Many thanks

 

[JeffM]

Your method looks fine to me. It is what I would do, but it certainly is not the only way to proceed. For example

[MATH]\dfrac{6}{2x - \frac{1}{x}} = 5 \implies \dfrac{6}{\frac{2x^2-1}{x}} = 5 \implies[/MATH]
[MATH]\dfrac{6x}{2x^2 - 1} = 5 \implies 6x = 10x^2 - 5[/MATH]
gets you to the same place.

I would NOT show my answer as you do.

[MATH]x = \dfrac{6 \pm \sqrt{36 - 4(10)(-5)}}{2 * 10} = \dfrac{6 \pm \sqrt{4 * 59}}{2 * 10} \implies[/MATH]
[MATH]x = \dfrac{3 \pm \sqrt{59}}{10} \approx 1.068 \text { or } -0.0468.[/MATH]
I would always show the exact answer and never show an approximation using an equal sign.

 

[Sabi]

Your method looks fine to me. It is what I would do, but it certainly is not the only way to proceed. For example

[MATH]\dfrac{6}{2x - \frac{1}{x}} = 5 \implies \dfrac{6}{\frac{2x^2-1}{x}} = 5 \implies[/MATH]
[MATH]\dfrac{6x}{2x^2 - 1} = 5 \implies 6x = 10x^2 - 5[/MATH]
gets you to the same place.

I would NOT show my answer as you do.

[MATH]x = \dfrac{6 \pm \sqrt{36 - 4(10}(-5)}}{2 * 10} = \dfrac{6 \pm \sqrt{4 * 59}}{2 * 10} \implies[/MATH]
[MATH]x = \dfrac{3 + \pm {59}}{10} \approx 1.068 \text { or } -0.0468.[/MATH]
I would always show the exact answer and never show an approximation using an equal sign.

Thanks for the reply, from the sound of it I'm just being paranoid. Logically as long as a method is correct and it proves out the results. How can it be a problem right. I will make sure I write my answers out properly now.

All the best.

 

[JeffM]

Remember that you can always put your results back into the original equation to check it.

[MATH]{6} \div \left ( 2 * \dfrac{3 \pm \sqrt{59}}{10} - \dfrac{1}{\dfrac{3 \pm \sqrt{59}}{10}} \right ) =[/MATH]
[MATH]6 \div {\dfrac{2 * \left ( \dfrac{3 \pm \sqrt{59}}{10} \right )^2- 1}{\dfrac{3 \pm \sqrt{59}}{10}}} =[/MATH]
[MATH]6 * \dfrac{3 \pm \sqrt{59}}{10} \div \left \{ 2 * \left ( \dfrac{3 \pm \sqrt{59}}{10} \right )^2- 1 \right \}=[/MATH]
[MATH]6 * \dfrac{3 \pm \sqrt{59}}{10} \div \left \{ 2 * \dfrac{9 \pm 6 \sqrt{59} + 59}{100} - 1 \right \} =[/MATH]
[MATH]6 * \dfrac{3 \pm \sqrt{59}}{10} \div \dfrac{18 \pm 12\sqrt{59} + 118 - 100}{100} =[/MATH]
[MATH]6 * \dfrac{3 \pm \sqrt{59}}{10} \div \dfrac{36 \pm 12 \sqrt{59}}{100} = [/MATH]
[MATH]\dfrac{6}{10} * \dfrac{100}{12} = \dfrac{1}{2} * 10 = 5.[/MATH]
Now you have proved that there are two correct answers. But you are correct: two methods that are both valid must give the same result.

 

[Deleted member 4993]

Thanks for the reply, from the sound of it I'm just being paranoid. Logically as long as a method is correct and it proves out the results. How can it be a problem right. I will make sure I write my answers out properly now.

All the best.

A bit of paranoia is good for inquisitive soul ......

But remember it is a "bit" - just a "bit" .... no drowning in paranoia is allowed....

 

[lookagain]

Hi guys,

Looking for a bit of clarification on a couple questions I have done.

After doing these questions. My teacher started talking about alternative methods, so now I'm questioning what what I've done. He made it sound like he's after a single value.

Input on answers/method would be great. Sorry it's a bit messy

Many thanks

View attachment 22940

Sabi, your b-values in each equation are negative. You substituted
a positive b-value in each one. That is incorrect. It just so happens
that a negative number squared equals a positive number. It's a
good practice to use parentheses to show values being substituted.

For instance,

\(\displaystyle x \ = \ \dfrac{-(-6) \pm \sqrt{(-6)^2 \ - \ 4(10)(-5) \ } \ }{2(10)}\)

You might elect to start at a step past this, but you must show correct
steps, which includes showing the correct values being substituted.