find the value of x

[chijioke]

√x - 6x√6x = 1, find the value of x
This is what I did.
$$\text{I multiplied both sides of the equation by} \sqrt{x}$$$$x-6=\sqrt{x}$$$$\text{ I took the square of both sides to obtain}$$$$x^2-12x+36=x$$$$\text{I then rearranged the terms.}$$$$x^2-11x+36=0$$$$\text{I discovered that the resulting quadratic equation cannot}$$$$\text{be factorized so I tried using the quadratic formula.}$$$$\frac{-11 \pm \sqrt{-23}}{2}$$$$\text{I am stuck here as} \pm \sqrt{-23}$$$$\text{is a complex number}$$$$\text{What do I need to do now?}$$

Please how do I create a line indent using latex?

 

[Aion]

You made a calculation error here.

$$\text{ I took the square of both sides to obtain}$$$$x^2-12x+36=x$$

Then, we subtract $x$ from both sides of the equation. To obtain

$$x^2-13x+36=0$$This equation can be factored. Keep in mind that squaring both sides may introduce extraneous solutions, so you must check all solutions in the original equation to ensure they are valid.

 

[Dr.Peterson]

√x - 6x√6x = 1, find the value of x

Just to be sure, please restate the equation that you are solving, so we can be sure the first step is correct.

But I agree with @Aion on the correction.

 

[The Highlander]

√x - 6x√6x = 1, find the value of x
This is what I did.
$$\text{I multiplied both sides of the equation by} \sqrt{x}$$$$x-6=\sqrt{x}$$$$\text{ I took the square of both sides to obtain}$$$$x^2-12x+36=x$$$$\text{I then rearranged the terms.}$$$$x^2-11x+36=0$$$$\text{I discovered that the resulting quadratic equation cannot}$$$$\text{be factorized so I tried using the quadratic formula.}$$$$\frac{-11 \pm \sqrt{-23}}{2}$$$$\text{I am stuck here as} \pm \sqrt{-23}$$$$\text{is a complex number}$$$$\text{What do I need to do now?}$$

Please how do I create a line indent using latex?

I'm not clear either.

Is your equation meant to be:- $$\sqrt{x}-6x\sqrt{6}x=1$$

(Which is what it looks like to me.)​

or is it meant to be:- $$\sqrt{x}-6x\sqrt{6x}=1$$
But, in either case, I don't see how multiplying by $\displaystyle \sqrt{x}$ gets you: $\displaystyle x-6=\sqrt{x}\;$

 

[chijioke]

$$\text{Sorry every body as the initial question was not well tendered.}$$$$\text{The actual problem is}$$$$\sqrt{x}-\frac{6}{\sqrt{x}}=1~\text{,find the value of x}$$$$\text{multiplying both sides of the equation by} \sqrt{x}$$$$\sqrt{x} \left(\sqrt{x}-\frac{6}{\sqrt{x}} \right)=1(\sqrt{x})$$$$\text{, we have}$$$$x-6=\sqrt{x}$$$$\text{taking the squares of both sides}$$$$(x-6)^2=(\sqrt{x})^2$$$$x^2-12x+36=x$$$$\text{rearranging the terms}$$$$x^2-12x-x+36=x$$$$x^2-13x+36=0$$$$\text{factoring and solving further}$$$$x^2-9x-4x+36=0$$$$x(x-9)-4x(x-9)=0$$$$(x-4)(x-9)=0$$$$\therefore x=4~~ \text{or}~~ x=9$$$$\text{The solution that satisfies the problem is}~~ x=9$$

 

[chijioke]

I am just okay now. Thank you all for your efforts in trying to assist.

 

[The Highlander]

I am just okay now. Thank you all for your efforts in trying to assist.

Hi @chijioke

Haven't seen you for some time. When you say: "I am just okay now." does that mean you have been unwell?
If so, then I am glad to know you are better now; welcome back.

Thank you for clearing up the issue of what the original equation was and your solution is fine right up to the point where you factorized the quadratic you had arrived at.

I'm afraid there is a line of very bad Algebra in there (even though you get to the correct result for the factors at the end!)

You wrote:-
$$\text{factoring and solving further}$$$$x^2-9x-4x+36=0$$

$\displaystyle x(x-9)-4x(x-9)=0$ ​

$$(x-4)(x-9)=0$$$$\therefore x=4~~ \text{or}~~ x=9$$
The line in the middle (I've marked with the "") is just plain wrong I'm afraid.

I don't know how you've been taught to factorize quadratics (or what you were trying to do when you came up with that wrong line) but, for a simple quadratic equation like this one, I would tell my students to, first of all, write down all the factor pairs of the $\displaystyle x^0$ term in a wee table like the one opposite.
(The $\displaystyle x^0$ term is the one at the end where, although $\displaystyle x^0$ does not appear, 36 is really the coefficient of $\displaystyle x^0$ but, since $\displaystyle x^0=1$ and anything multiplied by 1 is unchanged, $\displaystyle x^0$ isn't usually included; it is, however, still there and that's worth being aware of. Any quadratic equation is really as shown below.)
$$ax^2+bx^1+cx^0=0$$

where a, b & c are the coefficients of the three x terms.​

Once you have all the factor pairs (of 36 in this case) you simply look to find a pair that will add up to the coefficient of the $\displaystyle x^1$ term (ie: -13 in this case) and you can easily see that -4 & -9 fit the bill!
That allows you to immediately write down the factors of the quadratic as:-

$$(x-4)\text{ and }(x-9)$$
Usually, you would only need to list the positive factor pairs (I included the negative pairs in this instance just for the sake of clarity) and you would decide whether one or both of them needed to be negative based on how to get them to add up to the $\displaystyle x^1$ term and multiply together to get the $\displaystyle x^0$ term with the correct signs.

If you cannot find a pair of factors that give you the required results then you would proceed to the use of the quadratic equation.

Hope that helps.

 

[pka]

$\text{Sorry every body as the initial question was not well tendered.}$
$\text{The actual problem is}$
$\large{\sqrt{x}-\frac{6}{\sqrt{x}}=1}~\text{,find the value of x}$

I too thank you for that clarification.
But I would hope that seeing $\sqrt{x}-\Large{\frac{6}{\sqrt{x}}}=1$, a student would see at once that $x=9$.
The same might be said for $\left(\sqrt{x}-\Large{\frac{6}{\sqrt{x}}}\right)=\left(\Large{\frac{x-6}{\sqrt{x}}}\right)=1$
B.T.W. It should be clear in any case that a solution must be a positive real number.

 

[chijioke]

Hi @chijioke

Haven't seen you for some time. When you say: "I am just okay now." does that mean you have been unwell?

I have not been around. Not because I am unwell but because I have been tied down with other engagements. All the same, I am happy for at least coming around to visit.

You wrote:-
$$\text{factoring and solving further}$$$$x^2-9x-4x+36=0$$

$\displaystyle x(x-9)-4x(x-9)=0$ ​

Thanks for seeing and pointing out the wrong part of the work. I must confess that it is a typo.
I intended writing
$$x(x-9)-4(x-9)=0 ~~ \text{and not}~~ x(x-9)-4x(x-9)=0$$

 

[chijioke]

I too thank you for that clarification.
But I would hope that seeing $\sqrt{x}-\Large{\frac{6}{\sqrt{x}}}=1$, a student would see at once that $x=9$.
The same might be said for $\left(\sqrt{x}-\Large{\frac{6}{\sqrt{x}}}\right)=\left(\Large{\frac{x-6}{\sqrt{x}}}\right)=1$
B.T.W. It should be clear in any case that a solution must be a positive real number.

 

[lookagain]

$$x^2-12x+36=x$$$$\text{rearranging the terms}$$$$x^2-12x-x+36=x$$

That third line in the quote box should be

$\displaystyle x^2 - 12x - x + 36 = 0$

 

[The Highlander]

You appear to be unhappy about something in @pka's post (#8, above).
If you tell us what your problem is with it, we can perhaps explain.

 

[khansaheb]

Probably the statement:

a student would see at once that x=9x=9x=9.

expects a little too intense "see"ing. I had to think a bit to justify "seeing" x = 9. It did not happen "at once".

 

[The Highlander]

Probably the statement:

expects a little too intense "see"ing. I had to think a bit to justify "seeing" x = 9. It did not happen "at once".

@chijioke

Was that what was bothering you?