# Thread: Solve x for log (98+sqrt(x^3-x^2+12x+36))=2

1. ## Re: Solve x for log (98+sqrt(x^3-x^2+12x+36))=2

Originally Posted by SlipEternal
Suppose the original problem was

$\displaystyle \log_{10}(98+\sqrt{x^3-x^2-12x+36}) = 2$

(The coefficient of x is -12 instead of 12)

Then the problem simplifies to

$\displaystyle x^3-x^2-12x+32 = 0$

This factors as:

$\displaystyle (x+4)(x^2-5x+8) = 0$

That has $\displaystyle x=-4$ as a solution.
SSSSSSSSSS, I missed a small thing, I wrote +12x, instead of -12x.

I think It is the reason .

But tell me how did you take $\displaystyle (x+4)(x^2-5x+8) = 0$ to x= -4

2. ## Re: Solve x for log (98+sqrt(x^3-x^2+12x+36))=2

Originally Posted by NameIsHidden
SSSSSSSSSS, I missed a small thing, I wrote +12x, instead of -12x.

I think It is the reason .

But tell me how did you take $\displaystyle (x+4)(x^2-5x+8) = 0$ to x= -4
what value of x would make the (x+4) term equal 0?

What is 0 times anything?

3. ## Re: Solve x for log (98+sqrt(x^3-x^2+12x+36))=2

Originally Posted by romsek
what value of x would make the (x+4) term equal 0?

What is 0 times anything?
Sorry, I forgot that.

Either $\displaystyle x+4=0$

or

$\displaystyle x^2-5x+8=0$

If $\displaystyle x+4=0$, then we know $\displaystyle x=-4$.

What if $\displaystyle x^2-5x+8=0$. $\displaystyle x$ is an imaginary number. How did it become -4.

My maths teacher also did like this for one question.

It's something like (An expression)X(Another expression)=0.

He made the solution only if first expression=0. He did not make the solution if another equation is 0.

4. ## Re: Solve x for log (98+sqrt(x^3-x^2+12x+36))=2

A variable can only take one value at a time. If $\displaystyle x=-4$, the whole product becomes zero. That means it is a solution. The second equation cannot be solved over the reals, so the equation only has one real solution.

5. ## Re: Solve x for log (98+sqrt(x^3-x^2+12x+36))=2

Originally Posted by NameIsHidden
Sorry, I forgot that.

Either $\displaystyle x+4=0$

or

$\displaystyle x^2-5x+8=0$

If $\displaystyle x+4=0$, then we know $\displaystyle x=-4$.

What if $\displaystyle x^2-5x+8=0$. $\displaystyle x$ is an imaginary number. How did it become -4.

My maths teacher also did like this for one question.

It's something like (An expression)X(Another expression)=0.

He made the solution only if first expression=0. He did not make the solution if another equation is 0.
if the solution to $x^2-5x+8=0$ then clearly $x\neq -4$ as $(-4)^2-5(-4)+8=44$

The solution is satisfied if either factor is 0, so in this case either

$x=-4$ OR $x=\dfrac 1 2 \left(5 \pm \imath \sqrt 7\right)$

your professor probably ignored the other factor when he/she did what they did because they were only interested in real solutions. I don't know, I wasn't there.

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### log 10 root x2-12x 36=2 x=

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