JEE Novice - (1)

Algebra Level 2

$\Large\sqrt[4]{|x-3|^{x+1}} = \sqrt[3]{|x-3|^{x-2}}$

Other than the value $x=3$ , what integer values of $x$ satisfy the equation above?

2,11

4,11

2.4

2,4,11

2 solutions

Chew-Seong Cheong
May 26, 2015

$\begin{aligned} \sqrt [4] {|x-3|^{x+1}} & = \sqrt [3] {|x-3|^{x-2}} \\ |x-3|^{3x+3} & = |x-3|^{4x-8} \end{aligned}$

For $x \ne 3$ :

$\dfrac{|x-3|^{3x+3}}{|x-3|^{4x-8}} = |x-3|^{x-11} = 1$

For $LHS = 1$ there are following likely cases:

$\begin{cases} |x-3| = 1 & \Rightarrow \begin{cases} x = 2 \\ x = 4 \end{cases} \\ x - 11 = 0 & \Rightarrow \quad x = 11 \end{cases}$

Therefore, the solution set is $\boxed{\{2,4,11\}}$

Moderator note:

For clarity, you need to show that $x=3$ is not a solution so division by $|x-3|$ is acceptable.

Don't understand what do you mean about $|x-3| = -1$ . It can't be negative by definition.

Chew-Seong Cheong - 6 years ago

Oh my bad, I was thinking of something else.

Brilliant Mathematics Staff - 6 years ago

Sir, I don't understand why 3 is not a solution (as it lies in the domain and satisfies the equation). Can you please explain.

Abhijeet Verma - 6 years ago

When $x=3$ , then $RHS = \sqrt[3]{|x-3|^{x-2}} = \sqrt[3]{0^{-2}} = \sqrt[3]{\color{#D61F06}{\frac{1}{0}}}$ . Note that $\color{#D61F06}{\frac{1}{0}}$ is not defined.

Chew-Seong Cheong - 6 years ago

When $x = 3, x - 2= 1$ , it becomes $\sqrt[3]{0^1} = 0$ .

Pi Han Goh - 6 years ago

@Pi Han Goh – Yes, as highlighted by Abhijeet Verma too. But there were no $\{2,3,4,11\}$ option.

Chew-Seong Cheong - 6 years ago

@Chew-Seong Cheong – It seems that the question has been edited accordingly. I think you should edit your solution now to reflect the fact that $x=3$ is indeed a solution.

@Calvin Lin , I think the challenge master note needs to be edited too, since $x=3$ is a solution.

Prasun Biswas - 6 years ago

So we can't treat $\sqrt[3]{|x-3|^{x-2}}$ as $\sqrt [ 3 ]{ { \left| x-3 \right| }^{ 1 } }$ for x=3?

Abhijeet Verma - 6 years ago

@Abhijeet Verma – Yes, you are right. I guess the answer options given were wrong.

Chew-Seong Cheong - 6 years ago

The best way to go from the second line of your solution would be to solve the equation by taking everything to LHS and then using the zero product property later.

$|x-3|^{3x+3}=|x-3|^{4x-8}\\ \implies |x-3|^{3x+3}\cdot (1-|x-3|^{x-11})=0\\ \implies |x-3|^{3x+3}=0\quad\lor\quad |x-3|^{x-11}=1$

The first case (left one) has the obvious solution $x=3$ . Now, when dealing with the second case (right one), we can easily show by casework that it has solutions $x=2,4,11$ in reals (as you did in your solution).

Although, if we're to find all complex solutions, there are two more solutions, which are $x=(3+i),(3-i)$ where $i=\sqrt{-1}$ is the imaginary unit.

Prasun Biswas - 6 years ago

$3-i$ works too!

Pi Han Goh - 6 years ago

Cool! Let me add that. :)

Prasun Biswas - 6 years ago

Jesse Nieminen
May 27, 2015

|x - 3|^(3x+3) = |x - 3|^(4x-8)

1^n = 1

|x - 3| = 1

x = 2 or x = 4

3x+3=4x-8

x=11

Result: 2,4,11

JEE Novice - (1)

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