Is Wolfram the Key to Solving Tricky Equations on the Go?

[MSG100]

Solve |x^2-5|=4|x|.


I tried to rewrite it as:

(sqrt(x^2-5))^2=4*(sqrt(x))^2

Is this the right way to solve the equation?

 

[SammyS]

Solve |x^2-5|=4|x|.

I tried to rewrite it as:

(sqrt(x^2-5))^2=4*(sqrt(x))^2

Is this the right way to solve the equation?

No.

For one thing, each side of the original expression exists for all values of x.

For your equation, (sqrt(x^2-5))^2=4*(sqrt(x))^2 , :

The left hand side exists only if x ≥ √(5) or x ≤ -√(5).

The right hand side exists only if x ≥ 0 .

 

[haruspex]

(sqrt(x^2-5))^2=4*(sqrt(x))^2

Did you mean to apply those operations in the other order: square then square root?
That would work, but why bother square-rooting?

 

[MSG100]

I make a new attempt.

I put the equation in two possible cases.

Case 1: x^2-5=4x
x^2-4x-5=0
x=5 and x=-1

Case 2: x^2-5=4(-x)
x^2+4x-5=0
x=-5 and x=1

 

[Mark44]

I make a new attempt.

I put the equation in two possible cases.

Case 1: x^2-5=4x
x^2-4x-5=0
x=5 and x=-1

That should be x = 5 OR x = -1.
Also, you're omitting something important. In Case 1, your assumption (not stated) is that x ≥ 0. How does this assumption fit in with your solutions?

Case 2: x^2-5=4(-x)
x^2+4x-5=0
x=-5 and x=1

Same here - it should be x = -5 OR x = 1. The assumption in this case is that x < 0. How does this assumption fit in with the solutions from this case?

 

[MSG100]

Of course, I meant "or" not "and"

It's just x=5 that fit in case 1
In case 2 it's only x=-5 that fit.

How do you make the assumptions?

 

[Ray Vickson]

Solve |x^2-5|=4|x|.


I tried to rewrite it as:

(sqrt(x^2-5))^2=4*(sqrt(x))^2

Is this the right way to solve the equation?

Since absolute-value equations can be tricky, the first step in such cases should be to plot both sides; that is, draw graphs of y = |x^2-5| and y = 4|x|, to see where they cross. That will help you to keep things straight. Be sure to make the x-range wide enough to show all the possibilities.

 

[Mark44]

Of course, I meant "or" not "and"

It's just x=5 that fit in case 1
In case 2 it's only x=-5 that fit.

How do you make the assumptions?

When you change |x| to x, you are assuming that x ≥ 0. When you change |x| to -x, you are assuming that x < 0.

One other thing is that when you change |x² - 5| to x² - 5, you are assuming that x² - 5 ≥ 0, or equivalently, that x ≥ √5 or that x < -√5.

 

[MSG100]

They cross at x=-5, x=-1, x=1, x=5.

Now I'm lost. The x^2 term makes it harder to get it right.

Could someone show how this should be solved or make an exemple. I can't find any similar task with the x^2 term in it. I would be very grateful.

 

[MSG100]

I usually write it as three intevals, but with this I can't do that.

 

[Mark44]

I usually write it as three intevals, but with this I can't do that.

Make four cases (hence four intervals), since you have four different possibilities.

1. x ≥ √5
2. 0 ≤ x < √5
3. -√5 ≤ x < 0
4. x < -√5

Case 1. x ≥ √5
For the first case, since x ≥ √5, we can say with certainty that x > 0.
Then the original equation simplifies to
x² - 5 = 4x
==> x² - 4x = 5 = 0
==> x = 5 or x = -1
Since our assumption was that x ≥ √5, we discard x = -1.
Solution for case 1: x = 5

Now look at the other three cases.

 

[MSG100]

Case 2: 0 ≤ x < √5

x has to be positive. 0 or more, but less then √5

(here I'm not sure how to use the signs)

Should it be
-(x^2-5)=4x
or
(x^2-5)=-4x
Both gives x=-5 and x=1

 

[MSG100]

I did it like this:

Case 2:
-(x^2-5)=4x
Solution: x=1

Case 3:
-(x^2-5)=-4x
Solution: x=-1

Case 4:
-(x^2-5)=4x
Solution: x=-5

 

[haruspex]

You can avoid all this breaking into cases. Just square both sides of the original equation to get rid of the modulus signs. This gives you a quadratic in x². Solve. Final step is to check you have not introduced any extra solutions.

 

[ehild]

Did you mean to apply those operations in the other order: square then square root?
That would work, but why bother square-rooting?

Yes, $$|A|=\sqrt{A^2}$$
so
$$\sqrt{(x^2-5)^2}=4\sqrt{x^2}$$

eliminate the square root by squaring both sides of the equation, and solve as haruspex suggested.


ehild

 

[MSG100]

Okay, if I do so I'll get:

x^4-26x^2+25=0

and when I type it in Wolfram I get the solution:
x=1
x=-1

 

[MSG100]

I made a mistake when I wrote it in Wolfram.

Now I got it right!

 

[Mentallic]

You don't need a calculator to figure this out. Let $u=x^2$, then you'll have a quadratic in u that can be factorized. Once you have two solutions for u, solve for x.

 

[MSG100]

Yes, it's easy, I was outside with no pen or paper so I took some help from Wolfram.
Math becomes hard when your in a hurry.