Solve Y= (4th root of x) +1

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Archie Deetoo

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Solve y =(4th root of x) +1

I have graphed this function and can see that there are no real solutions for x. I would draw the graph here but don't know how to. It starts off at (0,1) and the x value increases rapidly as the y value increases slowly.
But then I tried to solve for x using algebra and found a real solution.
See below:

Let 0=(4th root of x) +1
-1=(4th root of x)
(-1)^4 = (4th root of x)^4
1 = x

So I have a solution of x =1 and yet the function does not cross the x axis at any point when I graph it.

Is someone able to show me why there is no solution on the graph but there is one when I use algebra. I am assuming that my algebra is wrong somewhere but can't see where.

Thanks a lot.

Zen
 
Archie Deetoo said:
Solve y =(4th root of x) +1

I have graphed this function and can see that there are no real solutions for x. I would draw the graph here but don't know how to. It starts off at (0,1) and the x value increases rapidly as the y value increases slowly.
But then I tried to solve for x using algebra and found a real solution.
See below:

Let 0=(4th root of x) +1
-1=(4th root of x)
(-1)^4 = (4th root of x)^4
1 = x

So I have a solution of x =1 and yet the function does not cross the x axis at any point when I graph it.

Is someone able to show me why there is no solution on the graph but there is one when I use algebra. I am assuming that my algebra is wrong somewhere but can't see where.
Click to expand...

Let's be clear about terminology. When you talk about "solving", I think you mean finding zeros (or y-intercepts) of the function y = x^(1/4) + 1, that is, solutions of the equation x^(1/4) + 1 = 0. You are right that this equation has no real solutions.

What you did in solving is to raise both sides to the 4th power, which introduces extraneous solutions. You have to check, and you find that x = 1 is not in fact a solution of the original equation.
 
Archie Deetoo said:
Solve y =(4th root of x) +1

I have graphed this function and can see that there are no real solutions for x. I would draw the graph here but don't know how to. It starts off at (0,1) and the x value increases rapidly as the y value increases slowly.
But then I tried to solve for x using algebra and found a real solution.
See below:

Let 0=(4th root of x) +1
-1=(4th root of x)
(-1)^4 = (4th root of x)^4
1 = x

So I have a solution of x =1 and yet the function does not cross the x axis at any point when I graph it.

Is someone able to show me why there is no solution on the graph but there is one when I use algebra. I am assuming that my algebra is wrong somewhere but can't see where.

Thanks a lot.

Zen
Click to expand...
Zen, when you raise numbers to an even power you might get erroneous roots.
Consider the equation x=-2 (SO X EQUALS -2 AN NOTHING ELSE)
Now square both sides and get x^2 =4
Solving x^2=4 we get x = +/- 2. Clearly x=2 is an erroneous solution!
 
Dr.Peterson said:
Let's be clear about terminology. When you talk about "solving", I think you mean finding zeros (or y-intercepts) of the function y = x^(1/4) + 1, that is, solutions of the equation x^(1/4) + 1 = 0. You are right that this equation has no real solutions.

What you did in solving is to raise both sides to the 4th power, which introduces extraneous solutions. You have to check, and you find that x = 1 is not in fact a solution of the original equation.
Click to expand...

Thank you very much, Doctor. Great help.
 
Jomo said:
Zen, when you raise numbers to an even power you might get erroneous roots.
Consider the equation x=-2 (SO X EQUALS -2 AN NOTHING ELSE)
Now square both sides and get x^2 =4
Solving x^2=4 we get x = +/- 2. Clearly x=2 is an erroneous solution!
Click to expand...

Thanks a lot for your help. Makes sense. Cheers!
 
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