Reduce to a quadratic formula?

[sia.wears.hats]

I was given this equation:

square root (x-5) - fourth root (x-5) = 12

I couldn't figure out how to do it, so I clicked the "help" button and it told me to reduce it to a quadratic formula first. How do I do this when the exponents are 1/2 and 1/4? I have to solve for x somehow.
The final answer is supposed to be 261, but I have no idea how to get it.

Any help is appreciated!!

 

[HallsofIvy]

Let \(\displaystyle y= \sqrt[4]{x- 5}\). Then \(\displaystyle \sqrt{x- 5}= (\sqrt[4]{x- 5})^2= y^2\).

 

[sia.wears.hats]

Let \(\displaystyle y= \sqrt[4]{x- 5}\). Then \(\displaystyle \sqrt{x- 5}= (\sqrt[4]{x- 5})^2= y^2\).

Thank you! That explained a ton! But once it's factored now, why isn't \(\displaystyle -3= \sqrt[4]{x-5}\) one of the solutions? I was told 261 was the only solution, and that 86 wasn't one.

 

[lookagain]

But once it's factored now, why isn't \(\displaystyle -3= \sqrt[4]{x-5}\) one of the solutions? I was told 261 was the only solution, and that 86 wasn't one.

The smallest potential real value that \(\displaystyle \ \sqrt[4]{x - 5}\) can be is 0, because it's an even root of a quantity.

It cannot equal -3.

 

[sia.wears.hats]

The smallest potential real value that \(\displaystyle \ \sqrt[4]{x - 5}\) can be is 0, because it's an even root of a quantity.

It cannot equal -3.

Oh! Okay, that makes complete sense! I feel silly for not realizing that. Thank you guys so much for your help!

 

[HallsofIvy]

I was given this equation:

square root (x-5) - fourth root (x-5) = 12

I couldn't figure out how to do it, so I clicked the "help" button and it told me to reduce it to a quadratic formula first. How do I do this when the exponents are 1/2 and 1/4? I have to solve for x somehow.
The final answer is supposed to be 261, but I have no idea how to get it.

Any help is appreciated!!

As I pointed out before, letting \(\displaystyle y= \sqrt[4]{x- 5}\) converts the equation to \(\displaystyle y^2- y= 12\) or \(\displaystyle y^2- y- 12= (y- 4)(y+ 3)= 0\) so y= 4 or y= -3. y= 4 gives \(\displaystyle \sqrt[4]{x- 5}= 4\), \(\displaystyle x- 5= 256\), \(\displaystyle x= 261\).

y= -3 gives \(\displaystyle \sqrt[4]{x- 5}= -3\) but that is impossible because any root of a non-negative number is non-negative while an even root of a negative number is imaginary. There is NO number whose fourth root is -3.

(Note: there are two numbers such that \(\displaystyle x^2= 16\), 4 and -4 but \(\displaystyle \sqrt{16}\) is defined as the positive one, 4, NOT -4. Similarly, there are two real numbers, -2 and 2, such that \(\displaystyle x^4= 16\) as well as two imaginary numbers, 2i and -2i, but \(\displaystyle \sqrt[4]{16}\) is, by definition, 2 only.)

 

[Ishuda]

Thank you! That explained a ton! But once it's factored now, why isn't \(\displaystyle -3= \sqrt[4]{x-5}\) one of the solutions? I was told 261 was the only solution, and that 86 wasn't one.

The solutions to the quadratic equation are
y = -3
and
y = 4.
Since
\(\displaystyle y= \sqrt[4]{x- 5}\)
y is defined as positive and thus y = -3 is not an allowed solution for the problem at hand.

As a BTW, at a higher level of mathematics you will find that for problems of this kind, you will need an extended definition of the root functions (square root, 4th root, xth root, ...) in which solutions such as a fourth root being negative is allowed, but for now the definition of fourth root is restricted to the positive real numbers.

 

[lookagain]

.
Since
\(\displaystyle y= \sqrt[4]{x- 5}\)
y is defined as positive

That gives non-negative values.

 

[Ishuda]

That gives non-negative values.

It was written that way just so you could comment on the question without repeating someone (but I'm surprised you missed the other one).