Solve Radical Equation: 5x√2x-3=4

[blackfriars]

solving this problem
not sure how to calculate this 5x√2x-3=4
i was planning to isolate radical by subtracting the 5x when i do this i end with a quadratic and no solutions
any help appreciated

 

[MarkFL]

solving this problem
not sure how to calculate this 5x√2x-3=4
i was planning to isolate radical by subtracting the 5x when i do this i end with a quadratic and no solutions
any help appreciated

I am going to assume the equation is:

$$5x\sqrt{2x-3}=4$$

Now, if we square both sides, we get:

$$25x^2(2x-3)=16$$

In standard form, this is:

$$50x^3-75x^2-16=0$$

Discarding the extraneous complex roots, we are left with:

$$x=\frac{1}{10}\left(5+\sqrt[3]{5\left(57-8\sqrt{41}\right)}+\sqrt[3]{5\left(57+8\sqrt{41}\right)}\right)\approx1.62168011109094$$

I used W|A to get the exact value of the root, which could be found using a cubic root formula, however, in practice I would use a numeric technique to approximate the root to a desired level of accuracy. :D

 

[blackfriars]

hi what is w/a and i was trying to use synthetic division , i do not understand your solution , sorry if it sounds dumb

 

[MarkFL]

hi what is w/a and i was trying to use synthetic division , i do not understand your solution , sorry if it sounds dumb

Did I interpret the problem correctly?

W|A is:

Wolfram Alpha

Synthetic/polynomial division isn't going to work here, because there are no rational roots. We have one irrational real root, and 2 extraneous complex roots.

Please don't feel dumb for asking questions if you don't understand something...it is only "dumb" to not understand something offered as help and not ask about it. :D

 

[blackfriars]

hi thanks, you did interpret the problem correctly
the solution you have given has me confused
i don't know how to go from the cubic equation to your solution
could you possibly show me the steps or guide me to an explanation this is the first time i have come across radical equations
thanks

 

[MarkFL]

hi thanks, you did interpret the problem correctly
the solution you have given has me confused
i don't know how to go from the cubic equation to your solution
could you possibly show me the steps or guide me to an explanation this is the first time i have come across radical equations
thanks

Well, you would have to look up methods for solving general cubic equations. These are in general very messy and cumbersome formulas (unlike the quadratic formula), and is why I would use a numeric technique instead.

Are you sure you copied the question correctly? This is a bad problem to present to students who have not been given general methods to solve cubics, or who have not studied numeric root finding techniques.

 

[blackfriars]

hi, yes the question is correct as it was given to use on a paper
we were shown how to do synthetic division when we were given the factor or the root, as this equation did not have a singular x term i wrote it as 50x^3-75x^2+0x-16=0
i tried to list all the factors of -16 and 50 it was a long process
i did not find grouping to work to find the factors
the solution you gave me i do not know how to get from the cubic to it
hope you understand my confusion thanks

 

[MarkFL]

As I explained, I used W|A to obtain the exact value of the root, however, here is a page explaining the "Cubic Formula" derivation:

Cubic Formula

As you can see it is quite the ordeal, and is why I would use a numeric technique, like the Newton-Raphson method instead. :D

 

[blackfriars]

hi Markf, well i have just used the Newton method, it was long, but the result i got was similar to yours

this is what i did i inputed 1.6216 as the x value in this eq 50x^3-75x^2-16=15.98... this is fine
but when i input the same value into the original eq i get something different 5(1.6216)+Sqroot2x-3=4 ido not get 4 but i get 8.6...
could you tell me the reason or is this the way it is meant to be
thanks

 

[MarkFL]

hi Markf, well i have just used the Newton method, it was long, but the result i got was similar to yours

this is what i did i inputed 1.6216 as the x value in this eq 50x^3-75x^2-16=15.98... this is fine
but when i input the same value into the original eq i get something different 5(1.6216)+Sqroot2x-3=4 ido not get 4 but i get 8.6...
could you tell me the reason or is this the way it is meant to be
thanks

The original equation is:

$$5x\sqrt{2x-3}=4$$

You are using instead:

$$5x+\sqrt{2x-3}=4$$

 

[blackfriars]

hi mark , really sorry , but i made the mistake the original equation was 5x+ sqroot2x-3=4
i did not cop the missing addition sign
sorry about that

 

[MarkFL]

hi mark , really sorry , but i made the mistake the original equation was 5x+ sqroot2x-3=4
i did not cop the missing addition sign
sorry about that

In that case, I wold isolate the radical:

$$\sqrt{2x-3}=4-5x$$

Square both sides:

$$2x-3=16-40x+25x^2$$

Write in standard form:

$$25x^2-42x+19=0$$

We find the discriminat to be negative, which means the two complex roots are extraneous, and therefore there is no solution to the original equation.

 

[blackfriars]

hi, yes this is what i initially did and when entering the data into my casio i kept getting an error message
this led me to believe there was no solution, but then again i was not 100% sure
thanks so much for your help