SOLUTION: if x^2 -16√(x) =12 then x-2√(x) = ??

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Question 1117121: if x^2 -16√(x) =12 then x-2√(x) = ??
Found 2 solutions by ankor@dixie-net.com, ikleyn:
Answer by ankor@dixie-net.com(22740) About Me  (Show Source):
You can put this solution on YOUR website!
if x%5E2+-+16sqrt%28x%29+=+12
plot this equation
y+=+x%5E2+-+16sqrt%28x%29+-+12
+graph%28+300%2C+200%2C+-6%2C+10%2C+-35%2C+50%2C+x%5E2-16%2Asqrt%28x%29-12%29+
x = 7.464
:
then x-2sqrt%28x%29+=+__
using calc
7.464-2%2Asqrt%287.464%29+=+2

Answer by ikleyn(52781) About Me  (Show Source):
You can put this solution on YOUR website!
.
Let me introduce new variable  u = sqrt%28x%29.


Then I can reformulate the problem in this EQUIVALENT way:


    If  u%5E4+-+16u = 12,  then find  u%5E2+-+2u.

Solution

If  u%5E4+-+16u = 12,   then

    u%5E4+-+16u+-+12 = 0.    (1)


The left side can be factored:

    u%5E4+-+16u+-+12 = %28u%5E2-2u-2%29.%28u%5E2%2B2u%2B6%29


into the product of two quadratic polynomials.  (You may check the validity of this decomposition on your own).


Thus the equation (1) is equivalent to

    %28u%5E2-2u-2%29.%28u%5E2%2B2u%2B6%29 = 0.     (2)


If "u" is the real root of the equation (1),  then "u" is the real root of the equation (2).

But the second polynomial (second multiplier)  %28u%5E2%2B2u%2B6%29  is positively defined quadratic function  %28u%2B1%29%5E2%2B5,  which has NO real roots.


Therefore, "u" is the root of the first trinomial  of (2),  i.e.

    u%5E2-2u-2 = 0,


Then  u%5B1%2C2%5D = %282+%2B-+sqrt+%28%28-2%29%5E2%2B4%2A2%29%29%2F2 = %282+%2B-+sqrt%2812%29%29%2F2 = 1+%2B-+sqrt%283%29.


Now, if  u = 1+%2Bsqrt%283%29,  then  u%5E2-2u = %281%2Bsqrt%283%29%29%5E2-2%2A%281%2Bsqrt%283%29%29 = 4+%2B+2%2Asqrt%283%29+-+2+-+2%2Asqrt%283%29 = 2.


     If  u = 1+-sqrt%283%29,  then  u%5E2-2u = %281-sqrt%283%29%29%5E2-2%2A%281-sqrt%283%29%29 = 4+-+2%2Asqrt%283%29+-+2+%2B+2%2Asqrt%283%29 = 2.

    
Thus in any case  u%5E2-2u = 2.


It is the answer to the problem question.


Answer.  If x is a real number such as  x%5E2+-+sqrt%28x%29 = 12,  then  x-2sqrt%28x%29 = 2.

Solved.