SOLUTION: How do I solve equations quadratic equations like this? 4 and then x under a radical minus 3=x.

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Question 982302: How do I solve equations quadratic equations like this? 4 and then x under a radical minus 3=x.
Found 2 solutions by MathLover1, solver91311:
Answer by MathLover1(20849) About Me  (Show Source):
You can put this solution on YOUR website!

if you mean this:
4sqrt%28x%29+-+3=x, then
4sqrt%28x%29+=x%2B3
%284sqrt%28x%29%29%5E2+=%28x%2B3%29%5E2
16x+=x%5E2%2B6x%2B9
0+=x%5E2-16x%2B6x%2B9
0+=x%5E2-10x%2B9
0+=x%5E2-x-9x%2B9
0+=%28x%5E2-x%29-%289x-9%29
0+=x%28x-1%29-9%28x-1%29
0+=%28x-9%29%28x-1%29
solutions:
if 0+=%28x-9%29=>x=9
if 0+=%28x-1%29=>x=1

but, if you mean this:
4sqrt%28x+-+3%29=x, then
%284sqrt%28x+-+3%29%29%5E2=x%5E2
16%28x+-+3%29=x%5E2
16x+-+48=x%5E2
0=x%5E2-16x%2B48
0=x%5E2-4x-12x%2B48
0=%28x%5E2-4x%29-%2812x-48%29
0=x%28x-4%29-12%28x-4%29
0+=+%28x-12%29%28x-4%29
solutions:
if 0+=+%28x-12%29=>x=12
if 0+=+%28x-4%29=>x=4

Answer by solver91311(24713) About Me  (Show Source):
You can put this solution on YOUR website!




Is not a quadratic. A quadratic is a polynomial of degree 2, whereas the given equation is not a polynomial at all because of the exponent that is not a positive integer. However, in order to solve this you have to make it into a quadratic.

If you get the term with the radical by itself on one side of the equation, then you can square both sides and eliminate the radical. Thus:





Then put it in standard form:



Now just factor and solve like any other factorable quadratic. Note that anytime you square an equation in the process of solving it, you take the chance of introducing an extraneous root, that is an answer that follows logically from the process of solving the equation but that is not actually an element of the solution set of your original problem. To make sure that you either do not have any extraneous roots or to be able to eliminate them if you do, you must check each of your roots in the original equation to see if you get a true statement when substituting a potential solution value.

John

My calculator said it, I believe it, that settles it