SOLUTION: which is a possible solution of 6 + sqrt(3x^4 + 1) = 10 - 2x^2 the answer is -sqrt(15) but i don't know the solution

Question 498215: which is a possible solution of
6 + sqrt(3x^4 + 1) = 10 - 2x^2
the answer is -sqrt(15) but i don't know the solution
Found 2 solutions by josmiceli, Theo:
Answer by josmiceli(19441) (Show Source): You can put this solution on YOUR website!

Square both sides

I can make the substitution

Completing the square:

Take the square root of both sides

and, since ,

and

---------------
check:

The needs to be
I would get the same result if I used the answer
so both are valid
I think the has to be the negative root,
or else the equation doesn't work
Answer by Theo(13342) (Show Source): You can put this solution on YOUR website!
the answer is:
possible solutions to the given equation are:
x = +/- 1
or:
x = +/- sqrt(15)
here's how this equation is solved:
original equation is:
6 + sqrt(3x^4 + 1) = 10 - 2x^2
subtract 6 from both sides of the equation to get:
sqrt(3x^4 + 1) = 4 - 2x^2
square both sides of the equation to get:
3x^4 + 1 = (4 - 2x^2)^2
simplify to get:
3x^4 + 1 = 4x^4 - 16x^2 + 16
subtract 3x^4 from both sides of the equation and subtract 1 from both sides of the equation to get:
0 = x^4 - 16x^2 + 15
this is the same as:
x^4 - 16x^2 + 15 = 0
let y = x^2 to get:
y^2 - 16y + 15 = 0
this is a quadratic equation that can be factored with:
(y - 1) * (y - 15) = 0
this gets:
y = 1
or:
y = 15
since y = x^2, then this is the same as:
x^2 = 1
or:
x^2 = 15
take the square root of both sides of these equations and you get:
x = +/- 1
x = +/- sqrt(15)
these are possible solutions of the original equation.
this does not mean that they are actual solutions to the original equation.
to see if we did the factoring right, we substitute in the revised equations that got us to this point.
that equation is:
3x^4 + 1 = (4 - 2x^2)^2
if we substitute + 1 for x in that equation, we get 0 as we should.
if we substitute - 1 for x in that equation, we get 0 as we should.
if we substitute sqrt(15) for x in that equation, we get 0 as we should.
if we substitute sqrt(15) for x in that equation, we get 0 as we should.
this means that our solution is good for the equation that we evaluated it at.
we need, however, to confirm that these are solutions to the original equation.
the original equation is:
6 + sqrt(3x^4 + 1) = 10 - 2x^2
if this equation is equal, then we can subtract (10-2x^2) from both sides of the equation to get:
6 + sqrt(3x^4 + 1) - (10 - 2x^2) = 0
if we substitute + 1 for x in that equation, we get 0 as we should.
if we substitute - 1 for x in that equation, we get 0 as we should.
if we substitute + sqrt(15) for x in that equation, we get 52 which we should not have gotten if sqrt(15) was a solution. apparently it is not.
if we substitute - sqrt(15) for x in that equation, we get 52 which we should not have gotten if sqrt(15) was a solution. apparently it is not.
while +/- sqrt(15) were possible solutions to the original equations, it turned out that they were not solutions to the original equations.
while +/- 1 were possible solutions to the original equations, it turned out that they were actually solutions to the original equations.
the answer is that the solutions to the original equation given is:
x = +/- 1.
if we graph the original equation and the revised equation, this is what we get:

you can see that the revised equation has roots at +/- 1 and +/- sqrt(15).
you can also see that the original equation only has roots at +/- 1.
that is why +/- 1 was a solution to the original equation, but +/- sqrt(15) was not.
in this graph, there are vertical lines at x = +/- sqrt(15) and x = +/- 1 to show you that the revised equation has roots in both places (y = 0) while the original equation only has roots at x = +/- 1 and does not have roots at y = +/- sqrt(15).
the fact that the graph of the revised equation and the graph of the original equation intersect at some other point besides the 0 point is meaningless to the problem.
the problem was to find the solution to the original equation.
if we had graphed the original equation by itself, we would only have seen roots at +/- 1.
that was sufficient to solve the problem.
the fact that we created the revised equation to solve the original equation was a matter of convenience to make a 4th degree equation equivalent to a second degree equation, which we know how to solve because it's a quadratic equation.
we got possible solutions, but there was no guarantee that a solution to the revised equation would also be a solution to the original equation. we did preserve the equality, but to say that these 2 equations are identical would not be correct.
you get a possible solution, but you still have to confirm that the possible solution is an actual solution by substituting that value for x in the original equation.
to summarize:
+/- 1 and +/- sqrt(15) were possible solutions to the original equation.
+/- 1 were the only solutions to the original equation.
that, to the best of my knowledge, is your answer.

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