Question 421527: Need help having trouble with my math on this one I know I cannot factor and have to use the quadratic formula to solve, I posted the problem wrong earlier.
6x^4-7x^2+2=0
This is what I have so far
-(-7)+√(-7)^2-4(6)(2)/2(6)
49+√1/12 or 49-√1/12
I am unsure if this is in its simplest form or even if my math is correct as I am having trouble checking it. Any help is greatly appreciated. Thanks!
Found 3 solutions by sudhanshu_kmr, Gogonati, Theo:
Answer by sudhanshu_kmr(1152)   (Show Source):
You can put this solution on YOUR website!
6x^4-7x^2+2=0
=> 6x^4 -4x^2 -3x^2 +2 =0
=> 2x^2(3x^2-2) -1(3x^2 -2) = 0
=> (2x^2-1)(3x^2-2) =0
for 2x^2 -1 = 0
=> x^2 =1/2 , x = 1/√2 , -1/√2
for 3x^2 -2 =0
=> x^2 = 2/3 , x= √2/√3 , -√2/√3
Answer by Gogonati(855)  (Show Source):
Answer by Theo(13342)  (Show Source):
You can put this solution on YOUR website! if you let y = x^2, then your equation becomes:
6y^2 - 7x + 2 = 0
this is a quadratic equation that can now be solved.
you are correct in assuming that it can't be factored, so i went to the quadratic formula.
the standard formula of a quadratic equation with the independed variable of y is:
ay^2 + by + c = 0
your equation is:
6y^2 - 7y + 2 = 0
that makes:
a = 6
b = -7
c = 2
from that we can derive:
2a = 12
b^2 = 49
4ac = 4*6*2 = 48
b^2 - 4ac = 1
the quadratic formula using y as the independent variable states that y = (-b +/- sqrt(b^2-4ac))/2a
that becomes x = (7 +/- sqrt(1))/12 which becomes (7 +/- 1)/12.
from that we get:
y = 8/12 or y = 6/12
your solution to the equation of 6y^2 - 7y + 2 is y = 8/12 or y = 6/12
you confirm that by substituting in the original equation to see if it's true.
when y = 8/12 and when y = 6/12, the equation 6y^2 - 7y + 2 does indeed equal to 0 so the solution is good for y.
you started by making y = x^2.
you now need to go back and solve for x.
you have x^2 = 8/12 and you have x^2 = 6/12, because x^2 is equivalent to y.
this leads to x = + or - sqrt (8/12) and x = + or - sqrt (6/12).
you confirm that by substituting in your original equation of 6x^4 - 7x^2 + 2 = 0.
for example:
when x = - sqrt(8/12), the equation becomes 6 * (-sqrt(8/12)^4 - 7 * (-sqrt(8/12)^2 + 2 = 0
you can use your calculator to confirm that this becomes 6 * .4444444444 - 7 * .666666667 + 2 = 0 which becomes 0 = 0
you need to store your intermediate results to minimize the impact of rounding.
first you take sqrt(8/12) and then make it minus and then raise it to the 4th power and then store it in memory location 1.
you will get .44444444 which, when stored, will be carried out many more decimal places.
next you take sqrt(8/12) and then make it minus and then raise it to the 2d power and then store it in memory location 2.
you will get .666666667 which, when stored, will be carried out many more decimal places.
you will then take your equation of 6x^4 - 7x^2 + 2 = 0 and translate it to:
6 * memory location 1 - 7 * memory location 2 + 2 = 0
once solved, you will see that the answer will be equal to 0, confirming your solution is correct.
the solution to your problem is x = +/- sqrt (8/12) or x = +/- sqrt (6/12).
you solved it by making y = x^2 and then solving for the quadratic equation of 6y^2 - 7y + 2 = 0
that got you y = 8/12 or y = 6/12
you then derived the final solution by taking y = x^2 and solving for x which led to x = +/0 sqrt(y).
it was hard for me to determine if that's what you did.
if so, you did it right.
if not, then you may want to re-look at the problem in light of what i just showed you.
|
|
|
