solve and check

refresh · #1 $Old$ November 11th, 2009, 11:43 AM

solve and check

21x^4 - 56x^2 + 35=0

ok this is what i did

7(3x^4 - 8x^2 + 5)

bigwave · #2 $Old$ November 11th, 2009, 12:10 PM

$7(3x^4 - 8x^2 + 5) = 0$

you can factor more

$7(3x^2 - 5)(x^2 - 1) = 0$

$x = \frac{+}{-}\sqrt {\frac{5}{3}} \ and \ x = \frac{+}{-}1$

earboth · #3 $Old$ November 11th, 2009, 12:14 PM

Quote:

Originally Posted by refresh $View Post$

solve and check

21x^4 - 56x^2 + 35=0

ok this is what i did

7(3x^4 - 8x^2 + 5)

Use substitution: $t = x^2$

Your equation becomes:

$21t^2-56t+35 = 0$

Now use the quadratic formula to solve this equation for t. Don't forget to re-substitute the results to calculate the corresponding x-values.

@ bigwave: There are 2 more solutions missing ...

refresh · #4 $Old$ November 11th, 2009, 12:20 PM

i use the quadratic equation and got

x= + and - square root 15/3
x= + and - 1

am i correct ?? when i plugged the square root of 15/3 back into the equation i got -46515 and it is suppose to be 0. what did i do wrong?? this is how i did it on my calculator

21(square root 15/3)^4 - 56(square root 15/3)^2 + 35 = -46515