Question 62877
7x^4-6x^2-1=0
7x^4-7x^2+1x^2-1=0
(7x^4-7x^2)+(1x^2-1)=0
7x^2(x^2-1)+1(x^2-1)=0
(x^2-1)(7x^2+1)=0
(x+1)(x-1)(7x^2+1)=0
The real soltutions are:
x+1=0 and x-1=0
x+1-1=0-1 and x-1+1=0+1
{{{highlight(x=-1)}}} and {{{highlight(x=1)}}}
If you are also supposed to find the complex solution:
7x^2+1=0
7x^2+1-1=0-1
7x^2=-1
7x^2/7=-1/7
x^2=-1/7
{{{sqrt(x^2)=+-sqrt(-1/7)}}}
{{{x=+-i/sqrt(7)}}}
{{{highlight(x=+-i*sqrt(7)/7)}}}
If you've never heard of complex solutions and imaginary numbers ignore the last one, 7x^2+1 is prime and will not yeild a real soltuion so you can ignore it if you aren't using complex numbers yet.
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Happy Calculating!!!