how would i solve 4x^2-1=8x by completing the square and have it in the form x=a+-radical b over c, where a, b and c have no common factors??

Question

anonymous · Answer

idk that one srrrry

anonymous · Answer

ok so first convert it into a quadratic in standard form so you get:
$$4x^2-8x-1=0$$ (i subtracted 8x from both sides)
now divide the whole equation by 4 so you get:$$x^2-2x-\frac{ 1 }{ 4 }=0$$
Now add 1/4 to both sides:
$$x^2-2x=\frac{ 1 }{ 4 }$$
now add half of the b-term squared to both sides (-1)^2 so 1:$$x^2-2x+1=\frac{ 5 }{ 4 }$$
(i added 1 to both sides). Now you can rewrite that as (x+half of the b-term)^2:$$(x-1)^2=\frac{ 5 }{ 4 }$$ now take the plus/minus square root so:$$\pm \sqrt{(x-1)^2}=\pm\sqrt{\frac{ 5 }{ 4 }}$$ so if you simplify that:$$x-1=\pm \frac{ \sqrt{5} }{ 2 }$$
now add 1 to both sides so:$$x=\pm \frac{ \sqrt{5} }{ 2 }+1$$
as you can see 2, sqrt(5), and 1 have no common factors but 1
so thats the answer!

THE ANSWER IS:
$$x=\pm \frac{ \sqrt{5} }{ 2 }+1$$

anonymous · Answer

$$\pm $$ means positive or negative and read my work carefully.