Answer:
[tex]\large \boxed{\sf \ \ \ 4 \ \ \ }[/tex]
Step-by-step explanation:
Hello, please find below my work.
[tex]x^2-4x-1=0 \ \ \text{add 1 to both parts of the equation}\\\\<=> x^2-4x-1+1=0+1=1\\\\<=> x^2-4x=1[/tex]
We know that for any a and x real numbers we can write
[tex](x-a)^2=x^2-2ax+a^2[/tex]
When we compare with the left part of the equation we can identify the term in x so that -4=-2a (multiply by -1) <=>4=2a (divide by 2) <=> a = 4/2 = 2
So we can write
[tex](x-2)^2=x^2-4x+2^2=x^2-4x+4[/tex]
So we have to add 4 to both sides of the equation to complete the square and it comes:
[tex]x^2-4x-1=0 \ \ \text{add 1 to both parts of the equation}\\\\<=> x^2-4x-1+1=0+1=1\\\\<=> x^2-4x=1 \ \boxed{\text{ add 4 to complete the square}} \\\\<=>x^2-4x\boxed{+4}=1+4=5\\\\<=>(x-2)^2=5 \ \text{ we take the root } \\ \\ <=>x-2=\pm \sqrt{5}\ \text{ we add 2 } \\ \\ <=> x = 2+\sqrt{5} \ \text{ or } \ x = 2-\sqrt{5}[/tex]
Hope this helps.
Do not hesitate if you need further explanation.
Thank you
