Answer:
[tex]x=\frac{1+\sqrt5}{2}[/tex] and [tex]x=\frac{1-\sqrt5}{2}[/tex]
and b=[tex]\frac{5}{4}[/tex]
Step-by-step explanation:
We are given that a quadratic equation
[tex]x^2-x-1=0[/tex]
We have to solve the equation by completing square and find the value of b.
[tex](x)^2-2\times x\times \frac{1}{2}+(\frac{1}{2})^2-(\frac{1}{2})^2-1=0[/tex]
[tex](x-\frac{1}{2})^2-\frac{1}{4}-1=0[/tex]
[tex](x-\frac{1}{2})^2-\frac{1-4}{4}=0[/tex]
[tex](x-\frac{1}{2})-\frac{5}{4}=0[/tex]
[tex](x-\frac{1}{2})^2-(\frac{\sqrt5}{2})^2=0[/tex]
[tex](x-\frac{1}{2})^2=(\frac{\sqrt5}{2})^2[/tex]
[tex]x-\frac{1}{2}=\frac{\sqrt5}{2}[/tex] and [tex]x-\frac{1}{2}=-\frac{\sqrt5}{2}[/tex]
[tex]x=\frac{\sqrt5}{2}+\frac{1}{2}=\frac{1+\sqrt5}{2}[/tex]
And [tex]x=\frac{1}{2}-\frac{\sqrt5}{2}=\frac{1-\sqrt5}{2}[/tex]
Therefore,[tex]x=\frac{1+\sqrt5}{2}[/tex] and [tex]x=\frac{1-\sqrt5}{2}[/tex]
and b=[tex]\frac{5}{4}[/tex]
