Sviatoslav solved the quadratic equation x^2-x-1=0 by completing the square. In the process, he came up with the equivalent equation (x+a)^2 = b, where a and b are constants. Wh…

Question

asked Aug 24, 2020 94.2k views

1 Answer

Erdemgc · Answer 1 · 2020-08-30T18:38:46+0000

Answer:

$x=(1+\sqrt5)/(2)$ and
$x=(1-\sqrt5)/(2)$

and b=

Explanation:

We are given that a quadratic equation

x^2-x-1=0

We have to solve the equation by completing square and find the value of b.

(x)^2-2* x* (1)/(2)+((1)/(2))^2-((1)/(2))^2-1=0

(x-(1)/(2))^2-(1)/(4)-1=0

(x-(1)/(2))^2-(1-4)/(4)=0

(x-(1)/(2))-(5)/(4)=0

$(x-(1)/(2))^2-((\sqrt5)/(2))^2=0$

$(x-(1)/(2))^2=((\sqrt5)/(2))^2$

$x-(1)/(2)=(\sqrt5)/(2)$ and
$x-(1)/(2)=-(\sqrt5)/(2)$

$x=(\sqrt5)/(2)+(1)/(2)=(1+\sqrt5)/(2)$

And
$x=(1)/(2)-(\sqrt5)/(2)=(1-\sqrt5)/(2)$

Therefore,
$x=(1+\sqrt5)/(2)$ and
$x=(1-\sqrt5)/(2)$

and b=