Respuesta :
Answer:
[tex]1+\sqrt{2}[/tex]
Step-by-step explanation:
If a quadratic equation is defined as
[tex]ax^2+bx+c=0[/tex] .... (1)
then the quadratic formula is
[tex]x=\dfrac{-b\pm \sqrt{b^2-4ac}}{2a}[/tex]
The given quadratic equation is
[tex]0=-x^2+2x+1[/tex]
It can we written as
[tex]-x^2+2x+1=0[/tex] .... (2)
On comparing (1) and (2) we get
[tex]a=-1,b=2,c=1[/tex]
Substitute these values in the quadratic formula.
[tex]x=\dfrac{-2\pm \sqrt{2^2-4(-1)(1)}}{2(-1)}[/tex]
[tex]x=\dfrac{-2\pm \sqrt{4+4}}{-2}[/tex]
[tex]x=\dfrac{-2\pm \sqrt{8}}{-2}[/tex]
[tex]x=\dfrac{-2\pm 2\sqrt{2}}{-2}[/tex]
Taking out common factors.
[tex]x=\dfrac{-2(1\pm \sqrt{2})}{-2}[/tex]
[tex]x=1\pm \sqrt{2}[/tex]
Two roots are
[tex]x=1+\sqrt{2}[/tex] and [tex]x=1-\sqrt{2}[/tex]
We know that
[tex]\sqrt{2}=1.41[/tex]
So
[tex]1<\sqrt{2}[/tex]
Therefore, root [tex]x=1+\sqrt{2}[/tex] is positive and [tex]x=1-\sqrt{2}[/tex] is negative.
