Respuesta :
When you have a quadratic in the form ax² + bx + c = 0, you can solve for x using the quadratic formula, which is as follows:
[tex]x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}[/tex]
In this case, a = 2, b = -2, and c = -1. Let's plug these in.
[tex]x=\frac{-(-2)\pm\sqrt{(-2)^2-4(2)(-1)}{2(2)}[/tex]
Simplify...
[tex]x=\frac{2\pmsqrt{4+8}}4[/tex]
[tex]x=\frac{2\pmsqrt{12}}4[/tex]
[tex]\sqrt{12}=\sqrt{2\times2\times3}=2\sqrt{3}[/tex]
[tex]x=\frac{2\pm2\sqrt{3}}4[/tex]
[tex]x=\frac{1\pm\sqrt{3}}2[/tex]
[tex]\boxed{x=\frac{1+\sqrt{3}}2\ or\ \frac{1-\sqrt{3}}2}[/tex]
[tex]x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}[/tex]
In this case, a = 2, b = -2, and c = -1. Let's plug these in.
[tex]x=\frac{-(-2)\pm\sqrt{(-2)^2-4(2)(-1)}{2(2)}[/tex]
Simplify...
[tex]x=\frac{2\pmsqrt{4+8}}4[/tex]
[tex]x=\frac{2\pmsqrt{12}}4[/tex]
[tex]\sqrt{12}=\sqrt{2\times2\times3}=2\sqrt{3}[/tex]
[tex]x=\frac{2\pm2\sqrt{3}}4[/tex]
[tex]x=\frac{1\pm\sqrt{3}}2[/tex]
[tex]\boxed{x=\frac{1+\sqrt{3}}2\ or\ \frac{1-\sqrt{3}}2}[/tex]
2x² - 2x - 1 = 0
x = -(-2) ± √((-2)² - 4(2)(-1))
2(2)
x = 2 ± √(4 + 8)
4
x = 2 ± √(12)
4
x = 2 ± 2√(3)
4
x = 1 ± √(3)
2
x = 1 + √(3) U x = 1 - √(3)
2 2
x = -(-2) ± √((-2)² - 4(2)(-1))
2(2)
x = 2 ± √(4 + 8)
4
x = 2 ± √(12)
4
x = 2 ± 2√(3)
4
x = 1 ± √(3)
2
x = 1 + √(3) U x = 1 - √(3)
2 2