Question

Use the quadratic formula to solve x2 - 2x - 3 = 0.
x = -3 or x = 1 x = -1 or x = 3
x = 1 or x = -2 x = -2 or x = -1

Answer 1

\(\bf \textit{quadratic formula}\\
{\color{blue}{ 1}}x^2{\color{red}{ -2}}x{\color{green}{ -3}}=0
\qquad \qquad
x= \cfrac{ - {\color{red}{ b}} \pm \sqrt { {\color{red}{ b}}^2 -4{\color{blue}{ a}}{\color{green}{ c}}}}{2{\color{blue}{ a}}}\)

Answer 2

?

Answer 3

im not understanding

Answer 4

\(\bf \textit{quadratic formula}\\
{\color{blue}{ 1}}x^2{\color{red}{ -2}}x{\color{green}{ -3}}=0
\qquad \qquad
x= \cfrac{ - {\color{red}{ (-2)}} \pm \sqrt { {\color{red}{ (-2)}}^2 -4{\color{blue}{ (1)}}{\color{green}{ (-3)}}}}{2{\color{blue}{ (1)}}}\)

Answer 5

x=1 and x=-2?

Answer 6

well \({\bf \textit{quadratic formula}\\
{\color{blue}{ 1}}x^2{\color{red}{ -2}}x{\color{green}{ -3}}=0
\qquad \qquad
x= \cfrac{ - {\color{red}{ (-2)}} \pm \sqrt { {\color{red}{ (-2)}}^2 -4{\color{blue}{ (1)}}{\color{green}{ (-3)}}}}{2{\color{blue}{ (1)}}}
\\ \quad \\
x=\cfrac{2\pm\sqrt{4+12}}{2}\implies x=\cfrac{2\pm\sqrt{16}}{2}\implies x=\cfrac{2\pm\sqrt{4^2}}{2}
\\ \quad \\
x=\cfrac{\cancel{ 2 }\pm \cancel{ 4 }}{\cancel{ 2 }}\implies x= 1\pm 2}\to
\large \begin{cases}
x=1+2\to &3\\
x=1-2\to &-1
\end{cases}\)