Question

Use the quadratic formula to solve
10v^2+8v-1= 0

Answer 1

(-2+-.5/sqrt(26)/5

Answer 2

\(\bf {\color{blue}{ 10}}v^2{\color{red}{ +8}}v{\color{green}{ -1}}= 0\qquad
\begin{array}{llll}
\textit{quadratic formula}\\ \quad \\
x= \cfrac{ - {\color{red}{ b}} \pm \sqrt { {\color{red}{ b}}^2 -4{\color{blue}{ a}}{\color{green}{ c}}}}{2{\color{blue}{ a}}}
\end{array} \)

Answer 3

I know that. I think I'm having trouble with the discriminant. Like simplifying the radical.

Answer 4

\(\bf {\color{blue}{ 10}}v^2{\color{red}{ +8}}v{\color{green}{ -1}}= 0\qquad
\begin{array}{llll}
\textit{quadratic formula}\\ \quad \\
x= \cfrac{ - {\color{red}{ 8}} \pm \sqrt { {\color{red}{ 8}}^2 -4{\color{blue}{ (10)}}{\color{green}{ (-1)}}}}{2{\color{blue}{ (10)}}}
\end{array} \\ \quad \\

x= \cfrac{ - 8 \pm \sqrt { 64 -(-40)}}{20}\)

Answer 5

I got that part. So it's 104 and I simplified it and got \[2\sqrt{26}\]

Answer 6

What do I do next?

Answer 7

\(\bf x= \cfrac{ - 8 \pm \sqrt { 64 -(-40)}}{20}\implies x= \cfrac{ - 8 \pm \sqrt { 104}}{20}\implies x= \cfrac{ - 8 \pm \sqrt { 2^2\cdot 26}}{20}\\ \quad \\

x= \cfrac{ - \cancel{8} \pm \cancel{2}\sqrt {26}}{\cancel{20}}

\implies x=\cfrac{-4\pm \sqrt{26}}{10}\)

Answer 8

How did you get 4 ?

Answer 9

\(\bf x= \cfrac{ \frac{- 8 \pm \sqrt {26}}{2}}{\frac{20}{2}}\implies
x=\cfrac{\frac{-8}{{\color{red}{ 2}}}\pm \frac{2\sqrt{26}}{{\color{red}{ 2}}}}{\frac{20}{{\color{red}{ 2}}}}

\implies x=\cfrac{-4\pm \sqrt{26}}{10}\)

Answer 10

another way to look at it would be, take common factor to the numerator, then cancel out