Question

i did it but it said incorrect any help thanks

Use the quadratic formula to solve the following equation. 2a-5=4/a

Answer 1

first multiply both sides of the equation by a

write your result in your next answer...

Answer 2

1/4 (5+/-r57)

Answer 3

\(\bf 2a-5=\cfrac{4}{a}\implies 2a^2-5a=4\implies {\color{blue}{ 2}}a^2{\color{red}{ -5}}a{\color{green}{ -4}}=0
\\ \quad \\

\textit{quadratic formula}\\

\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 4

2a-5=4/a
2a^2-5a-4=0

Answer 5

so it is not a=14(5+57−−√),14(5−57−−√)

Answer 6

plug in the values... what would they give you?

Answer 7

@wildfire95 looks correct to me

Answer 8

1/4 (5+/-r57)

your first equation looks right (assuming r57 means sqrt(57)

Answer 9

\[a=\frac{ 5\pm \sqrt{\left( -5 \right)^2-4*2*-4} }{ 2*2 }=\frac{ 5\pm \sqrt{25+32} }{4 }\]
\[=\frac{ 5\pm \sqrt{57} }{ 4 }\]

Answer 10

\(\bf 2a-5=\cfrac{4}{a}\implies 2a^2-5a=4\implies {\color{blue}{ 2}}a^2{\color{red}{ -5}}a{\color{green}{ -4}}=0
\\ \quad \\

\textit{quadratic formula}\\

\qquad \qquad
x= \cfrac{ - {\color{red}{ -5}} \pm \sqrt { {\color{red}{ -5}}^2 -4{\color{blue}{ (2)}}{\color{green}{ (-4)}}}}{2{\color{blue}{ (2)}}}\implies \cfrac{5\pm\sqrt{57}}{4}\)