Question

solve the following problem by using the Quadratic Formula

5w^2 - 3

Answer 1

i don't know how to and the second term, in order to apply the cuadratic formula

Answer 2

how to find *

Answer 3

\(\bf 5w^2 - 3\implies 5w^2 +0w- 3\\
\textit{quadratic formula}\\
{\color{blue}{ 5}}w^2{\color{red}{ +0}}w{\color{green}{ -3}}
\qquad \qquad
w= \cfrac{ - {\color{red}{ b}} \pm \sqrt { {\color{red}{ b}}^2 -4{\color{blue}{ a}}{\color{green}{ c}}}}{2{\color{blue}{ a}}}\)

Answer 4

what about : \[(\sqrt{5}w + \sqrt{3}) (\sqrt{5}w - \sqrt{3})\]
can I do this ?

Answer 5

you could factor it, yes, however -> "olve the following problem by using the \(\bf Quadratic\ Formula\)"

Answer 6

but that would be an imaginary solution

Answer 7

yes.. and factoring will give the same anyhow...but you're asked to use the quadratic formula

Answer 8

ok so that woulde all in this case:

5w^2-3 = 0

Answer 9

yes

Answer 10

"solving" a quadratic equation, means just finding the x-intercepts and thus setting y = 0

Answer 11

ok Im going to do the whole process and then call yo just to check it out

Answer 12

ok

Answer 13

i was thinking if instead just...

w^2 = 3
\[w =\pm \sqrt{3}\]

Answer 14

i have to get the same result using the quadratic ?

Answer 15

yeap, simplifying or factoring, or using the quadratic, will yield the same result

Answer 16

i meant : \[w = \pm \sqrt{\frac{ 3 }{ 5 }}\]

Answer 17

ok

Answer 18

\[x = \frac{ -(0) \pm \sqrt{0^{2}-4(5)(-3)} }{ 2(5) }\]

Answer 19

\[x =\frac{ \pm \sqrt{-4(-3)(5)} }{ 10 }\]

Answer 20

\[x =\frac{ \pm \sqrt{60} }{ 10 }\]