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3 years ago

Find the roots of the quadratic equation w+w²/3=0

Mathematics

1 answer:

Daniel [21]3 years ago

8 0

I assume that the equation you mean is below:

$\large \boxed{w + \frac{ {w}^{2} }{3} = 0}$

To find roots for this equation, we have to get rid of the denominator. We can do by multiplying both sides by 3.

$\large{w(3) + \frac{ {w}^{2} }{3} (3) = 0(3)} \\ \large{3w + {w}^{2} = 0}$

Factor w-term out (common factor)

$\large{w(3 + w) = 0} \\ \large{w = 0 \: \: \: or \: \: \: 3 + w = 0} \\ \large{w = 0, - 3}$

Answer

The roots of quadratic equation are 0,-3

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Step-by-step explanation:

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3 years ago

Explain the steps necessary to convert a quadratic function in standard form to vertex form

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Answer:

See below.

Step-by-step explanation:

Here's an example to illustrate the method:

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First divide the first 2 terms by the coefficient of x^2 , which is 3:

= 3(x^2 - 2x) + 10

Now divide the -2 ( in -2x) by 2 and write the x^2 - 2x in the form

(x - b/2)^2 - b/2)^2 (where b = 2) , which will be equal to x^2 - 2x in a different form.

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= 3 [(x - 1)^2 -1 ] + 10

= 3(x - 1)^2 - 3 + 10

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In general form the vertex form of:

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This is not easy to commit to memory so I suggest the best way to do these conversions is to remember the general method.

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3 years ago

Read 2 more answers

Avon Edwards worked 45.5 hours at $5 per hour. He made $100 in tips but no bonuses. Anything over 40 hours is paid time-and-a-ha

posledela

Answer:

1. D

2. B

Step-by-step explanation:

Question 1:

We can get Avon's earnings using the equation $W=HR+1.5VR+B+T$

Where W is his earnings

HR is the regular hours (40 hours in this case)

V is overtime hours (hours over 40, 45.5 - 40 = 5.5 hours)

B is bonus (no bonus)

T is tips ( $100 tips, given)

and R is the rate (which is $5 per hour)

Substituting the given info into the equation, we get:

$W=HR+1.5VR+B+T\\W=(40)(5)+1.5(5.5)(5)+0+100\\W=341.25$

So avon's earnings are $341.25

Answer choice D is right.

Question 2:

The other employee needs to work $x$ hours @ $15 per hour to equate or surpass $428. So we can set up the equation shown below and solve for x:

$(15)(x)=428\\x=\frac{428}{15}\\x=28.53$

Rounded to 1 decimal place, this is about 28.5 hours

Answer choice B is right.

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This 1 seems really complicated

Fofino [41]

The solution to this system set is: "x = 4" , "y = 0" ; or write as: [4, 0] .
________________________________________________________
Given:
________________________________________________________
y = - 4x + 16 ;

4y − x + 4 = 0 ;
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"Solve the system using substitution" .
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First, let us simplify the second equation given, to get rid of the "0" ;

→ 4y − x + 4 = 0 ;

Subtract "4" from each side of the equation ;

→ 4y − x + 4 − 4 = 0 − 4 ;

→ 4y − x = -4 ;
________________________________________________________
So, we can now rewrite the two (2) equations in the given system:
________________________________________________________

y = - 4x + 16 ; ===> Refer to this as "Equation 1" ;

4y − x = -4 ; ===> Refer to this as "Equation 2" ;
________________________________________________________
Solve for "x" and "y" ; using "substitution" :
________________________________________________________
We are given, as "Equation 1" ;

→ " y = - 4x + 16 " ;
_______________________________________________________
→ Plug in this value for [all of] the value[s] for "y" into {"Equation 2"} ;

to solve for "x" ; as follows:
_______________________________________________________
Note: "Equation 2" :

→ " 4y − x = - 4 " ;
_________________________________________________
Substitute the value for "y" {i.e., the value provided for "y"; in "Equation 1}" ;
for into the this [rewritten version of] "Equation 2" ;
→ and "rewrite the equation" ;

→ as follows:
_________________________________________________

→ " 4 (-4x + 16) − x = -4 " ;
_________________________________________________
Note the "distributive property" of multiplication :
_________________________________________________

a(b + c) = ab + ac ; AND:

a(b − c) = ab − ac .
_________________________________________________
As such:

We have:

→ " 4 (-4x + 16) − x = - 4 " ;
_________________________________________________
AND:

→ "4 (-4x + 16) " = (4* -4x) + (4 *16) = " -16x + 64 " ;
_________________________________________________
Now, we can write the entire equation:

→ " -16x + 64 − x = - 4 " ;

Note: " - 16x − x = -16x − 1x = -17x " ;

→ " -17x + 64 = - 4 " ; Solve for "x" ;

Subtract "64" from EACH SIDE of the equation:

→ " -17x + 64 − 64 = - 4 − 64 " ;

to get:

→ " -17x = -68 " ;

Divide EACH side of the equation by "-17" ;
to isolate "x" on one side of the equation; & to solve for "x" ;

→ -17x / -17 = -68/ -17 ;

to get:

→ x = 4 ;
______________________________________
Now, Plug this value for "x" ; into "{Equation 1"} ;

which is: " y = -4x + 16" ; to solve for "y".
______________________________________

→ y = -4(4) + 16 ;

= -16 + 16 ;

→ y = 0 .
_________________________________________________________
The solution to this system set is: "x = 4" , "y = 0" ; or write as: [4, 0] .
_________________________________________________________
Now, let us check our answers—as directed in this very question itself ;
_________________________________________________________
→ Given the TWO (2) originally given equations in the system of equation; as they were originally rewitten;

→ Let us check;

→ For EACH of these 2 (TWO) equations; do these two equations hold true {i.e. do EACH SIDE of these equations have equal values on each side} ; when we "plug in" our obtained values of "4" (for "x") ; and "0" for "y" ??? ;

→ Consider the first equation given in our problem, as originally written in the system of equations:

→ " y = - 4x + 16 " ;

→ Substitute: "4" for "x" and "0" for "y" ; When done, are both sides equal?

→ "0 = ? -4(4) + 16 " ?? ; → "0 = ? -16 + 16 ?? " ; → Yes! ;

{Actually, that is how we obtained our value for "y" initially.}.

→ Now, let us check the other equation given—as originally written in this very question:

→ " 4y − x + 4 = ?? 0 ??? " ;

→ Let us "plug in" our obtained values into the equation;

{that is: "4" for the "x-value" ; & "0" for the "y-value" ;

→ to see if the "other side of the equation" {i.e., the "right-hand side"} holds true {i.e., in the case of this very equation—is equal to "0".}.

→ " 4(0) − 4 + 4 = ? 0 ?? " ;

→ " 0 − 4 + 4 = ? 0 ?? " ;

→ " - 4 + 4 = ? 0 ?? " ; Yes!
_____________________________________________________
→ As such, from "checking [our] answer (obtained values)" , we can be reasonably certain that our answer [obtained values] :
_____________________________________________________
→ "x = 4" and "y = 0" ; or; write as: [0, 4] ; are correct.
_____________________________________________________
Hope this lenghty explanation is of help! Best wishes!
_____________________________________________________

7 0

3 years ago

Find the roots of the quadratic equation w+w²/3=0​

Find the roots of the quadratic equation w+w²/3=0