Question

hi

Answer 1

what is your question about this quadratic equation

Answer 2

x(x-4)=0
x=0 or x-4=0
x=0 or x=4

Answer 3

find the GCF
what is common in both terms ?

Answer 4

i need help using the quadratic formula for this problem........

Answer 5

oaky should mention that
*using the qudratic formula*

Answer 6

sorry lol.........

Answer 7

\[\huge\rm x=\frac{ -b \pm \sqrt{b^2-4ac} }{ 2a }\] plugin a ,b c values \[\huge\rm Ax^2+Bx+C=0\]

a=leading coefficient
b= middle term
c= constant term

Answer 8

\[\frac{ 4\sqrt{16 - 0} }{ 2}\]

Answer 9

a= ?
b=?
c= ?

Answer 10

actually nvm listen to @Nnesha

Answer 11

damn i think i just gave her the answer ... sorry about that @Nnesha

Answer 12

\[\huge\rm \color{reD}{A}x^2+\color{blue}{B}x+\color{green}{C}=0\]
\[\huge\rm \color{reD}{1}x^2\color{blue}{-4}x=0\]a ,b ,c values are
a= leading coefficient
b=middle term
c=constant term

Answer 13

so what is a , b and c
in this equation 1x2 -4x = 0 ??

Answer 14

this is wat i got so far,,,

\[\frac{ (-4±\sqrt{-4^2-4*0*1}) }{ 2*1}\]

Answer 15

i know how to substitute and all..its the solving part......... @Nnesha

Answer 16

put the parentheses (-4)^2

Answer 17

i forgot that

Answer 18

oops ok next.........

Answer 19

-4^2 isn't same as (-4)^2 and there is negative sign in the formula and b value is also negative

\[\huge\rm \frac{\color{ReD}{-} (-4)±\sqrt{(-4)^2-(4*0*1)} }{ 2*1}\]

Answer 20

\[\frac{ (−(-4)±\sqrt{16−4∗0∗1}) }{ 2 }\]

Answer 21

yep :)

Answer 22

-(-4) distribute :)

Answer 23

\[\frac{ 4±\sqrt{1} }{ 2 }\]

Answer 24

???

Answer 25

hmmm

Answer 26

then??...

Answer 27

\[\frac{ 4\pm1 }{ 2 }\]

Answer 28

\[\frac{ 4+1 }{ 2 }=\frac{ 5 }{ 2 }\]

Answer 29

how did you get one under the square root ?

Answer 30

\[\frac{ 4-1 }{ 2 }=\frac{ 3 }{ 2 }\]

Answer 31

\(\color{blue}{\text{Originally Posted by}}\) @plzzhelpme
\[\frac{ 4±\sqrt{1} }{ 2 }\]
\(\color{blue}{\text{End of Quote}}\)
here how did you get one ? :)

Answer 32

i subtracted 4 from 16 = 12
then 12*0*1=1

Answer 33

ohhh i did that part wrong.......

Answer 34

its 15??

Answer 35

ye ^^ try again
\[\sqrt{16-(4 \times 0 \times 1)}\]

Answer 36

no not 15

Answer 37

when ou multiply a number by 0 what will you get ?

Answer 38

4 times 0 ?
x times 0 = ??
chocolates times 0 ?

Answer 39

0

Answer 40

yes so 4 times 0 times 1 = 0
\[\sqrt{16-(4 \times 0 \times 1)}\]
\[\sqrt{16-0}\]

Answer 41

16

Answer 42

sqrt{16} = ?

Answer 43

\(\color{blue}{\text{Originally Posted by}}\) @plzzhelpme
\[\frac{ 4±\sqrt{1} }{ 2 }\]
\(\color{blue}{\text{End of Quote}}\)
so it's
\[x=\frac{ 4±\sqrt{16} }{ 2 }\]
take square root of 16 and then solve!

Answer 44

oh ok.....thx.....

Answer 45

got the answer ?

Answer 46

ya 0 and 4

Answer 47

:=)

Answer 48

@Nnesha how would u check the solution when we only have : x=4 ?

Answer 49

plug it into the equation if both sides are equal then yes 4 is a solution of that equation
and if both sides are not equal then nope 4 isn't a solution of that equation

Answer 50

i got this
https://www.symbolab.com/solver/step-by-step/4%5E%7B2%7D-4%5Cleft(4%5Cright)%3D0/?origin=enterkey
@Nnesha

Answer 51

let 's use our own pretty hands
4^32-4(4) = 0
16-16=0
both sides are equal r?

Answer 52

hurry up! gtg ;P

Answer 53

why 4^32 ?? @Nnesha

Answer 54

how ??

Answer 55

ohh my bad *backspace* button isn't working only on this page so yea

Answer 56

ok ya thats 0

Answer 57

\(\color{blue}{\text{Originally Posted by}}\) @Nnesha
let 's use our own pretty hands
4^32-4(4) = 0
16-16=0
both sides are equal r?
\(\color{blue}{\text{End of Quote}}\)
correction
4^2 -4(4)=0

Answer 58

wiill give you 16-16=0
(16-16) =0
so 0=0 both sides are equal so yes 4 is a solution

Answer 59

yayyy thx soo much........