anonymous
  • anonymous
hi
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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anonymous
  • anonymous
what is your question about this quadratic equation
anonymous
  • anonymous
x(x-4)=0 x=0 or x-4=0 x=0 or x=4
Nnesha
  • Nnesha
find the GCF what is common in both terms ?

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anonymous
  • anonymous
i need help using the quadratic formula for this problem........
Nnesha
  • Nnesha
oaky should mention that *using the qudratic formula*
anonymous
  • anonymous
sorry lol.........
Nnesha
  • Nnesha
\[\huge\rm x=\frac{ -b \pm \sqrt{b^2-4ac} }{ 2a }\] plugin a ,b c values \[\huge\rm Ax^2+Bx+C=0\] a=leading coefficient b= middle term c= constant term
anonymous
  • anonymous
\[\frac{ 4\sqrt{16 - 0} }{ 2}\]
Nnesha
  • Nnesha
a= ? b=? c= ?
anonymous
  • anonymous
actually nvm listen to @Nnesha
anonymous
  • anonymous
damn i think i just gave her the answer ... sorry about that @Nnesha
Nnesha
  • Nnesha
\[\huge\rm \color{reD}{A}x^2+\color{blue}{B}x+\color{green}{C}=0\] \[\huge\rm \color{reD}{1}x^2\color{blue}{-4}x=0\]a ,b ,c values are a= leading coefficient b=middle term c=constant term
Nnesha
  • Nnesha
so what is a , b and c in this equation 1x2 -4x = 0 ??
anonymous
  • anonymous
this is wat i got so far,,, \[\frac{ (-4±\sqrt{-4^2-4*0*1}) }{ 2*1}\]
anonymous
  • anonymous
i know how to substitute and all..its the solving part......... @Nnesha
Nnesha
  • Nnesha
put the parentheses (-4)^2
anonymous
  • anonymous
i forgot that
anonymous
  • anonymous
oops ok next.........
Nnesha
  • Nnesha
-4^2 isn't same as (-4)^2 and there is negative sign in the formula and b value is also negative \[\huge\rm \frac{\color{ReD}{-} (-4)±\sqrt{(-4)^2-(4*0*1)} }{ 2*1}\]
anonymous
  • anonymous
\[\frac{ (−(-4)±\sqrt{16−4∗0∗1}) }{ 2 }\]
Nnesha
  • Nnesha
yep :)
Nnesha
  • Nnesha
-(-4) distribute :)
anonymous
  • anonymous
\[\frac{ 4±\sqrt{1} }{ 2 }\]
anonymous
  • anonymous
???
Nnesha
  • Nnesha
hmmm
anonymous
  • anonymous
then??...
anonymous
  • anonymous
\[\frac{ 4\pm1 }{ 2 }\]
anonymous
  • anonymous
\[\frac{ 4+1 }{ 2 }=\frac{ 5 }{ 2 }\]
Nnesha
  • Nnesha
how did you get one under the square root ?
anonymous
  • anonymous
\[\frac{ 4-1 }{ 2 }=\frac{ 3 }{ 2 }\]
Nnesha
  • Nnesha
\(\color{blue}{\text{Originally Posted by}}\) @plzzhelpme \[\frac{ 4±\sqrt{1} }{ 2 }\] \(\color{blue}{\text{End of Quote}}\) here how did you get one ? :)
anonymous
  • anonymous
i subtracted 4 from 16 = 12 then 12*0*1=1
anonymous
  • anonymous
ohhh i did that part wrong.......
anonymous
  • anonymous
its 15??
Nnesha
  • Nnesha
ye ^^ try again \[\sqrt{16-(4 \times 0 \times 1)}\]
Nnesha
  • Nnesha
no not 15
Nnesha
  • Nnesha
when ou multiply a number by 0 what will you get ?
Nnesha
  • Nnesha
4 times 0 ? x times 0 = ?? chocolates times 0 ?
anonymous
  • anonymous
0
Nnesha
  • Nnesha
yes so 4 times 0 times 1 = 0 \[\sqrt{16-(4 \times 0 \times 1)}\] \[\sqrt{16-0}\]
anonymous
  • anonymous
16
Nnesha
  • Nnesha
sqrt{16} = ?
Nnesha
  • Nnesha
\(\color{blue}{\text{Originally Posted by}}\) @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!
anonymous
  • anonymous
oh ok.....thx.....
Nnesha
  • Nnesha
got the answer ?
anonymous
  • anonymous
ya 0 and 4
Nnesha
  • Nnesha
:=)
anonymous
  • anonymous
@Nnesha how would u check the solution when we only have : x=4 ?
Nnesha
  • Nnesha
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
anonymous
  • anonymous
i got this https://www.symbolab.com/solver/step-by-step/4%5E%7B2%7D-4%5Cleft(4%5Cright)%3D0/?origin=enterkey @Nnesha
Nnesha
  • Nnesha
let 's use our own pretty hands 4^32-4(4) = 0 16-16=0 both sides are equal r?
Nnesha
  • Nnesha
hurry up! gtg ;P
anonymous
  • anonymous
why 4^32 ?? @Nnesha
Nnesha
  • Nnesha
how ??
Nnesha
  • Nnesha
ohh my bad *backspace* button isn't working only on this page so yea
anonymous
  • anonymous
ok ya thats 0
Nnesha
  • Nnesha
\(\color{blue}{\text{Originally Posted by}}\) @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
Nnesha
  • Nnesha
wiill give you 16-16=0 (16-16) =0 so 0=0 both sides are equal so yes 4 is a solution
anonymous
  • anonymous
yayyy thx soo much........

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