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If the roots of the quadratic equation $\frac12x^2+99x+c=0$ are $x=-99+\sqrt{8001}$ and $x=-99-\sqrt{8001}$, then what is the value of $c$?

Lightning  Jun 14, 2018
 #1
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+1

Can somebody translate this into proper LaTex?

Guest Jun 14, 2018
 #2
avatar+84 
+1

If the roots of the quadratic equation \(12x^2+99x+c=0\) are \(x = -99+\sqrt{8001}\) and \(x=-99-\sqrt{8001}\), then what is the value of \(c\)?

ChowMein  Jun 15, 2018
 #3
avatar+92806 
+1

\(12x^2+99x+c=0\)

 

\(x = -99\pm\sqrt{8001}\)

 

I think you have written the question wrong. I think both those answers are suppposed to be over 24.

Melody  Jun 15, 2018
 #4
avatar+867 
+2

If the roots of the quadratic equation \(\frac12x^2+99x+c=0 \) are \(x=-99+\sqrt{8001}\) and \(x=-99-\sqrt{8001},\) then what is the value of \(c\)?

 

In the factored form of a quadratic equation: \((x-b)(x-c)=0\) The two solutions are \(x_1=b;x_2=c\)
\([x-(-99-\sqrt{8001})][x-(-99+\sqrt{8001})]=0\) We used the logic above the reach this conclusion. 
\((x+99+\sqrt{8001})(x+99-\sqrt{8001})=0\) Simplifying
\(x^2+198x+1800=0\) Finalizing
\(\frac12x^2+99x+900=0 \) Dividing by 2 to get the form the question presents

 

\(c=\boxed{900}\)

 

I hope this helped,

 

Gavin

GYanggg  Jun 15, 2018
 #5
avatar+92806 
0

ahhh 

1/2    not   12

We have another forensic mathematician here. :)

Thanks Gavin :)

Melody  Jun 15, 2018

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