+0

+1
485
5
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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 $c$?

Jun 14, 2018

#1
+1

Can somebody translate this into proper LaTex?

Jun 14, 2018
#2
+117
+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
+101768
+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.

Jun 15, 2018
#4
+985
+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

Jun 15, 2018
#5
+101768
0

ahhh

1/2    not   12

We have another forensic mathematician here. :)

Thanks Gavin :)

Melody  Jun 15, 2018