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Over [174]
3 years ago
15

-3x^{2}-21x-54 what are the zeros? (Solutions)

Mathematics
1 answer:
Irina-Kira [14]3 years ago
4 0
The only way to solve if it is equal to something
assuming that the teacher wanted you to make it equal to zero do
0=-3x^2-21x-54

remember if we can do
xy=0 then assume x and y=0

so factor

0=-3x^2-21x-54
undistribute the -3
0=-3(x^2+7x+18)
remember 0 times anything=0 so
x^2+7x+18 must equal zero
use quadratice formula which is

if you have
ax^2+bx+c=0 then
x=\frac{-b+/- \sqrt{b^{2}-4ac} }{2a}

x^2+7x+18
a=1
b=7
c=18

x=\frac{-7+/- \sqrt{7^{2}-4(1)(18)} }{2(1)}
x=\frac{-7+/- \sqrt{49-72} }{2}
x=\frac{-7+/- \sqrt{-23} }{2}
i=√-1
x=\frac{-7+/- i\sqrt{23} }{2}



the zerose would be
x=\frac{-7+ i\sqrt{23} }{2} or \frac{-7- i\sqrt{23} }{2}




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  (a)  ΔARS ≅ ΔAQT

Step-by-step explanation:

The theorem being used to show congruence is ASA. In one of the triangles, the angles are 1 and R, and the side between them is AR. The triangle containing those angles and that side is ΔARS.

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