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Stells [14]
3 years ago
12

Determine the type and number of solutions of −4x^2 − 4x − 1 = 0

Mathematics
1 answer:
user100 [1]3 years ago
4 0
The question is about types and number of soultions so you must use b^2-4ac which is the experssion under the root in the quadratic formula
so
if
b^2-4ac=0 then there is one real soultion
if
b^2-4ac<0 then there are two imaginary soultions that differ in the sign between them
if
b^2-4ac>0 then there are two real soultions
so a polynomial is written in the form
ax^2+bx+c=0
so a=-4
b=-4
c=-1
so b^2-4ac=16-16=0
so one real soultion is the answer you are looking
type:real
number:1
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Can someone pls help??
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Answer:

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distance = sqrt ((x2-x1)^2+(y2-y1)^2)

We know Point B and the x coordinate of point A as well as the distance

15 = sqrt ((-4-5)^2+(-2-y1)^2)

15 = sqrt ((-9)^2+(-2-y1)^2)

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Square each side

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225 =  81+(-2-y1)^2

Subtract 81 from each side

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Take the square root of each side

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Add 2 to each side

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Multiply by -1

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The y coordinates for A are either 10 or -14

Now move B to -7,-2

15 = sqrt ((-7-5)^2+(-2-y1)^2)

15 = sqrt ((-12)^2+(-2-y1)^2)

15 = sqrt (144+(-2-y1)^2)

Square each side

15^2 =sqrt (144+(-2-y1)^2)^2

225 =  144+(-2-y1)^2

Subtract 144 from each side

225-144 = 144-144 +(-2-y1)^2

81 = (-2-y1)^2

Take the square root of each side

±sqrt 81 = sqrt(-2-y1)^2

±9 = (-2-y1)

Add 2 to each side

2±9 = 2-2-y1

2±9 = -y1

11 = -y1  or -7 = -y1

Multiply by -1

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