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2 years ago

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

1 answer:

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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Oduvanchick [21]

Consider triangle DEF with vertices D(1,1,1), E(1,-5,2) and F(-2,2,7).

1. Find

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Then

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2. Find

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Then

$\cos \angle E=\dfrac{0\cdot (-2)+6\cdot 7+(-1)\cdot 5}{\sqrt{0^2+6^2+(-1)^2}\cdot \sqrt{(-3)^2+7^2+5^2}}=\dfrac{37}{\sqrt{37} \cdot \sqrt{83} }=\sqrt{\dfrac{37}{83}}.$

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