Two solutions were found :
x =(4-√-64)/-10=2/-5+4i/5= -0.4000-0.8000i
x =(4+√-64)/-10=2/-5-4i/5= -0.4000+0.8000i
Step by step solution :
Step 1 :
Equation at the end of step 1 :
((0 - 5x2) - 4x) - 4 = 0
Step 2 :
Step 3 :
Pulling out like terms :
3.1 Pull out like factors :
-5x2 - 4x - 4 = -1 • (5x2 + 4x + 4)
Trying to factor by splitting the middle term
3.2 Factoring 5x2 + 4x + 4
The first term is, 5x2 its coefficient is 5 .
The middle term is, +4x its coefficient is 4 .
The last term, "the constant", is +4
Step-1 : Multiply the coefficient of the first term by the constant 5 • 4 = 20
Step-2 : Find two factors of 20 whose sum equals the coefficient of the middle term, which is 4 .
Observation : No two such factors can be found !!
Conclusion : Trinomial can not be factored
Equation at the end of step 3 :
-5x2 - 4x - 4 = 0
Step 4 :
Parabola, Finding the Vertex :
4.1 Find the Vertex of y = -5x2-4x-4
For any parabola,Ax2+Bx+C,the x -coordinate of the vertex is given by -B/(2A) . In our case the x coordinate is -0.4000
Plugging into the parabola formula -0.4000 for x we can calculate the y -coordinate :
y = -5.0 * -0.40 * -0.40 - 4.0 * -0.40 - 4.0
or y = -3.200
Parabola, Graphing Vertex and X-Intercepts :
Root plot for : y = -5x2-4x-4
Axis of Symmetry (dashed) {x}={-0.40}
Vertex at {x,y} = {-0.40,-3.20}
Function has no real roots
Solve Quadratic Equation by Completing The Square
4.2 Solving -5x2-4x-4 = 0 by Completing The Square .
Multiply both sides of the equation by (-1) to obtain positive coefficient for the first term:
5x2+4x+4 = 0 Divide both sides of the equation by 5 to have 1 as the coefficient of the first term :
x2+(4/5)x+(4/5) = 0
Subtract 4/5 from both side of the equation :
x2+(4/5)x = -4/5
Add 4/25 to both sides of the equation :
On the right hand side we have :
-4/5 + 4/25 The common denominator of the two fractions is 25 Adding (-20/25)+(4/25) gives -16/25
So adding to both sides we finally get :
x2+(4/5)x+(4/25) = -16/25
Adding 4/25 has completed the left hand side into a perfect square :
x2+(4/5)x+(4/25) =
(x+(2/5)) • (x+(2/5)) =
(x+(2/5))2
Things which are equal to the same thing are also equal to one another. Since
x2+(4/5)x+(4/25) = -16/25 and
x2+(4/5)x+(4/25) = (x+(2/5))2
then, according to the law of transitivity,
(x+(2/5))2 = -16/25
Note that the square root of
(x+(2/5))2 is
(x+(2/5))2/2 =
(x+(2/5))1 =
x+(2/5)
Now, applying the Square Root Principle to Eq. #4.2.1 we get:
x+(2/5) = √ -16/25
Subtract 2/5 from both sides to obtain:
x = -2/5 + √ -16/25
Since a square root has two values, one positive and the other negative
x2 + (4/5)x + (4/5) = 0
has two solutions:
x = -2/5 + √ 16/25 • i
or
x = -2/5 - √ 16/25 • i
Note that √ 16/25 can be written as
√ 16 / √ 25 which is 4 / 5
Solve Quadratic Equation using the Quadratic Formula
4.3 Solving -5x2-4x-4 = 0 by the Quadratic Formula .
According to the Quadratic Formula, x , the solution for Ax2+Bx+C = 0 , where A, B and C are numbers, often called coefficients, is given by :
- B ± √ B2-4AC
x = ————————
2A
In our case, A = -5
B = -4
C = -4
Accordingly, B2 - 4AC =
16 - 80 =
-64
Applying the quadratic formula :
4 ± √ -64
x = —————
-10
In the set of real numbers, negative numbers do not have square roots. A new set of numbers, called complex, was invented so that negative numbers would have a square root. These numbers are written (a+b*i)
Both i and -i are the square roots of minus 1
Accordingly,√ -64 =
√ 64 • (-1) =
√ 64 • √ -1 =
± √ 64 • i
Can √ 64 be simplified ?
Yes! The prime factorization of 64 is
2•2•2•2•2•2
To be able to remove something from under the radical, there have to be 2 instances of it (because we are taking a square i.e. second root).
√ 64 = √ 2•2•2•2•2•2 =2•2•2•√ 1 =
± 8 • √ 1 =
± 8
So now we are looking at:
x = ( 4 ± 8i ) / -10
Two imaginary solutions :
x =(4+√-64)/-10=2/-5-4i/5= -0.4000+0.8000i
or:
x =(4-√-64)/-10=2/-5+4i/5= -0.4000-0.8000i
Two solutions were found :
x =(4-√-64)/-10=2/-5+4i/5= -0.4000-0.8000i
x =(4+√-64)/-10=2/-5-4i/5= -0.4000+0.8000i
<em>hope i helped</em>
<em>-Rin:)</em>
and inner radius 
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