my equation is f(x)= x^2-16x-100 and i am using the complete the box method im really stuck

Question

my equation is f(x)= x^2-16x-100 and i am using the complete the box method im really stuck

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3 months 2021-10-18T07:10:14+00:00 1 Answer 0 views 0

x2-16x+100=0

Two solutions were found :

x =(16-√-144)/2=8-6i= 8.0000-6.0000i

x =(16+√-144)/2=8+6i= 8.0000+6.0000i

Step by step solution :

Step  1  :

Trying to factor by splitting the middle term

1.1     Factoring  x2-16x+100

The first term is,  x2  its coefficient is  1 .

The middle term is,  -16x  its coefficient is  -16 .

The last term, “the constant”, is  +100

Step-1 : Multiply the coefficient of the first term by the constant   1 • 100 = 100

Step-2 : Find two factors of  100  whose sum equals the coefficient of the middle term, which is   -16 .

-100    +    -1    =    -101

-50    +    -2    =    -52

-25    +    -4    =    -29

-20    +    -5    =    -25

-10    +    -10    =    -20

-5    +    -20    =    -25

For tidiness, printing of 12 lines which failed to find two such factors, was suppressed

Observation : No two such factors can be found !!

Conclusion : Trinomial can not be factored

Equation at the end of step  1  :

x2 – 16x + 100  = 0

Step  2  :

Parabola, Finding the Vertex :

2.1      Find the Vertex of   y = x2-16x+100

Parabolas have a highest or a lowest point called the Vertex .   Our parabola opens up and accordingly has a lowest point (AKA absolute minimum) .   We know this even before plotting  “y”  because the coefficient of the first term, 1 , is positive (greater than zero).

Each parabola has a vertical line of symmetry that passes through its vertex. Because of this symmetry, the line of symmetry would, for example, pass through the midpoint of the two  x -intercepts (roots or solutions) of the parabola. That is, if the parabola has indeed two real solutions.

Parabolas can model many real life situations, such as the height above ground, of an object thrown upward, after some period of time. The vertex of the parabola can provide us with information, such as the maximum height that object, thrown upwards, can reach. For this reason we want to be able to find the coordinates of the vertex.

For any parabola,Ax2+Bx+C,the  x -coordinate of the vertex is given by  -B/(2A) . In our case the  x  coordinate is   8.0000

Plugging into the parabola formula   8.0000  for  x  we can calculate the  y -coordinate :

y = 1.0 * 8.00 * 8.00 – 16.0 * 8.00 + 100.0

or   y = 36.000

Parabola, Graphing Vertex and X-Intercepts :

Root plot for :  y = x2-16x+100

Axis of Symmetry (dashed)  {x}={ 8.00}

Vertex at  {x,y} = { 8.00,36.00}

Function has no real roots

Solve Quadratic Equation by Completing The Square

2.2     Solving   x2-16x+100 = 0 by Completing The Square .

Subtract  100  from both side of the equation :

x2-16x = -100

Now the clever bit: Take the coefficient of  x , which is  16 , divide by two, giving  8 , and finally square it giving  64

Add  64  to both sides of the equation :

On the right hand side we have :

-100  +  64    or,  (-100/1)+(64/1)

The common denominator of the two fractions is  1   Adding  (-100/1)+(64/1)  gives  -36/1

So adding to both sides we finally get :

x2-16x+64 = -36

Adding  64  has completed the left hand side into a perfect square :

x2-16x+64  =

(x-8) • (x-8)  =

(x-8)2

Things which are equal to the same thing are also equal to one another. Since

x2-16x+64 = -36 and

x2-16x+64 = (x-8)2

then, according to the law of transitivity,

(x-8)2 = -36

We’ll refer to this Equation as  Eq. #2.2.1

The Square Root Principle says that When two things are equal, their square roots are equal.

Note that the square root of

(x-8)2   is

(x-8)2/2 =

(x-8)1 =

x-8

Now, applying the Square Root Principle to  Eq. #2.2.1  we get:

x-8 = √ -36

Add  8  to both sides to obtain:

x = 8 + √ -36

In Math,  i  is called the imaginary unit. It satisfies   i2  =-1. Both   i   and   -i   are the square roots of   -1

Since a square root has two values, one positive and the other negative

x2 – 16x + 100 = 0

has two solutions:

x = 8 + √ 36 •  i

or

x = 8 – √ 36 •  i

2.3     Solving    x2-16x+100 = 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   =     1

B   =   -16

C   =  100

Accordingly,  B2  –  4AC   =

256 – 400 =

-144

16 ± √ -144

x  =    ——————

2

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,√ -144  =

√ 144 • (-1)  =

√ 144  • √ -1   =

±  √ 144  • i

Can  √ 144 be simplified ?

Yes!   The prime factorization of  144   is

2•2•2•2•3•3

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).

√ 144   =  √ 2•2•2•2•3•3   =2•2•3•√ 1   =

±  12 • √ 1   =

±  12

So now we are looking at:

x  =  ( 16 ± 12i ) / 2

Two imaginary solutions :

x =(16+√-144)/2=8+6i= 8.0000+6.0000i

or:

x =(16-√-144)/2=8-6i= 8.0000-6.0000i

Two solutions were found :

x =(16-√-144)/2=8-6i= 8.0000-6.0000i

x =(16+√-144)/2=8+6i= 8.0000+6.0000i

Step-by-step explanation: