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Zinaida [17]
2 years ago
10

B2 - 2b - 8 = 0 factoring

Mathematics
1 answer:
Paladinen [302]2 years ago
5 0

Answer:

Step-by-step explanation:

Step-1 : Multiply the coefficient of the first term by the constant   1 • -8 = -8

Step-2 : Find two factors of  -8  whose sum equals the coefficient of the middle term, which is   -2 .

     -8    +    1    =    -7

     -4    +    2    =    -2    That's it

Step-3 : Rewrite the polynomial splitting the middle term using the two factors found in step 2 above,  -4  and  2

                    b2 - 4b + 2b - 8

Step-4 : Add up the first 2 terms, pulling out like factors :

                   b • (b-4)

             Add up the last 2 terms, pulling out common factors :

                   2 • (b-4)

Step-5 : Add up the four terms of step 4 :

                   (b+2)  •  (b-4)

            Which is the desired factorization

Equation at the end of step

1

:

 (b + 2) • (b - 4)  = 0

STEP

2

:

Theory - Roots of a product

2.1    A product of several terms equals zero.

When a product of two or more terms equals zero, then at least one of the terms must be zero.

We shall now solve each term = 0 separately

In other words, we are going to solve as many equations as there are terms in the product

Any solution of term = 0 solves product = 0 as well.

Solving a Single Variable Equation:

2.2      Solve  :    b+2 = 0

Subtract  2  from both sides of the equation :

                     b = -2

Solving a Single Variable Equation:

2.3      Solve  :    b-4 = 0

Add  4  to both sides of the equation :

                     b = 4

Supplement : Solving Quadratic Equation Directly

Solving    b2-2b-8  = 0   directly

Earlier we factored this polynomial by splitting the middle term. let us now solve the equation by Completing The Square and by using the Quadratic Formula

Parabola, Finding the Vertex:

3.1      Find the Vertex of   y = b2-2b-8

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,Ab2+Bb+C,the  b -coordinate of the vertex is given by  -B/(2A) . In our case the  b  coordinate is   1.0000  

Plugging into the parabola formula   1.0000  for  b  we can calculate the  y -coordinate :

 y = 1.0 * 1.00 * 1.00 - 2.0 * 1.00 - 8.0

or   y = -9.000

Parabola, Graphing Vertex and X-Intercepts :

Root plot for :  y = b2-2b-8

Axis of Symmetry (dashed)  {b}={ 1.00}

Vertex at  {b,y} = { 1.00,-9.00}

b -Intercepts (Roots) :

Root 1 at  {b,y} = {-2.00, 0.00}

Root 2 at  {b,y} = { 4.00, 0.00}

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2 years ago
Consider the initial value problem: 2ty′=8y, y(−1)=1. Find the value of the constant C and the exponent r so that y=Ctr is the s
VikaD [51]

The correct question is:

Consider the initial value problem

2ty' = 8y, y(-1) = 1

(a) Find the value of the constant C and the exponent r such that y = Ct^r is the solution of this initial value problem.

b) Determine the largest interval of the form a < t < b on which the existence and uniqueness theorem for first order linear differential equations guarantees the existence of a unique solution.

c) What is the actual interval of existence for the solution obtained in part (a) ?

Step-by-step explanation:

Given the differential equation

2ty' = 8y

a) We need to find the value of the constant C and r, such that y = Ct^r is a solution to the differential equation together with the initial condition y(-1) = 1.

Since Ct^r is a solution to the initial value problem, it means that y = Ct^r satisfies the said problem. That is

2tdy/dt - 8y = 0

Implies

2td(Ct^r)/dt - 8(Ct^r) = 0

2tCrt^(r - 1) - 8Ct^r = 0

2Crt^r - 8Ct^r = 0

(2r - 8)Ct^r = 0

But Ct^r ≠ 0

=> 2r - 8 = 0 or r = 8/2 = 4

Now, we have r = 4, which implies that

y = Ct^4

Applying the initial condition y(-1) = 1, we put y = 1 when t = -1

1 = C(-1)^4

C = 1

So, y = t^4

b) Let y = F(x,y)................(1)

Suppose F(x, y) is continuous on some region, R = {(x, y) : x_0 − δ < x < x_0 + δ, y_0 −ę < y < y_0 + ę} containing the point (x_0, y_0). Then there exists a number δ1 (possibly smaller than δ) so that a solution y = f(x) to (1) is defined for x_0 − δ1 < x < x_0 + δ1.

Now, suppose that both F(x, y)

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the unique solution to (1) for x_0 − δ2 < x < x_0 + δ2.

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So, the solution to differential equation exists everywhere, apart from t = 0.

The interval is (-infinity, 0) n (0, infinity)

n - means intersection.

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<u>Step-by-step explanation:</u>

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GCF (240, 150) = 2 x 3 x 5

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QveST [7]
Let the width of the yard  be w.
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Recall, length l = w + 18, l = 9 + 18 = 27

Hence width, w = 9, length,l = 27

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Area of rectangular yard =  243 square feet.           
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2 years ago
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maks197457 [2]
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