a) Factorise : 32 a^4 + 500a
Answers
Answer:
500+32a3=0
Three solutions were found :
a =(10-√-300)/8=(5-5i√ 3 )/4= 1.2500-2.1651i
a =(10+√-300)/8=(5+5i√ 3 )/4= 1.2500+2.1651i
a = -5/2 = -2.500
Step by step solution :
Step 1 :
Equation at the end of step 1 :
500 + 25a3 = 0
Step 2 :
Step 3 :
Pulling out like terms :
3.1 Pull out like factors :
32a3 + 500 = 4 • (8a3 + 125)
Trying to factor as a Sum of Cubes :
3.2 Factoring: 8a3 + 125
Theory : A sum of two perfect cubes, a3 + b3 can be factored into :
(a+b) • (a2-ab+b2)
Proof : (a+b) • (a2-ab+b2) =
a3-a2b+ab2+ba2-b2a+b3 =
a3+(a2b-ba2)+(ab2-b2a)+b3=
a3+0+0+b3=
a3+b3
Check : 8 is the cube of 2
Check : 125 is the cube of 5
Check : a3 is the cube of a1
Factorization is :
(2a + 5) • (4a2 - 10a + 25)
Trying to factor by splitting the middle term
3.3 Factoring 4a2 - 10a + 25
The first term is, 4a2 its coefficient is 4 .
The middle term is, -10a its coefficient is -10 .
The last term, "the constant", is +25
Step-1 : Multiply the coefficient of the first term by the constant 4 • 25 = 100
Step-2 : Find two factors of 100 whose sum equals the coefficient of the middle term, which is -10 .
-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 3 :
4 • (2a + 5) • (4a2 - 10a + 25) = 0
Step 4 :
Theory - Roots of a product :
4.1 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.
Equations which are never true :
4.2 Solve : 4 = 0
This equation has no solution.
A a non-zero constant never equals zero.
Solving a Single Variable Equation :
4.3 Solve : 2a+5 = 0
Subtract 5 from both sides of the equation :
2a = -5
Divide both sides of the equation by 2:
a = -5/2 = -2.500
Parabola, Finding the Vertex :
4.4 Find the Vertex of y = 4a2-10a+25
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, 4 , 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,Aa2+Ba+C,the a -coordinate of the vertex is given by -B/(2A) . In our case the a coordinate is 1.2500
Plugging into the parabola formula 1.2500 for a we can calculate the y -coordinate :
y = 4.0 * 1.25 * 1.25 - 10.0 * 1.25 + 25.0
or y = 18.750
Parabola, Graphing Vertex and X-Intercepts :
Root plot for : y = 4a2-10a+25
Axis of Symmetry (dashed) {a}={ 1.25}
Vertex at {a,y} = { 1.25,18.75}
Function has no real roots
Step-by-step explanation:
two things are equal, their square roots are equal.
Note that the square root of
(a-(5/4))2 is
(a-(5/4))2/2 =
(a-(5/4))1 =
a-(5/4)
Now, applying the Square Root Principle to Eq. #4.5.1 we get:
a-(5/4) = √ -75/16
Add 5/4 to both sides to obtain:
a = 5/4 + √ -75/16
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
a2 - (5/2)a + (25/4) = 0
has two solutions:
a = 5/4 + √ 75/16 • i
or
a = 5/4 - √ 75/16 • i
Note that √ 75/16 can be written as
√ 75 / √ 16 which is √ 75 / 4
Solve Quadratic Equation using the Quadratic Formula
4.6 Solving 4a2-10a+25 = 0 by the Quadratic Formula .
According to the Quadratic Formula, a , the solution for Aa2+Ba+C = 0 , where A, B and C are numbers, often called coefficients, is given by :
- B ± √ B2-4AC
a = ————————
2A
In our case, A = 4
B = -10
C = 25
Accordingly, B2 - 4AC =
100 - 400 =
-300
Applying the quadratic formula :
10 ± √ -300
a = ——————
8
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,√ -300 =
√ 300 • (-1) =
√ 300 • √ -1 =
±