P2 + 5P+8 solve equation
Answers
Answer:
p2-5p+8=0
Two solutions were found :
p =(5-√-7)/2=(5-i√ 7 )/2= 2.5000-1.3229i
p =(5+√-7)/2=(5+i√ 7 )/2= 2.5000+1.3229i
Step-by-step explanation:
Step 1 :
Trying to factor by splitting the middle term
1.1 Factoring p2-5p+8
The first term is, p2 its coefficient is 1 .
The middle term is, -5p its coefficient is -5 .
The last term, "the constant", is +8
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 -5 .
-8 + -1 = -9
-4 + -2 = -6
-2 + -4 = -6
-1 + -8 = -9
1 + 8 = 9
2 + 4 = 6
4 + 2 = 6
8 + 1 = 9
Observation : No two such factors can be found !!
Conclusion : Trinomial can not be factored
Solve Quadratic Equation by Completing The Square
2.2 Solving p2-5p+8 = 0 by Completing The Square .
Subtract 8 from both side of the equation :
p2-5p = -8
Now the clever bit: Take the coefficient of p , which is 5 , divide by two, giving 5/2 , and finally square it giving 25/4
Add 25/4 to both sides of the equation :
On the right hand side we have :
-8 + 25/4 or, (-8/1)+(25/4)
The common denominator of the two fractions is 4 Adding (-32/4)+(25/4) gives -7/4
So adding to both sides we finally get :
p2-5p+(25/4) = -7/4
Adding 25/4 has completed the left hand side into a perfect square :
p2-5p+(25/4) =
(p-(5/2)) • (p-(5/2)) =
(p-(5/2))2
Things which are equal to the same thing are also equal to one another. Since
p2-5p+(25/4) = -7/4 and
p2-5p+(25/4) = (p-(5/2))2
then, according to the law of transitivity,
(p-(5/2))2 = -7/4
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
(p-(5/2))2 is
(p-(5/2))2/2 =
(p-(5/2))1 =
p-(5/2)
Now, applying the Square Root Principle to Eq. #2.2.1 we get:
p-(5/2) = √ -7/4
Add 5/2 to both sides to obtain:
p = 5/2 + √ -7/4
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
p2 - 5p + 8 = 0
has two solutions:
p = 5/2 + √ 7/4 • i
or
p = 5/2 - √ 7/4 • i
Note that √ 7/4 can be written as
√ 7 / √ 4 which is √ 7 / 2
Solve Quadratic Equation using the Quadratic Formula
2.3 Solving p2-5p+8 = 0 by the Quadratic Formula .
According to the Quadratic Formula, p , the solution for Ap2+Bp+C = 0 , where A, B and C are numbers, often called coefficients, is given by :
- B ± √ B2-4AC
p = ————————
2A
In our case, A = 1
B = -5
C = 8
Accordingly, B2 - 4AC =
25 - 32 =
-7
Applying the quadratic formula :
5 ± √ -7
p = —————
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,√ -7 =
√ 7 • (-1) =
√ 7 • √ -1 =
± √ 7 • i
√ 7 , rounded to 4 decimal digits, is 2.6458
So now we are looking at:
p = ( 5 ± 2.646 i ) / 2
Two imaginary solutions :
p =(5+√-7)/2=(5+i√ 7 )/2= 2.5000+1.3229i
or:
p =(5-√-7)/2=(5-i√ 7 )/2= 2.5000-1.3229i
Two solutions were found :
p =(5-√-7)/2=(5-i√ 7 )/2= 2.5000-1.3229i
p =(5+√-7)/2=(5+i√ 7 )/2= 2.5000+1.3229i
Answer:
Answer in above picture
