Answer :
Use the formula (b/2)^2 in order to create a new term to complete the square.
(x+2)^2 + 11
(x+2)^2 + 11
Root plot for : y = x2-4x+15
Axis of Symmetry (dashed) {x}={ 2.00}
Vertex at {x,y} = { 2.00,11.00}
Function has no real roots
2.2 Solving x2-4x+15 = 0 by Completing The Square .
Subtract 15 from both side of the equation :
x2-4x = -15
Now the clever bit: Take the coefficient of x , which is 4 , divide by two, giving 2 , and finally square it giving 4
Add 4 to both sides of the equation :
On the right hand side we have :
-15 + 4 or, (-15/1)+(4/1)
The common denominator of the two fractions is 1 Adding (-15/1)+(4/1) gives -11/1
So adding to both sides we finally get :
x2-4x+4 = -11
Adding 4 has completed the left hand side into a perfect square :
x2-4x+4 =
(x-2) • (x-2) =
(x-2)2
Things which are equal to the same thing are also equal to one another. Since
x2-4x+4 = -11 and
x2-4x+4 = (x-2)2
then, according to the law of transitivity,
(x-2)2 = -11
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-2)2 is
(x-2)2/2 =
(x-2)1 =
x-2
Now, applying the Square Root Principle to Eq. #2.2.1 we get:
x-2 = √ -11
Add 2 to both sides to obtain:
x = 2 + √ -11
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 - 4x + 15 = 0
has two solutions:
x = 2 + √ 11 • i
or
x = 2 - √ 11 • i
2.3 Solving x2-4x+15 = 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 = -4
C = 15
Accordingly, B2 - 4AC =
16 - 60 =
-44
Applying the quadratic formula :
4 ± √ -44
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,√ -44 =
√ 44 • (-1) =
√ 44 • √ -1 =
± √ 44 • i
Can √ 44 be simplified ?
Yes! The prime factorization of 44 is
2•2•11
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).
√ 44 = √ 2•2•11 =
± 2 • √ 11
√ 11 , rounded to 4 decimal digits, is 3.3166
So now we are looking at:
x = ( 4 ± 2 • 3.317 i ) / 2
Two imaginary solutions :
x =(4-√-44)/2=2-i√ 11 = 2.0000-3.3166i
Two solutions were found : x =(4-√-44)/2=2-i√ 11 = 2.0000-3.3166i x =(4+√-44)/2=2+i√ 11 = 2.0000+3.3166i