17.7k views
5 votes
Solve this quadratic equation by completing the square.

X^2+4x+15

2 Answers

5 votes
Use the formula (b/2)^2 in order to create a new term to complete the square.
(x+2)^2 + 11
4 votes

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

Solve Quadratic Equation by Completing The Square

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

Solve Quadratic Equation using the Quadratic Formula

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 or:
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
User Jesan Fafon
by
7.8k points

No related questions found

Welcome to QAmmunity.org, where you can ask questions and receive answers from other members of our community.

9.4m questions

12.2m answers

Categories