Solve the quadratic equation by completing the square.

Mathematics

2 answers:

Rudik [331]3 years ago

6 0

Solution
1. x =(12-√228)/2=6-√ 57 = -1.550
2. x =(12+√228)/2=6+√ 57 = 13.550
H3LL0 this is How to get this^ is down there i suggest you take some notesL0L

The first term is, x2 its coefficient is 1 .
The middle term is, -12x its coefficient is -12 .
The last term, "the constant", is -21

Multiply the coefficient of the first term by the constant 1 • -21 = -21

Find two factors of -21 whose sum equals the coefficient of the middle term, which is -12 .

-21 + 1 = -20
-7 + 3 = -4
-3 + 7 = 4
-1 + 21 = 20

Find the Vertex of y = x2-12x-21
For any parabola,Ax2+Bx+C,the x -coordinate of the vertex is given by -B/(2A) is 6.0000 Plugging into the parabola formula 6.0000. for x we can calculate the y -coordinate :
 y = 1.0 * 6.00 * 6.00 - 12.0 * 6.00 - 21.0
or y = -57.000 Root plot for : y = x2-12x-21
Axis of Symmetry (dashed) {x}={ 6.00}
Vertex at {x,y} = { 6.00,-57.00}
x -Intercepts (Roots) :
Root 1 at {x,y} = {-1.55, 0.00}
Root 2 at {x,y} = {13.55, 0.00}

read the squar 2.2 Solving x2-12x-21 = 0 by Completing The Square .

Add 21 to both side of the equation : 
 x2-12x = 21 Now the clever bit: Take the coefficient of x , which is 12 , divide by two, giving 6 , and finally square it giving 36
Adding 36 has completed the left hand side into a perfect square :
 x2-12x+36 =
 (x-6) • (x-6) =
 (x-6)2
Since
 x2-12x+36 = 57 and
 x2-12x+36 = (x-6)2
then, according to the law of transitivity,
 (x-6)2 = 57
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.
the square root of (x-6)2 is
 (x-6)2/2 =
 (x-6)1 =
 x-6
Now, applying the Square Root Principle to Eq. #2.2.1 we get:
 x-6 = √ 57
Add 6 to both sides to obtain:
 x = 6 + √ 57
Since a square root has two values, one positive and the other negative
 x2 - 12x - 21 = 0
 has two solutions:
 x = 6 + √ 57 
 or
 x = 6 - √ 57
using the Quadratic Formula

2.3 Solving x2-12x-21 = 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 = -12
 C = -21

Accordingly, B2 - 4AC =
 144 - (-84) =
 228

Applying the quadratic formula :

 12 ± √ 228 
 x = ——————

2
 The prime factorization of 228 is
 2•2•3•19
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
√ 228 = √ 2•2•3•19 =
 ± 2 • √ 57 

 √ 57 , rounded to 4 decimal digits, is 7.5498
 So now we are looking at:
 x = ( 12 ± 2 • 7.550 ) / 2

Two real solutions:

 x =(12+√228)/2=6+√ 57 = 13.550 

or:

 x =(12-√228)/2=6-√ 57 = -1.550