solve the following quadratic equation by factorization . y2+2√3y-9=0
Answers
Answer:
Step-by-step explanation:
2.2 Solving y2-3y+9 = 0 by Completing The Square .
Subtract 9 from both side of the equation :
y2-3y = -9
Now the clever bit: Take the coefficient of y , which is 3 , divide by two, giving 3/2 , and finally square it giving 9/4
Add 9/4 to both sides of the equation :
On the right hand side we have :
-9 + 9/4 or, (-9/1)+(9/4)
The common denominator of the two fractions is 4 Adding (-36/4)+(9/4) gives -27/4
So adding to both sides we finally get :
y2-3y+(9/4) = -27/4
Adding 9/4 has completed the left hand side into a perfect square :
y2-3y+(9/4) =
(y-(3/2)) • (y-(3/2)) =
(y-(3/2))2
Things which are equal to the same thing are also equal to one another. Since
y2-3y+(9/4) = -27/4 and
y2-3y+(9/4) = (y-(3/2))2
then, according to the law of transitivity,
(y-(3/2))2 = -27/4
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
(y-(3/2))2 is
(y-(3/2))2/2 =
(y-(3/2))1 =
y-(3/2)
Now, applying the Square Root Principle to Eq. #2.2.1 we get:
y-(3/2) = √ -27/4
Add 3/2 to both sides to obtain:
y = 3/2 + √ -27/4
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
y2 - 3y + 9 = 0
has two solutions:
y = 3/2 + √ 27/4 • i
or
y = 3/2 - √ 27/4 • i
Note that √ 27/4 can be written as
√ 27 / √ 4 which is √ 27 / 2
Solve Quadratic Equation using the Quadratic Formula
2.3 Solving y2-3y+9 = 0 by the Quadratic Formula .
According to the Quadratic Formula, y , the solution for Ay2+By+C = 0 , where A, B and C are numbers, often called coefficients, is given by :
- B ± √ B2-4AC
y = ————————
2A
In our case, A = 1
B = -3
C = 9
Accordingly, B2 - 4AC =
9 - 36 =
-27
Applying the quadratic formula :
3 ± √ -27
y = —————
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,√ -27 =
√ 27 • (-1) =
√ 27 • √ -1 =
± √ 27 • i
Can √ 27 be simplified ?
Yes! The prime factorization of 27 is
3•3•3
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).
√ 27 = √ 3•3•3 =
± 3 • √ 3
√ 3 , rounded to 4 decimal digits, is 1.7321
So now we are looking at:
y = ( 3 ± 3 • 1.732 i ) / 2
Two imaginary solutions :
y =(3+√-27)/2=(3+3i√ 3 )/2= 1.5000+2.5981i
or:
y =(3-√-27)/2=(3-3i√ 3 )/2= 1.5000-2.5981i