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How do i factor 3x²-18x+1=7x-7

User Alphy
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1 Answer

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Answer: 0.3

Step-by-step explanation: Solving 3x2-25x+8 = 0 by Completing The Square .

Divide both sides of the equation by 3 to have 1 as the coefficient of the first term :

x2-(25/3)x+(8/3) = 0

Subtract 8/3 from both side of the equation :

x2-(25/3)x = -8/3

Now the clever bit: Take the coefficient of x , which is 25/3 , divide by two, giving 25/6 , and finally square it giving 625/36

Add 625/36 to both sides of the equation :

On the right hand side we have :

-8/3 + 625/36 The common denominator of the two fractions is 36 Adding (-96/36)+(625/36) gives 529/36

So adding to both sides we finally get :

x2-(25/3)x+(625/36) = 529/36

Adding 625/36 has completed the left hand side into a perfect square :

x2-(25/3)x+(625/36) =

(x-(25/6)) • (x-(25/6)) =

(x-(25/6))2

Things which are equal to the same thing are also equal to one another. Since

x2-(25/3)x+(625/36) = 529/36 and

x2-(25/3)x+(625/36) = (x-(25/6))2

then, according to the law of transitivity,

(x-(25/6))2 = 529/36

We'll refer to this Equation as Eq. #4.2.1

The Square Root Principle says that When two things are equal, their square roots are equal.

Note that the square root of

(x-(25/6))2 is

(x-(25/6))2/2 =

(x-(25/6))1 =

x-(25/6)

Now, applying the Square Root Principle to Eq. #4.2.1 we get:

x-(25/6) = √ 529/36

Add 25/6 to both sides to obtain:

x = 25/6 + √ 529/36

Since a square root has two values, one positive and the other negative

x2 - (25/3)x + (8/3) = 0

has two solutions:

x = 25/6 + √ 529/36

or

x = 25/6 - √ 529/36

Note that √ 529/36 can be written as √ 529 / √ 36 which is 23 / 6

Solving 3x2-25x+8 = 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 = 3

B = -25

C = 8

Accordingly, B2 - 4AC =

625 - 96 =

529

Applying the quadratic formula :

25 ± √ 529

x = ——————

6

Can √ 529 be simplified ?

Yes! The prime factorization of 529 is

23•23

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).

√ 529 = √ 23•23 =

± 23 • √ 1 =

± 23

So now we are looking at:

x = ( 25 ± 23) / 6

Two real solutions:

x =(25+√529)/6=(25+23)/6= 8.000

or:

x =(25-√529)/6=(25-23)/6= 0.333

User Anar Salimkhanov
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