Question

find the roots of 3x2-5x-7 by square rooot method

Answer 1

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

x2+(5/3)x-(7/3) = 0

Add 7/3 to both side of the equation :

x2+(5/3)x = 7/3

Now the clever bit: Take the coefficient of x , which is 5/3 , divide by two, giving 5/6 , and finally square it giving 25/36

Add 25/36 to both sides of the equation :

On the right hand side we have :

7/3 + 25/36 The common denominator of the two fractions is 36 Adding (84/36)+(25/36) gives 109/36

So adding to both sides we finally get :

x2+(5/3)x+(25/36) = 109/36

Adding 25/36 has completed the left hand side into a perfect square :

x2+(5/3)x+(25/36) =

(x+(5/6)) • (x+(5/6)) =

(x+(5/6))2

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

x2+(5/3)x+(25/36) = 109/36 and

x2+(5/3)x+(25/36) = (x+(5/6))2

then, according to the law of transitivity,

(x+(5/6))2 = 109/36

We'll refer to this Equation as Eq. #3.2.1

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

Note that the square root of

(x+(5/6))2 is

(x+(5/6))2/2 =

(x+(5/6))1 =

x+(5/6)

Now, applying the Square Root Principle to Eq. #3.2.1 we get:

x+(5/6) = √ 109/36

Subtract 5/6 from both sides to obtain:

x = -5/6 + √ 109/36

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

x2 + (5/3)x - (7/3) = 0

has two solutions:

x = -5/6 + √ 109/36

or

x = -5/6 - √ 109/36

Note that √ 109/36 can be written as

√ 109 / √ 36 which is √ 109 / 6