Question

factorise n^2-43n+432 by completing the square method​

Answer 1

Answer:

Solving n2+43n+432 = 0 by Completing The Square .

Subtract 432 from both side of the equation :

n2+43n = -432

Now the clever bit: Take the coefficient of n , which is 43 , divide by two, giving 43/2 , and finally square it giving 1849/4

Add 1849/4 to both sides of the equation :

On the right hand side we have :

-432 + 1849/4 or, (-432/1)+(1849/4)

The common denominator of the two fractions is 4 Adding (-1728/4)+(1849/4) gives 121/4

So adding to both sides we finally get :

n2+43n+(1849/4) = 121/4

Adding 1849/4 has completed the left hand side into a perfect square :

n2+43n+(1849/4) =

(n+(43/2)) • (n+(43/2)) =

(n+(43/2))2

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

n2+43n+(1849/4) = 121/4 and

n2+43n+(1849/4) = (n+(43/2))2

then, according to the law of transitivity,

(n+(43/2))2 = 121/4

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

(n+(43/2))2 is

(n+(43/2))2/2 =

(n+(43/2))1 =

n+(43/2)

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

n+(43/2) = √ 121/4

Subtract 43/2 from both sides to obtain:

n = -43/2 + √ 121/4

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

n2 + 43n + 432 = 0

has two solutions:

n = -43/2 + √ 121/4

or

n = -43/2 - √ 121/4

Note that √ 121/4 can be written as

√ 121 / √ 4 which is 11 / 2