Question

a=1/2+√3,
b=1/2-√3

find the value of a^2+b^2-14ab​

Answer 1

Step-by-step explanation:

We have to find a²+b²-14ab

This can be done in two values either by direct solving or using identities

The latter will be more preferable, as it will save time and ensure a better accuracy,

Given expression to be solved :

a²+b²-14ab which can be written as a²+b²-2ab-12ab as ( -2ab-12ab=-14ab )

We know , (a-b)² = a²+b²-2ab

This implies from above that a²+b²-2ab-12ab can be written as (a-b)²-12ab,

Now calculate the values of a-b and ab

Given, a=1/2+√3, b=1/2-√3

a-b = 1/2+√3 - (1/2-√3) = 1/2+√3 - 1/2+√3 ----(by using distributive law for multiplying -1 in (1/2-√3))

Finally, a-b = 2√3 ,--(1)

(a-b)² = (2√3)²

(a-b)² = 4 * 3 = 12

Now calculating 12ab,

Given, a=1/2+√3, b=1/2-√3

ab = (1/2+√3)(1/2-√3) which again greets the use of identity (x+y)(x-y) = x²-y²

Here x =1/2 and y = √3

Therefore, ab = (1/2)² - (√3)² = 1/4 - 3 = - 11/4

12ab = 12 * 11/4 = - 33

Now calculating a²+b²-14ab = (a-b)²-12ab = 12-(-33) = 12 + 33 = 45

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