Question

if a=root3+root2/root3-root2 and b=root3-root2/root3+root2 find the value of a^2+b^2​

Answer 1

Answer:

a=(√3+√2)/(√3-√2)

b=(√3-√2)/(√3+√2)

a.b=(√3+√2)/(√3-√2).(√3-√2)/(√3+√2)=1

a+b=(√3+√2)/(√3-√2)+(√3-√2)/(√3+√2)

=(√3+√2)^2+(√3-√2)^2/{(√3)^2-(√2)^2}

=(3+2+2√6+3+2-2√6)/(3-2)

=10

so,

a^2+b^2

=(a+b)^2-2ab

=(10)^2-2.1

=100-2

=98

Answer 2

Concept :

The word "square root" is translated as "radix" in a commentary on the eleventh book of Euclid's Elements that was translated from Arabic into Latin. R became a symbol that was frequently used to denote "root," but it was also occasionally used to denote the first degree of the unknowable quantity, x. Squares are the integers that are produced when a value is multiplied by itself. In contrast, a number's square root is a value that, when multiplied by itself, returns the original value. Additionally, the subject matter of an equation influences how important it is to find solutions.

Given:

$a=root3+root2/root3-root2\\ b=root3-root2/root3+root2$

Find:

$a^2+b^2$

Solution:

According to the problem,

$a=(√3+√2)/(√3-√2)\\b=(√3-√2)/(√3+√2)\\a.b=(√3+√2)/(√3-√2).(√3-√2)/(√3+√2)=1\\a+b=(√3+√2)/(√3-√2)+(√3-√2)/(√3+√2)$

$=(√3+√2)^2+(√3-√2)^2/{(√3)^2-(√2)^2}\\=(3+2+2√6+3+2-2√6)/(3-2)\\=10$

So,

$a^2+b^2=(a+b)^2-2ab=(10)^2-2.1\\=100-2\\=98$

Hence the value is $98$

SPJ2