Answer :
Certainly! Let's solve the expression:
[tex]\[ E = \sqrt{5 + 2 \sqrt{6}} \cdot \sqrt{5 - 2 \sqrt{6}} \][/tex]
First, let's define two terms for simplicity:
[tex]\[ a = \sqrt{5 + 2 \sqrt{6}} \][/tex]
[tex]\[ b = \sqrt{5 - 2 \sqrt{6}} \][/tex]
We need to find the product [tex]\( E = a \cdot b \)[/tex].
To find this, we will consider the product of the expressions under the square roots:
[tex]\[ E^2 = (\sqrt{5 + 2 \sqrt{6}} \cdot \sqrt{5 - 2 \sqrt{6}})^2 \][/tex]
When the square roots are multiplied and squared, we get:
[tex]\[ E^2 = (5 + 2 \sqrt{6})(5 - 2 \sqrt{6}) \][/tex]
Recall the algebraic identity for the difference of squares:
[tex]\[ (x + y)(x - y) = x^2 - y^2 \][/tex]
Here, [tex]\( x = 5 \)[/tex] and [tex]\( y = 2 \sqrt{6} \)[/tex]. Applying this identity:
[tex]\[ E^2 = 5^2 - (2 \sqrt{6})^2 \][/tex]
Calculate each square separately:
[tex]\[ 5^2 = 25 \][/tex]
[tex]\[ (2 \sqrt{6})^2 = 4 \cdot 6 = 24 \][/tex]
So we have:
[tex]\[ E^2 = 25 - 24 = 1 \][/tex]
Taking the square root of both sides, we find:
[tex]\[ E = \sqrt{1} = 1 \][/tex]
Thus, the result of the original expression is:
[tex]\[ E = 1 \][/tex]
Upon verification with numerical calculations:
- We have [tex]\( \sqrt{5 + 2 \sqrt{6}} \approx 3.146264369941972 \)[/tex]
- and [tex]\( \sqrt{5 - 2 \sqrt{6}} \approx 0.31783724519578294 \)[/tex].
Their product is:
[tex]\[ 3.146264369941972 \times 0.31783724519578294 \approx 1.000000000000002 \][/tex]
Considering the small numerical tolerance or rounding errors, we can confidently conclude:
[tex]\[ E = 1 \][/tex]
[tex]\[ E = \sqrt{5 + 2 \sqrt{6}} \cdot \sqrt{5 - 2 \sqrt{6}} \][/tex]
First, let's define two terms for simplicity:
[tex]\[ a = \sqrt{5 + 2 \sqrt{6}} \][/tex]
[tex]\[ b = \sqrt{5 - 2 \sqrt{6}} \][/tex]
We need to find the product [tex]\( E = a \cdot b \)[/tex].
To find this, we will consider the product of the expressions under the square roots:
[tex]\[ E^2 = (\sqrt{5 + 2 \sqrt{6}} \cdot \sqrt{5 - 2 \sqrt{6}})^2 \][/tex]
When the square roots are multiplied and squared, we get:
[tex]\[ E^2 = (5 + 2 \sqrt{6})(5 - 2 \sqrt{6}) \][/tex]
Recall the algebraic identity for the difference of squares:
[tex]\[ (x + y)(x - y) = x^2 - y^2 \][/tex]
Here, [tex]\( x = 5 \)[/tex] and [tex]\( y = 2 \sqrt{6} \)[/tex]. Applying this identity:
[tex]\[ E^2 = 5^2 - (2 \sqrt{6})^2 \][/tex]
Calculate each square separately:
[tex]\[ 5^2 = 25 \][/tex]
[tex]\[ (2 \sqrt{6})^2 = 4 \cdot 6 = 24 \][/tex]
So we have:
[tex]\[ E^2 = 25 - 24 = 1 \][/tex]
Taking the square root of both sides, we find:
[tex]\[ E = \sqrt{1} = 1 \][/tex]
Thus, the result of the original expression is:
[tex]\[ E = 1 \][/tex]
Upon verification with numerical calculations:
- We have [tex]\( \sqrt{5 + 2 \sqrt{6}} \approx 3.146264369941972 \)[/tex]
- and [tex]\( \sqrt{5 - 2 \sqrt{6}} \approx 0.31783724519578294 \)[/tex].
Their product is:
[tex]\[ 3.146264369941972 \times 0.31783724519578294 \approx 1.000000000000002 \][/tex]
Considering the small numerical tolerance or rounding errors, we can confidently conclude:
[tex]\[ E = 1 \][/tex]
