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Sagot :
To expand and simplify [tex]\((7 + \sqrt{6})^2\)[/tex], we use the formula for the square of a binomial, which is [tex]\((a + b)^2 = a^2 + 2ab + b^2\)[/tex].
Here, [tex]\(a = 7\)[/tex] and [tex]\(b = \sqrt{6}\)[/tex].
1. Calculate [tex]\(a^2\)[/tex]:
[tex]\[ a^2 = 7^2 = 49 \][/tex]
2. Calculate [tex]\(2ab\)[/tex]:
[tex]\[ 2ab = 2 \cdot 7 \cdot \sqrt{6} = 14\sqrt{6} \][/tex]
3. Calculate [tex]\(b^2\)[/tex]:
[tex]\[ b^2 = (\sqrt{6})^2 = 6 \][/tex]
Now, combine these terms:
[tex]\[ (7 + \sqrt{6})^2 = a^2 + 2ab + b^2 = 49 + 14\sqrt{6} + 6 \][/tex]
Simplify the constant terms:
[tex]\[ 49 + 6 = 55 \][/tex]
Thus, the expanded and simplified form of [tex]\((7 + \sqrt{6})^2\)[/tex] is:
[tex]\[ 55 + 14\sqrt{6} \][/tex]
Therefore, [tex]\(b = 55\)[/tex] and [tex]\(c = 14\)[/tex], so the answer in the form [tex]\(b + c \sqrt{6}\)[/tex] is:
[tex]\[ 55 + 14\sqrt{6} \][/tex]
Here, [tex]\(a = 7\)[/tex] and [tex]\(b = \sqrt{6}\)[/tex].
1. Calculate [tex]\(a^2\)[/tex]:
[tex]\[ a^2 = 7^2 = 49 \][/tex]
2. Calculate [tex]\(2ab\)[/tex]:
[tex]\[ 2ab = 2 \cdot 7 \cdot \sqrt{6} = 14\sqrt{6} \][/tex]
3. Calculate [tex]\(b^2\)[/tex]:
[tex]\[ b^2 = (\sqrt{6})^2 = 6 \][/tex]
Now, combine these terms:
[tex]\[ (7 + \sqrt{6})^2 = a^2 + 2ab + b^2 = 49 + 14\sqrt{6} + 6 \][/tex]
Simplify the constant terms:
[tex]\[ 49 + 6 = 55 \][/tex]
Thus, the expanded and simplified form of [tex]\((7 + \sqrt{6})^2\)[/tex] is:
[tex]\[ 55 + 14\sqrt{6} \][/tex]
Therefore, [tex]\(b = 55\)[/tex] and [tex]\(c = 14\)[/tex], so the answer in the form [tex]\(b + c \sqrt{6}\)[/tex] is:
[tex]\[ 55 + 14\sqrt{6} \][/tex]
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