Answer :
To find the product [tex]\((x \sqrt{7} - 3 \sqrt{8})(x \sqrt{7} - 3 \sqrt{8})\)[/tex], we will expand this expression and simplify it.
First, observe that [tex]\((x \sqrt{7} - 3 \sqrt{8})(x \sqrt{7} - 3 \sqrt{8})\)[/tex] is in the form of [tex]\((A - B)^2\)[/tex], where:
- [tex]\(A = x \sqrt{7}\)[/tex]
- [tex]\(B = 3 \sqrt{8}\)[/tex]
We know the formula for the square of a binomial:
[tex]\[ (A - B)^2 = A^2 - 2AB + B^2 \][/tex]
Let's apply this formula step-by-step.
1. Calculate [tex]\(A^2\)[/tex]:
[tex]\[ A^2 = (x \sqrt{7})^2 = x^2 \cdot (\sqrt{7})^2 = x^2 \cdot 7 = 7x^2 \][/tex]
2. Calculate [tex]\(B^2\)[/tex]:
[tex]\[ B^2 = (3 \sqrt{8})^2 = 3^2 \cdot (\sqrt{8})^2 = 9 \cdot 8 = 72 \][/tex]
3. Calculate [tex]\(2AB\)[/tex]:
[tex]\[ 2AB = 2 \cdot (x \sqrt{7}) \cdot (3 \sqrt{8}) = 2 \cdot 3 \cdot x \cdot \sqrt{7} \cdot \sqrt{8} = 6x \cdot \sqrt{7 \cdot 8} \][/tex]
Simplify [tex]\(\sqrt{7 \cdot 8}\)[/tex]:
[tex]\[ \sqrt{56} = \sqrt{4 \cdot 14} = \sqrt{4} \cdot \sqrt{14} = 2 \sqrt{14} \][/tex]
So,
[tex]\[ 2AB = 6x \cdot 2 \sqrt{14} = 12x \sqrt{14} \][/tex]
4. Substitute the components back into the binomial square formula:
[tex]\[ (x \sqrt{7} - 3 \sqrt{8})^2 = 7x^2 - 2AB + 72 = 7x^2 - 12x \sqrt{14} + 72 \][/tex]
Now, compare this result with the given choices:
1. [tex]\(7 x^2 + 72\)[/tex]
2. [tex]\(7 x^2 - 12 x \sqrt{14} + 72\)[/tex]
3. [tex]\(7 x^2 - 12 x \sqrt{14} - 72\)[/tex]
4. [tex]\(7 x^2 - 72\)[/tex]
The correct choice is:
[tex]\[ \boxed{7 x^2 - 12 x \sqrt{14} + 72} \][/tex]
Thus, the product [tex]\((x \sqrt{7} - 3 \sqrt{8})(x \sqrt{7} - 3 \sqrt{8})\)[/tex] simplifies to [tex]\(\boxed{7 x^2 - 12 x \sqrt{14} + 72}\)[/tex].
First, observe that [tex]\((x \sqrt{7} - 3 \sqrt{8})(x \sqrt{7} - 3 \sqrt{8})\)[/tex] is in the form of [tex]\((A - B)^2\)[/tex], where:
- [tex]\(A = x \sqrt{7}\)[/tex]
- [tex]\(B = 3 \sqrt{8}\)[/tex]
We know the formula for the square of a binomial:
[tex]\[ (A - B)^2 = A^2 - 2AB + B^2 \][/tex]
Let's apply this formula step-by-step.
1. Calculate [tex]\(A^2\)[/tex]:
[tex]\[ A^2 = (x \sqrt{7})^2 = x^2 \cdot (\sqrt{7})^2 = x^2 \cdot 7 = 7x^2 \][/tex]
2. Calculate [tex]\(B^2\)[/tex]:
[tex]\[ B^2 = (3 \sqrt{8})^2 = 3^2 \cdot (\sqrt{8})^2 = 9 \cdot 8 = 72 \][/tex]
3. Calculate [tex]\(2AB\)[/tex]:
[tex]\[ 2AB = 2 \cdot (x \sqrt{7}) \cdot (3 \sqrt{8}) = 2 \cdot 3 \cdot x \cdot \sqrt{7} \cdot \sqrt{8} = 6x \cdot \sqrt{7 \cdot 8} \][/tex]
Simplify [tex]\(\sqrt{7 \cdot 8}\)[/tex]:
[tex]\[ \sqrt{56} = \sqrt{4 \cdot 14} = \sqrt{4} \cdot \sqrt{14} = 2 \sqrt{14} \][/tex]
So,
[tex]\[ 2AB = 6x \cdot 2 \sqrt{14} = 12x \sqrt{14} \][/tex]
4. Substitute the components back into the binomial square formula:
[tex]\[ (x \sqrt{7} - 3 \sqrt{8})^2 = 7x^2 - 2AB + 72 = 7x^2 - 12x \sqrt{14} + 72 \][/tex]
Now, compare this result with the given choices:
1. [tex]\(7 x^2 + 72\)[/tex]
2. [tex]\(7 x^2 - 12 x \sqrt{14} + 72\)[/tex]
3. [tex]\(7 x^2 - 12 x \sqrt{14} - 72\)[/tex]
4. [tex]\(7 x^2 - 72\)[/tex]
The correct choice is:
[tex]\[ \boxed{7 x^2 - 12 x \sqrt{14} + 72} \][/tex]
Thus, the product [tex]\((x \sqrt{7} - 3 \sqrt{8})(x \sqrt{7} - 3 \sqrt{8})\)[/tex] simplifies to [tex]\(\boxed{7 x^2 - 12 x \sqrt{14} + 72}\)[/tex].
