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QuestionCalculate the value of (23−2)2+(3+2)2(2 \sqrt{3}-\sqrt{2})^{2}+(\sqrt{3}+\sqrt{2})^{2}.

Studdy Solution

STEP 1

Assumptions1. We are dealing with real numbers. . The operations of addition, subtraction, multiplication, and square root are defined in the usual way.

STEP 2

First, we need to expand the squares in the expression. The square of a binomial (a−b)2(a-b)^2 is given by a2−2ab+b2a^2 -2ab + b^2 and the square of a binomial (a+b)2(a+b)^2 is given by a2+2ab+b2a^2 +2ab + b^2.
(2−2)2=(2)2−2⋅2⋅2+(2)2 (2 \sqrt{}-\sqrt{2})^{2} = (2 \sqrt{})^2 -2 \cdot2 \sqrt{} \cdot \sqrt{2} + (\sqrt{2})^2 (+2)2=()2+2⋅⋅2+(2)2 (\sqrt{}+\sqrt{2})^{2} = (\sqrt{})^2 +2 \cdot \sqrt{} \cdot \sqrt{2} + (\sqrt{2})^2

STEP 3

Now, we calculate the squares and the products in the expressions.
(23−2)2=⋅3−2⋅26+2 (2 \sqrt{3}-\sqrt{2})^{2} = \cdot3 -2 \cdot2 \sqrt{6} +2 (3+2)2=3+2⋅26+2 (\sqrt{3}+\sqrt{2})^{2} =3 +2 \cdot2 \sqrt{6} +2

STEP 4

implify the expressions.
(23−2)2=12−46+2=14−46 (2 \sqrt{3}-\sqrt{2})^{2} =12 -4 \sqrt{6} +2 =14 -4 \sqrt{6} (3+2)2=3+46+2=+46 (\sqrt{3}+\sqrt{2})^{2} =3 +4 \sqrt{6} +2 = +4 \sqrt{6}

STEP 5

Now, add the two simplified expressions.
(23−2)2+(3+2)2=(14−4)+(5+4) (2 \sqrt{3}-\sqrt{2})^{2} + (\sqrt{3}+\sqrt{2})^{2} = (14 -4 \sqrt{}) + (5 +4 \sqrt{})

STEP 6

implify the final expression.
(23−2)2+(3+2)2=14+5=19 (2 \sqrt{3}-\sqrt{2})^{2} + (\sqrt{3}+\sqrt{2})^{2} =14 +5 =19 So, (23−2)2+(3+2)2=19(2 \sqrt{3}-\sqrt{2})^{2} + (\sqrt{3}+\sqrt{2})^{2} =19.

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