Question - Simplifying an Algebraic Expression with Binomial Expansion
mathalgebraic expression simplificationbinomial expansion stepssimplify algebraic termsmathematics binomial theorem
Solution:
The image shows an algebraic expression to simplify:\[ 2^2 \cdot (2^{-\frac{1}{2}} + 1)^2 - (2^{\frac{3}{2}} + 1)^2 \]Let's simplify step by step:1. Simplify the terms inside the parentheses first:\[ (2^{-\frac{1}{2}} + 1)^2 = ( \frac{1}{\sqrt{2}} + 1 )^2 \]\[ (2^{\frac{3}{2}} + 1)^2 = ( \sqrt{8} + 1 )^2 = (2\sqrt{2} + 1)^2 \]2. Apply the binomial expansion to the squared terms:\[ ( \frac{1}{\sqrt{2}} + 1 )^2 = \left(\frac{1}{\sqrt{2}}\right)^2 + 2 \cdot \left(\frac{1}{\sqrt{2}}\right) \cdot 1 + 1^2 \]\[ (2\sqrt{2} + 1)^2 = (2\sqrt{2})^2 + 2 \cdot (2\sqrt{2}) \cdot 1 + 1^2 \]3. Simplify each term:\[ \left(\frac{1}{\sqrt{2}}\right)^2 = \frac{1}{2} \]\[ 2 \cdot \left(\frac{1}{\sqrt{2}}\right) = 2 \cdot \frac{1}{\sqrt{2}} = \sqrt{2} \]\[ (2\sqrt{2})^2 = 8 \]\[ 2 \cdot (2\sqrt{2}) = 4\sqrt{2} \]4. Substitute back into the equation:\[ ( \frac{1}{2} + \sqrt{2} + 1 )^2 = ( \frac{3}{2} + \sqrt{2} )^2 \]\[ (8 + 4\sqrt{2} + 1 )^2 = (9 + 4\sqrt{2})^2 \]5. Now multiply by the factor outside the parentheses in the original expression, which is $$2^2 = 4$$, and subtract the squared terms:\[ 4 \cdot ( \frac{3}{2} + \sqrt{2} )^2 - (9 + 4\sqrt{2})^2 \]\[ = 4 \cdot (\frac{9}{4} + 3\sqrt{2} + 2) - (81 + 8\sqrt{2} + 16) \]\[ = (9 + 12\sqrt{2} + 8) - (81 + 8\sqrt{2} + 16) \]6. Combine like terms:\[ 9 + 12\sqrt{2} + 8 - 81 - 8\sqrt{2} - 16 \]\[ = (9 + 8 - 81 - 16) + (12\sqrt{2} - 8\sqrt{2}) \]\[ = (-70) + 4\sqrt{2} \]So the simplified expression is:\[ -70 + 4\sqrt{2} \]
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