Respuesta :
Answer: [tex]\boldsymbol{-25+18\sqrt{2}}[/tex]
The expression is in the form [tex]a+b\sqrt{2}[/tex] where a = -25 and b = 18
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Work Shown:
Let [tex]x = 3\sqrt{2}[/tex] and then square both sides
[tex]x = 3\sqrt{2}\\\\x^2 = \left(3\sqrt{2}\right)^2\\\\x^2 = \left(3\right)^2*\left(\sqrt{2}\right)^2\\\\x^2 = 9*2\\\\x^2 = 18\\\\[/tex]
We'll use both x and x^2 later on.
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The denominator is in the form 4+x. We'll multiply top and bottom by 4-x to then use the difference of squares rule. This will allow us to eliminate the square root in the denominator.
[tex]\frac{8-3\sqrt{2}}{4+3\sqrt{2}}\\\\\frac{8-x}{4+x}\\\\\frac{(8-x)(4-x)}{(4+x)(4-x)}\\\\\frac{32-8x-4x+x^2}{16-x^2} \ \text{ difference of squares rule}\\\\\frac{32-12x+x^2}{16-x^2}\\\\[/tex]
[tex]\frac{32-12*3\sqrt{2}+18}{16-18} \ \text{ plug in } x = 3\sqrt{2} \text{ and } x^2 = 18\\\\\frac{50-36\sqrt{2}}{-2}\\\\\frac{-2(-25+18\sqrt{2})}{-2}\\\\\boldsymbol{-25+18\sqrt{2}}[/tex]
Therefore,
[tex]\frac{8-3\sqrt{2}}{4+3\sqrt{2}}=\boldsymbol{-25+18\sqrt{2}}[/tex]
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Checking the answer:
Use a calculator to find that,
[tex]\frac{8-3\sqrt{2}}{4+3\sqrt{2}} \approx 0.45584412271571 \\\\-25+18\sqrt{2} \approx 0.45584412271571[/tex]
We get the same decimal approximation, which helps confirm the correct answer.
Or you could use the idea that if M = N, then M-N = 0 to subtract the original expression M and the final result N. You should get 0 or very close to it.
