Question

Find the value of a and b: 3+√2/3−√2 = +√2

Answer 1

Step-by-step explanation:

Given -

• 3 + √2/3 - √2 = a + b√2

To Find -

• Value of a and b

Rationalising the denominator :-

→ 3+√2/3-√2 × 3+√2/3+√2 = a + b√2

→ 9+3√2+3√2+2/9-2 = a + b√2

→ 11 + 6√2/7 = a + b√2

→ 11/7 + 6√2/7 = a + b√2

After comparing 11 + 6√2/7 with a + b√2, we get :

Hence,

The value of a is 11/7 and b is 6/7

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Answer 2

$\large{\underline{\bf{\pink{Answer:-}}}}$

✰$\sf \pink{a = \frac{11}{7}, \:and\:b=\frac{6}{7}}$

$\large{\underline{\bf{\blue{Explanation:-}}}}$

$\large{\underline{\bf{\green{Given:-}}}}$

✰ $\frac{3+\sqrt{2}}{3-\sqrt{2}}=a+b\sqrt{2}$

$\large{\underline{\bf{\green{To\:Find:-}}}}$

✰ we need to find the value of a and b.

$\huge{\underline{\bf{\red{Solution:-}}}}$

Rationalising the denominator

$\frac{3+\sqrt{2}}{3-\sqrt{2}}=a+b\sqrt{2} \\ \\ : \implies \sf\frac{3 + \sqrt{2} }{3 - \sqrt{2} } \times \frac{3 + \sqrt{2} }{3 + \sqrt{2} } = a + b \sqrt{2} \\ \\ : \implies \sf \: by \: using \: (a + b) {}^{2} = {a}^{2} + {b}^{2} + 2ab \\ \\\sf \: \: \: \: \: \: and \: {a}^{2} - {b}^{2} = ( a+b ) + (a - b) \\ \\: \implies \sf \frac{9 + 2 + 6 \sqrt{2} }{9 - 2} = a + b \sqrt{2} \\ \\</p><p>: \implies \sf \frac{11 + 6 \sqrt{2} }{7} = a + b \sqrt{2} \\ \\ : \implies \sf \frac{ 11 + 6 \sqrt{2} }{7}=a + b \sqrt{2} \\ \\: \implies \sf \frac{11 }{7} + \frac{6 + \sqrt{2} }{7} = a + b \sqrt{2} \\ \\ : \implies \sf \pink{a = \frac{11}{7}, \:and\:b=\frac{6}{7}}$

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