Question

if a is equals to 1 by 7 minus 4 root 3 and b is equals to 1 by 7 + 4 root 3 then find the value of a square + b square​

Answer 1

It was a easy question

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Answer 2

The value is 194.

$a^2 + b^2 = 194$

Step-by-step explanation:

We are given the following in the question:

$a = \dfrac{1}{7-4\sqrt3}\\\\b = \dfrac{1}{7+4\sqrt3}$

We have to find the value of $a^2 + b^2$

$a + b = \displaystyle\frac{1}{7-4\sqrt3} + \frac{1}{7+4\sqrt3}\\\\= \frac{7+4\sqrt3 + 7-4\sqrt3}{(7-4\sqrt3)(7+4\sqrt3)} = \frac{14}{49-48} = 14\\\\\text{Identity: }a^2-b^2 = (a+b)(a-b)$

$ab = \displaystyle\frac{1}{7 - 4\sqrt3}\times \frac{1}{7 + 4\sqrt3} ==\frac{1}{(7 - 4\sqrt3)(7 + 4\sqrt3)} = \frac{1}{49-48}= 1$

$a^2 + b^2 = (a+b)^2 - 2ab\\a^2 + b^2 = (14)^2 - 2(1) = 194$

Thus, the value is 194.

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