Question 113

If $$a=\frac{2+\sqrt{3}}{2-\sqrt{3}}$$ and $$b=\frac{2-\sqrt{3}}{2+\sqrt{3}}$$, then the value of $$a^2+b^2+a \times b$$ is

Solution

$$a=\frac{2+\sqrt{3}}{2-\sqrt{3}}$$ on rationalising we will get a = $$(2 + \surd3)^2$$

$$b=\frac{2-\sqrt{3}}{2+\sqrt{3}}$$ on rationalizing we will get b = $$(2 - \surd3)^2$$

now putting values of a and b in , $$a^2+b^2+a \times b$$

$$a^2+b^2+a \times b$$ = 195

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