frac{14}{3+sqrt{2}}=? | vedclass

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Question

(A) To solve $\frac{14}{3+\sqrt{2}}$,we rationalize the denominator by multiplying the numerator and denominator by the conjugate $(3-\sqrt{2})$.$\fra

Vedclass · Accepted Answer

$3.172$

Vedclass · Answer

(A) To solve $\frac{14}{3+\sqrt{2}}$,we rationalize the denominator by multiplying the numerator and denominator by the conjugate $(3-\sqrt{2})$.$\frac{14}{3+\sqrt{2}} = \frac{14(3-\sqrt{2})}{(3+\sqrt{2})(3-\sqrt{2})}$Using the identity $(a+b)(a-b) = a^2 - b^2$,we get:$= \frac{14(3-\sqrt{2})}{3^2 - (\sqrt{2})^2} = \frac{14(3-\sqrt{2})}{9-2} = \frac{14(3-\sqrt{2})}{7}$$= 2(3-\sqrt{2})$Given $\sqrt{2} \approx 1.414$,we substitute the value:$= 2(3 - 1.414) = 2(1.586) = 3.172$

$\frac{14}{3+\sqrt{2}}=?$

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