Question

rationalize the denominator 5 minus 3 root 14 upon 7 + 2 root 14​

Answer 1

The rationalization of $\dfrac{5-3\sqrt{14}}{7+2\sqrt{14}}$ is equal to 31 + $\dfrac{31\sqrt{14}}{7}$.

Step-by-step explanation:

We have,

$\dfrac{5-3\sqrt{14}}{7+2\sqrt{14}}$

To find, the rationalization of $\dfrac{5-3\sqrt{14}}{7+2\sqrt{14}}$ = ?

∴ $\dfrac{5-3\sqrt{14}}{7+2\sqrt{14}}$

The rationalizing numerator and denominator, we get

$=\dfrac{5-3\sqrt{14}}{7+2\sqrt{14}}\times \dfrac{7-2\sqrt{14}}{7-2\sqrt{14}}$

Using the algebraic identity,

$a^{2} -b^{2} =(a+b)(a-b)$

$=\dfrac{(5-3\sqrt{14})(7-2\sqrt{14})}{7^2-(2\sqrt{14})^2}$

$=\dfrac{35-10\sqrt{14}-21\sqrt{14}+6\times 14}{49-4(14)}$

$=\dfrac{35-31\sqrt{14}+84}{49-56}$

$=\dfrac{119-31\sqrt{14}}{-7}$

$=\dfrac{119}{-7}+\dfrac{-31\sqrt{14}}{-7}$

= 31 + $\dfrac{31\sqrt{14}}{7}$

∴  The rationalization of $\dfrac{5-3\sqrt{14}}{7+2\sqrt{14}}$ = 31 + $\dfrac{31\sqrt{14}}{7}$

Thus, the rationalization of $\dfrac{5-3\sqrt{14}}{7+2\sqrt{14}}$ is equal to 31 + $\dfrac{31\sqrt{14}}{7}$.

Answer 2

Answer:

5-3√14 \ 7+2√14

5-3√14 \ 7+2√14×7-2√14 / 7-2√14

(5-3√14) (7-2√14) / (7)^2-(2√14)^2

35-10√14-21√14+6×14 / 35-31√14+84 / -7

119-31√14 / -7

so, friends ans is finish here

#hope you would be agree with my answer

#please like and follow me tooooo