Question

If 5+3√3/7+4√3 = a+b√3 then a-b =​

Answer 1

Answer:

11 and - 6 is the value of a and b if

$</p><p> \bold{\frac{(5+2 \sqrt{3})}{(7+4\sqrt{3})}=a-b \sqrt{3}.}</p><p></p><p>(7+4</p><p></p><p>3</p><p></p><p>)</p><p></p><p>(5+2</p><p></p><p>3</p><p></p><p>)</p><p></p><p>=a−b</p><p></p><p>3</p><p></p><p>.$

Given:

$\frac{(5+2 \sqrt{3})}{(7+4 \sqrt{3})}=a-b \sqrt{3}</p><p></p><p>(7+4</p><p></p><p>3</p><p></p><p>)</p><p></p><p>(5+2</p><p></p><p>3</p><p></p><p>)</p><p></p><p>=a−b</p><p></p><p>3</p><p>$

To find:

The value of a and b.=?

Solution:

To find the value of a and b, rationalize the given fraction with

$7- 4\sqrt{3}7−4</p><p></p><p>3$

By multiplying it both in numerator and denominator we get:

$\frac{(5+2 \sqrt{3})}{(7+4 \sqrt{3})} \cdot \frac{(7-4 \sqrt{3})}{(7-4 \sqrt{3})}=\frac{5 \times 7-5 \times 4 \sqrt{3}+2 \sqrt{3} \times 7-2 \sqrt{3} \times 4 \sqrt{3}}{7^{2}-4^{2} \cdot 3}</p><p></p><p>(7+4</p><p></p><p>3</p><p></p><p>)</p><p></p><p>(5+2</p><p></p><p>3</p><p></p><p>)</p><p></p><p>⋅</p><p></p><p>(7−4</p><p></p><p>3</p><p></p><p>)</p><p></p><p>(7−4</p><p></p><p>3</p><p></p><p>)</p><p></p><p>=</p><p></p><p>7</p><p></p><p>2</p><p></p><p>−4</p><p></p><p>2</p><p></p><p>⋅3</p><p></p><p>5×7−5×4</p><p></p><p>3</p><p></p><p>+2</p><p></p><p>3</p><p></p><p>×7−2</p><p></p><p>3</p><p></p><p>×4</p><p>$

$</p><p></p><p>$

$\frac{5 \times 7-5 \times 4 \sqrt{3}+2 \sqrt{3} \times 7-2 \sqrt{3} \times 4 \sqrt{3}}{7^{2}-4^{2} \cdot 3}</p><p></p><p></p><p>7</p><p></p><p>2</p><p></p><p>−4</p><p></p><p>2</p><p></p><p>⋅3</p><p></p><p>5×7−5×4</p><p></p><p>3</p><p></p><p>+2</p><p></p><p>3</p><p></p><p>×7−2</p><p></p><p>3</p><p></p><p>×4</p><p></p><p>3$

Subtracting twenty four from thirty five and

$-20\sqrt{3} + 14\sqrt{3}20</p><p></p><p>3</p><p></p><p>+14</p><p></p><p>3$

separately we get:

$\frac{[35-20 \sqrt{3}+14 \sqrt{3}-24]}{[49-48]}=11-6 \sqrt{3}</p><p></p><p>[49−48]</p><p></p><p>[35−20</p><p></p><p>3</p><p></p><p>+14</p><p></p><p>3</p><p></p><p>−24]</p><p></p><p>=11−6</p><p></p><p>3$

Therefore, the value of the rationalization is

$11-6 \sqrt{3}11−6</p><p></p><p>3$

Now equating

$\bold{11-6 \sqrt{3} with a-b \sqrt{3}}11−6</p><p></p><p>3</p><p></p><p></p><p></p><p>3$

,we get the value of

$\bold{a = 11}a=11 and \bold{b = - 6.}b=−6.$