Question

5+2√3/7+4√3=a-b√3. Find the value of a and b.

Answer 1

11 and - 6 is the value of a and b if $\bold{\frac{(5+2 \sqrt{3})}{(7+4 \sqrt{3})}=a-b \sqrt{3}.}$

Given:

$\frac{(5+2 \sqrt{3})}{(7+4 \sqrt{3})}=a-b \sqrt{3}$

To find:

The value of a and b.=?

Solution:

To find the value of a and b, rationalize the given fraction with $7- 4\sqrt{3}$. By multiplying it both in numerator and denominator we get:

$\frac{(5+2 \sqrt{3})}{(7+4 \sqrt{3})} \cdot \frac{(7-4 \sqrt{3})}{(7-4 \sqrt{3})}=\frac{5 \times 7-5 \times 4 \sqrt{3}+2 \sqrt{3} \times 7-2 \sqrt{3} \times 4 \sqrt{3}}{7^{2}-4^{2} \cdot 3}$

Solving the denominator in form of $(a^2-b^2)$ to get the value of the denominator

$\frac{5 \times 7-5 \times 4 \sqrt{3}+2 \sqrt{3} \times 7-2 \sqrt{3} \times 4 \sqrt{3}}{7^{2}-4^{2} \cdot 3}$

Subtracting twenty four from thirty five and -$20\sqrt{3} + 14\sqrt{3}$ separately we get:

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

Therefore, the value of the rationalization is $11-6 \sqrt{3}$

Now equating $\bold{11-6 \sqrt{3} with a-b \sqrt{3}}$ ,we get the value of $\bold{a = 11}$ and $\bold{b = - 6.}$

Answer 2

Answer:

$value \: of \: a = 11 \\and \: b = 6$

Step-by-step explanation:

$LHS = \frac{5+2\sqrt{3}}{7+4\sqrt{3}}$

Multiply numerator and denominator by (7-4√3), we get

$= \frac{(5+2\sqrt{3})(7-4\sqrt{3})}{(7+4\sqrt{3})(7-4\sqrt{3})}$

$= \frac{35-20\sqrt{3}+14\sqrt{3}-24}{7^{2}-\left(4\sqrt{3}\right)^{2}}$

/* By algebraic identity:

(x+y)(x-y) = x²-y² */

$=\frac{11-6\sqrt{3}}{49-48}$

$= 11-6\sqrt{3}--(1)$

$RHS = a-b\sqrt{3} --(2)\\(given)$

Compare (1) and (2), we get

a = 11 , b = 6

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