Find the values of a if : 5+2√3 7+4 √3 = a - 6√3
Please answer 17 Q
Answers
Answer:
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}.}
(7+4
3
)
(5+2
3
)
=a−b
3
.
Given:
\frac{(5+2 \sqrt{3})}{(7+4 \sqrt{3})}=a-b \sqrt{3}
(7+4
3
)
(5+2
3
)
=a−b
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}7−4
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}
(7+4
3
)
(5+2
3
)
⋅
(7−4
3
)
(7−4
3
)
=
7
2
−4
2
⋅3
5×7−5×4
3
+2
3
×7−2
3
×4
3
Solving the denominator in form of (a^2-b^2)(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}
7
2
−4
2
⋅3
5×7−5×4
3
+2
3
×7−2
3
×4
3
Subtracting twenty four from thirty five and -20\sqrt{3} + 14\sqrt{3}20
3
+14
3
separately we get:
\frac{[35-20 \sqrt{3}+14 \sqrt{3}-24]}{[49-48]}=11-6 \sqrt{3}
[49−48]
[35−20
3
+14
3
−24]
=11−6
3
Therefore, the value of the rationalization is 11-6 \sqrt{3}11−6
3
Now equating \bold{11-6 \sqrt{3} with a-b \sqrt{3}}11−6
3
witha−b
3
,we get the value of \bold{a = 11}a=11 and \bold{b = - 6.}b=−6.
Step-by-step explanation:
- hope it helps you!
