prime factorise and rationalise denominator
Answers
Step-by-step explanation:
Given 14 / √108 - √96 + √192 - √54
- = 14 / √2^2 x 3^3 - √2^5 x 3 + √2^6 x 3 - √2 x 3^3
- = 14 / √2^2 x 3^2 x 3 - √2^4 x 2 x 3 + √2^6 x 3 - √2 x 3^2 x 3
- = 14 / 2 x 3 √3 – 2 x 2 √6 + 8 √3 - 3√6
- = 14 / 6 √3 - 4√6 + 8√3 - 3√6
- = 14 / 14 √3 - 7√6
- = 14 / 7 (2 √3 - √6)
- = 2 / 2√3 - √6
- Now rationalizing the denominator we get
- = 2 / 2√3 - √6 x 2√3 + √6 / 2√3 + √6
- = 4 √3 + 2√6 / 12 – 6 ( since (a + b)(a – b) = a^2 – b^2 in the denominator)
- = 4√3 + 2√6 / 6
- = 2 (2√3 + √6) / 6
- = 2√3 + √6 / 3 will be the required answer
Reference link will be
https://brainly.in/question/15841574
Answer:
Given :
\frac{14}{ \sqrt{108} - \sqrt{96} + \sqrt{192} - \sqrt{54} }
108
−
96
+
192
−
54
14
To find :
Rationalise the denominator.
Solution :
Firstly we will prime factorise all the elements of denominator, so
\frac{14}{ \sqrt{108} - \sqrt{96} + \sqrt{192} - \sqrt{54} }
108
−
96
+
192
−
54
14
this will become,
\frac{14}{ \sqrt{ {2}^{2} \times {3}^{3} } - \sqrt{ {2}^{5} \times 3} + \sqrt{ {2}^{6} \times 3 } - \sqrt{2 \times {3}^{3} } }
2
2
×3
3
−
2
5
×3
+
2
6
×3
−
2×3
3
14
When we take even powers of all the elements outside the square roots,
we will get,
\frac{14}{ 6\sqrt{ 3 } - 4\sqrt{ 6} + 8\sqrt{ 3 } - 3\sqrt{6} }
6
3
−4
6
+8
3
−3
6
14
On further solving the denominator,
we get
\frac{14}{ 14\sqrt{ 3 } - 7\sqrt{ 6} }
14
3
−7
6
14
On dividing numerator and denominator by 7,
we get
\frac{2}{ 2\sqrt{ 3 } - \sqrt{ 6} }
2
3
−
6
2
then, to rationalise the denominator, we multiply numerator and denominator with :
2 \sqrt{3 } + \sqrt{6}2
3
+
6
so,we will get,
\begin{lgathered}\\ \frac{2}{ 2\sqrt{ 3 } - \sqrt{ 6} } \times \frac{2 \sqrt{3} + \sqrt{6} }{2 \sqrt{3} + \sqrt{6} }\end{lgathered}
2
3
−
6
2
×
2
3
+
6
2
3
+
6
\begin{lgathered}= \frac{4{\sqrt {3}} + 2{\sqrt {6}}} {12 - 6}\\ = \frac{4{\sqrt {3}} + 2{\sqrt {6}}} {6} \\ = \bf \frac{2 \sqrt{3} + \sqrt{6} }{3}\end{lgathered}
=
12−6
4
3
+2
6
=
6
4
3
+2
6
=
3
2
3
+
6
So,
Rationalisation of given expression is,
\begin{lgathered}\\ \bf \: = \frac{2 \sqrt{3} + \sqrt{6} }{3}\end{lgathered}
=
3
2
3
+
6