Simplify the following: 4√12 × 7√6
Solution:
Given, the expression is \(\sqrt[4]{12}\times \sqrt[7]{6}\)
We have to simplify the expression.
\(\sqrt[4]{12}=(12)^{\frac{1}{4}}=(2\times 2\times3 )^{\frac{1}{4}}\)
= \((2)^{\frac{1}{4}}\times (2)^{\frac{1}{4}}\times (3)^{\frac{1}{4}}\)
We know \(a^{m}.a^{n}=a^{m+n}\)
= \((2)^{({\frac{1}{4}}+\frac{1}{4})}\times (3)^{\frac{1}{4}}\)
= \((2)^{({\frac{1}{2}})}\times (3)^{\frac{1}{4}}\)
\(\sqrt[7]{6}=(6)^{\frac{1}{7}}=(2\times 3)^{\frac{1}{7}}\)
= \((2)^{\frac{1}{7}}\times (3)^{\frac{1}{7}}\)
Now, \(\sqrt[4]{12}\times \sqrt[7]{6}\) = \([(2)^{({\frac{1}{2}})}\times (3)^{\frac{1}{4}}]\times [(2)^{\frac{1}{7}}\times (3)^{\frac{1}{7}}]\)
= \((2)^{\frac{1}{2}}\times (3)^{\frac{1}{4}}\times (2)^{\frac{1}{7}}\times (3)^{\frac{1}{7}}=(2)^{\frac{1}{2}+\frac{1}{7}}\times(3)^{\frac{1}{4}+\frac{1}{7}}=(2)^{\frac{9}{14}}\times (3)^{\frac{11}{28}}\)
= \(\sqrt[14]{2^{9}}\times \sqrt[28]{3^{11}}\)
We know \(\sqrt[n]{a}=\sqrt[mn]{a^{m}}\)
= \(\sqrt[28]{2^{18}}\times \sqrt[28]{3^{11}}\)
= \(\sqrt[28]{2^{18}\times 3^{11}}\)
✦ Try This: Simplify the expression: \(\sqrt[4]{16}\times \sqrt[7]{8}\)
☛ Also Check: NCERT Solutions for Class 9 Maths Chapter 1
NCERT Exemplar Class 9 Maths Exercise 1.3 Problem 9(iii)
Simplify the following: 4√12 × 7√6
Summary:
Irrational numbers are the set of real numbers that cannot be expressed in the form of a fraction, p/q where p and q are integers. The simplified form of the expression is \(\sqrt[28]{2^{18}\times 3^{11}}\)
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