Question

How do you simplify ${\left( {\dfrac{9}{{16}}} \right)^{ - \dfrac{1}{2}}}$?

Answer 1

Hint: Here the power of the given term is negative so we will find the reciprocal of the given number. Then we will apply the exponent on each term of the fraction to get the required answer. The negative exponent rule states that if the power is negative then the denominator and numerator value can interchange.

Complete step by step solution:
The value given to us is,
${\left( {\dfrac{9}{{16}}} \right)^{ - \dfrac{1}{2}}}$
As the power is negative so we can take reciprocal of the term inside the bracket and remove the negative sign as,
$ \Rightarrow {\left( {\dfrac{9}{{16}}} \right)^{ - \dfrac{1}{2}}} = {\left( {\dfrac{{16}}{9}} \right)^{\dfrac{1}{2}}}$
Now as the power means we have to find the square root of the term we will proceed forward as,
\[ \Rightarrow {\left( {\dfrac{9}{{16}}} \right)^{ - \dfrac{1}{2}}} = \sqrt {\dfrac{{16}}{9}} \]
We know that 16 is the square of 4 and 9 is the square of 3. So, we can write above equation as:
\[ \Rightarrow {\left( {\dfrac{9}{{16}}} \right)^{ - \dfrac{1}{2}}} = \sqrt {\dfrac{{{4^2}}}{{{3^2}}}} \]
Simplifying the expression, we get
\[ \Rightarrow {\left( {\dfrac{9}{{16}}} \right)^{ - \dfrac{1}{2}}} = \dfrac{4}{3}\]

So the simplified value of ${\left( {\dfrac{9}{{16}}} \right)^{ - \dfrac{1}{2}}}$ is $\dfrac{4}{3}$.

Note:
A negative power means that we have to perform division with the base. It means how many times we have to divide the number. Negative power is the opposite of a positive power as in positive power we multiply the number that much times whereas in negative power we divide the number that much time. A square root is used to find out what number multiplied with itself gives the number inside the square root symbol. All non-negative real numbers have a unique non-negative square root called the principal square root. The number whose square root we are finding is also known as radicand.