Question

Which among the following is NOT a perfect square?

• A. $9$
• B. $900$
• C. $90$

Answer 1

The correct option is C $90$

Let's take a look at each choice one-by-one:

$\bullet \underset{–}{9}$
We can write $9$ as:

$\Rightarrow 9=3\times 3$
$\Rightarrow 9=3^2$
$\Rightarrow \sqrt{9}=\sqrt{3^2}$ (Taking square root both the sides)

$\Rightarrow 9^{\frac{1}{2}}=(3^2{)}^{\frac{1}{2}}$
$\Rightarrow 9^{\frac{1}{2}}=3^{2\times \frac{1}{2}}(\because (a^m{)}^n=a^{m\times n})$
$\Rightarrow 9^{\frac{1}{2}}=3$
$\Rightarrow \sqrt{9}=3$

$\therefore 9$ is a perfect square.$\underset{–}{}$

$\bullet \underset{–}{900}$
We can write $900$ as:

$900=30\times 30$
$\Rightarrow 900={30}^2$
$\Rightarrow \sqrt{900}=\sqrt{{30}^2}$ (Taking square root both the sides)

$\Rightarrow {900}^{\frac{1}{2}}=({30}^2{)}^{\frac{1}{2}}$
$\Rightarrow {900}^{\frac{1}{2}}={30}^{2\times \frac{1}{2}}(\because (a^m{)}^n=a^{m\times n})$
$\Rightarrow {900}^{\frac{1}{2}}=30$
$\Rightarrow \sqrt{900}=30$

$\therefore 900$ is a perfect square. $\underset{–}{}$

$\bullet \underset{–}{90}$
We can write $90$ as:

$90=9\times 10$
$\Rightarrow 90=3\times 3\times 5\times 2$
$\Rightarrow 90=3^2\times 5\times 2$
$\Rightarrow \sqrt{90}=\sqrt{3^2\times 5\times 2}$ (Taking square root both the sides)
$\Rightarrow {90}^{\frac{1}{2}}=(3^2\times 5\times 2{)}^{\frac{1}{2}}$

$\Rightarrow {90}^{\frac{1}{2}}=(3^2{)}^{\frac{1}{2}}\times (5\times 2{)}^{\frac{1}{2}}$ $(\because (a\times b{)}^n=a^n\times b^n)$

$\Rightarrow {90}^{\frac{1}{2}}=(3{)}^{2\times \frac{1}{2}}\times (5\times 2{)}^{\frac{1}{2}}$
$(\because (a^m{)}^n=a^{m\times n})$

$\Rightarrow \sqrt{90}=3\times \sqrt{10}$

We see in the above equation that $\sqrt{10}=\sqrt{5\times 2}$ would not give any integer, i.e., $10$ is NOT a perfect square.

Hence, we can conclude that $90$ is also NOT a perfect square.

$\therefore$ Option (c.) is the correct answer.