Question

Find the positive square root of $36{\mathrm{x}}^2+60\mathrm{x}+25$

Answer 1

Solution:

Step 1: Factorize the given polynomial:

Let $\mathrm{p}\left(\mathrm{x}\right)=36{\mathrm{x}}^2+60\mathrm{x}+25$

The given polynomial can be rewritten as:

$$\begin{gathered} \mathrm{p}\left(\mathrm{x}\right)={\left(6\mathrm{x}\right)}^2+2\left(6\mathrm{x}\right)\left(5\right)+{\left(5\right)}^2 \\ \Rightarrow \mathrm{p}\left(\mathrm{x}\right)={(6\mathrm{x}+5)}^2............\left(1\right)\mathbf{[}\mathbf{\because }{\mathbf{(}\mathbf{a}\mathbf{+}\mathbf{b}\mathbf{)}}^{\mathbf{2}}\mathbf{=}{\mathbf{a}}^{\mathbf{2}}\mathbf{+}\mathbf{2}\mathbf{ab}\mathbf{+}{\mathbf{b}}^{\mathbf{2}}\mathbf{]} \end{gathered}$$

Step 2: Find the positive square root of $\mathrm{p}\left(\mathrm{x}\right)$:

$$\begin{gathered} \sqrt{\mathrm{p}\left(\mathrm{x}\right)}=\pm \sqrt{{(6\mathrm{x}+5)}^2}\left[\mathrm{Taking}\mathrm{square}\mathrm{root}\mathrm{of}\mathrm{equation}\left(1\right)\right] \\ =\pm (6\mathrm{x}+5) \end{gathered}$$

Final answer: Hence, the positive square root of $36{\mathrm{x}}^2+60\mathrm{x}+25\mathrm{is}\mathbf{}\mathbf{(}\mathbf{6}\mathbf{x}\mathbf{+}\mathbf{5}\mathbf{)}.$