Chapter 7.1, Problem 1WE

(a)

To determine

To Solve:

  $x^2=3$

Answer to Problem 1WE

  $x=\pm \sqrt{3}$

Explanation of Solution

Concept Used:

If you take square root, then since the index of the radical is 2, which is even , there are 2 roots for the equation , one is postitive and the other one is negative:

  $\begin{array}{l}{x}^2=a\\ \Rightarrow x=\pm a\end{array}$

Calculation:

Given:

  $\begin{array}{l}{x}^2=3\\ \Rightarrow x=\pm \sqrt{3}\mathrm{..............}[\text{take square root each side]}\end{array}$

Since 3 does not have any perfect square root, it is in the simplest form.

So,

  $x=\pm \sqrt{3}$

(b)

To determine

To Solve:

  ${\left(x-1\right)}^2=3$

Answer to Problem 1WE

  $x=\pm \sqrt{3}$

Explanation of Solution

Concept Used:

If you take square root, then since the index of the radical is 2, which is even , there are 2 roots for the equation , one is postitive and the other one is negative:

  $\begin{array}{l}{x}^2=a\\ \Rightarrow x=\pm a\end{array}$

Calculation:

Given:

  $\begin{array}{l}{\left(x-1\right)}^2=3\\ \Rightarrow x-1=\pm \sqrt{3}\mathrm{..............}[\text{take square root each side]}\\ \Rightarrow x=1\pm \sqrt{3}\mathrm{................}[\text{add 1 each side]}\end{array}$

Since 3 does not have any perfect square root, it is in the simplest form.

So,

  $x=1\pm \sqrt{3}$

(c)

To determine

To Solve:

  ${\left(2x-1\right)}^2=3$

Answer to Problem 1WE

  $x=\frac{1\pm \sqrt{3}}{2}$

Explanation of Solution

Concept Used:

If you take square root, then since the index of the radical is 2, which is even , there are 2 roots for the equation , one is postitive and the other one is negative:

  $\begin{array}{l}{x}^2=a\\ \Rightarrow x=\pm a\end{array}$

Calculation:

Given:

  $\begin{array}{l}{\left(2x-1\right)}^2=3\\ \Rightarrow 2x-1=\pm \sqrt{3}\mathrm{..............}[\text{take square root each side]}\\ \Rightarrow 2x=1\pm \sqrt{3}\mathrm{................}[\text{add 1 each side]}\\ \Rightarrow \frac{2x}{2}=\frac{1\pm \sqrt{3}}{2}\mathrm{..............}[\text{divide each side by 2]}\\ \Rightarrow x=\frac{1\pm \sqrt{3}}{2}\end{array}$

Since 3 does not have any perfect square root, it is in the simplest form.

So,

  $x=\frac{1\pm \sqrt{3}}{2}$