Chapter 1.3, Problem 71PS

Solution

To find: the value of $x$ .

The value of $x=-10\pm 10\sqrt{2}$ .

Given:

Side of square garden are $10\text{ feet long}\text{.}$

Concept Used:

Algebraic identity: $a^2+2ab+b^2={\left(a+b\right)}^2$ .

Properties of square root:

  $\begin{array}{c}{a}^2=b\\ a=\pm \sqrt{b}\end{array}$

Calculation:

Side of square garden are $10\text{ feet long}\text{.}$

Formula of Area of square $=\text{side }\times \text{ side}$ .

Area of square $=10\times 10{\text{feet}}^2$

  $=100{\text{ feet}}^2$

Area of square garden need to be doubled:

New area of square garden is $2\times 100=200{\text{ feet}}^2$

New side of square after adding the same distance of $x$ to the sides is $x+10\text{ feet long}\text{.}$

  $\left(x+10\right)\times \left(x+10\right)=200$

Solving the above equation:

  $\begin{array}{c}{\left(x+10\right)}^2=200\\ \left(x+10\right)=\pm \sqrt{200}\\ x+10=\pm 10\sqrt{2}\left\{\because 200=2\times 10\times 10\right\}\\ x=-10\pm 10\sqrt{2}\end{array}$

Therefore, the value of $x=-10\pm 10\sqrt{2}$ .

Conclusion:

The value of $x=-10\pm 10\sqrt{2}$ .