Chapter MPS, Problem 5.5MPS

(a)

To determine

To Find: The quadratic equation representing Sandy’s closet

Answer to Problem 5.5MPS

  $\left(x^2-22x+57\right)=0$

Explanation of Solution

Given:

  $\begin{array}{l}Initialsizeofroom=10feetX12Feet=120Sq.ft\\ \dim ensiondeductedoneachside=x\\ Newareaofthecloset=63Sq.ft\end{array}$

Formula Used:

  $\begin{array}{l}y\left(x\right)=a{x}^2+bx+c=\left(x+c\right)\left(x+d\right)\end{array}$

Calculation:

  $\begin{array}{l}Newlegthoftheroom=10-x\\ Newwidthoftheroom=12-x\\ Areaoftheroom=\left(10-x\right)\left(12-x\right)=63\\ \text{⇒}120-10x-12x+x^2=63\\ \text{⇒}{x}^2-22x+120-63=0\\ \text{⇒}\left(x^2-22x+57\right)=0\end{array}$

Conclusion:

Thus the required equation is $\left(x^2-22x+57\right)=0$ .

(b)

To determine

To Find: The new dimensions of Sandy’s closet.

Answer to Problem 5.5MPS

  $\begin{array}{l}Newlengthis10-3=7feet\\ Newwidthis12-3=9feet\end{array}$

Explanation of Solution

Given:

  $\begin{array}{l}Initialsizeofroom=10feetX12Feet=120Sq.ft\\ \dim ensiondeductedoneachside=x\\ Newareaofthecloset=63Sq.ft\end{array}$

Formula Used:

  $\begin{array}{l}y\left(x\right)=a{x}^2+bx+c=\left(x+c\right)\left(x+d\right)\end{array}$

Calculation:

  $\begin{array}{l}Newlegthoftheroom=10-x\\ Newwidthoftheroom=12-x\\ Areaoftheroom=\left(10-x\right)\left(12-x\right)=63\\ \text{⇒}120-10x-12x+x^2=63\\ \text{⇒}{x}^2-22x+120-63=0\\ \text{⇒}\left(x^2-22x+57\right)=0\end{array}$

  $\begin{array}{l}\left(x^2-22x+57\right)=0\\ Byu\sin gthemiddletermsplitmethod\\ \left(x^2-19x-3x+57\right)=0\\ \left(x\left(x-19\right)-3\left(x-19\right)\right)=0\\ \left(x-3\right)\left(x-19\right)=0\\ x=3orx=19\\ asweknowthatthesmallersideis10sox\le 10\\ sox=3\end{array}$

Conclusion:

Thus the required new dimensions are

  $\begin{array}{l}\text{ New length is }10-3=7\text{ feet}\\ \text{New width is }12-3=9\text{ feet}\end{array}$