Chapter 9.6, Problem 49PPS

(a)

To determine

The equation for the area of the section with text

Answer to Problem 49PPS

The equation for the area of the section with text is $49x^2-215x+500$

Explanation of Solution

Given:

Bartolo is making a poster for the dance and want to cover three fourths of the area with text

  $\begin{array}{l}\text{Full}\text{poster}\\ lengthl=20in\\ heighth=25in\\ \text{text}\text{section}\\ l=\left(20-2x\right)\\ h=\left(25-7x\right)\end{array}$

  

Concept Used:

Area of square $\left(length\times height\right)$

  $2^x$

Calculation:

In order to find the equation for the area of the section with text simplify further as shown below:

  $\begin{array}{c}Area=l\times b\\ =\left(20-2x\right)\left(25-7x\right)\\ =20\left(25-7x\right)-7x\left(25-7x\right)\\ =500-140x-175x+49x^2\\ =49x^2-215x+500\end{array}$

Thus, the equation for the area of the section with text is $49x^2-215x+500$

(b)

To determine

Solve the equation $49x^2-215x+500=0$ by using the Quadratic Formula

Answer to Problem 49PPS

  $x=\frac{215}{98}-\frac{\sqrt{517525}}{98},\text{ and }x=\frac{215}{98}+\frac{\sqrt{51775}}{98}$

Explanation of Solution

Given:

The equation $49x^2-215x+500=0$

Concept Used:

Quadratic formula: The roots of a quadratic equation $a{x}^2+bx+c=0;\text{ where }a\ne 0$ are:

  $x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

Calculation:

In order to solve the given equation $49x^2-215x+500=0$ , since it is in the descending order of exponents, so comparing with the general equation $a{x}^2+bx+c=0$ , the value of the coefficient a, b, and c is given by

  $a=49,b=-215,\text{ and }c=500$

Now, above quadratic equation $49y^2-215y+500=0$ can be solved using quadratic formula as shown below,

  $\begin{array}{c}x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}\\ =\frac{-\left(-215\right)\pm \sqrt{{\left(-215\right)}^2-4\left(49\right)\left(500\right)}}{2\left(49\right)}\\ =\frac{215\pm \sqrt{46225-98000}}{98}\\ =\frac{215\pm \sqrt{-51775}}{98}\\ =\frac{215\pm \sqrt{-51775}}{98}\\ =\frac{215\pm \sqrt{-51775}}{98}\\ x=\frac{215}{98}-\frac{\sqrt{517525}}{98},\text{ and }x=\frac{215}{98}+\frac{\sqrt{51775}}{98}\end{array}$

Thus, the solutions of the given question are, $x=\frac{215}{98}-\frac{\sqrt{517525}}{98},\text{ and }x=\frac{215}{98}+\frac{\sqrt{51775}}{98}$ .

(c)

To determine

The margins of the poster

Answer to Problem 49PPS

  $x=\frac{215}{98}+\frac{\sqrt{51775}}{98}$

Explanation of Solution

The margins of the poster is $x=\frac{215}{98}+\frac{\sqrt{51775}}{98}$