Chapter 9.7, Problem 21PPS

(a)

To determine

Solve the equations.

Answer to Problem 21PPS

  $\left(x,y\right)=\left(1,29\right),\left(-\frac{1}{5},-\frac{23}{5}\right)$

Explanation of Solution

Given Information: $y=-5x^2+32x+2$ and $y=28x+1$

Formula Used: Factorization

Calculation:

Franklin Road is modelled $y=-5x^2+32x+2$

and

Jefferson Street is modelled $y=28x+1$

So,

  $y=-5x^2+32x+2$ and $y=28x+1$

Now, equate

  $5x^2-32x-2+28x+1=0$

So, simplify

  $5x^2-4x-1=0$

Now, factorize

  $5x^2-5x+x-1=0$

So,

  $5x\left(x-1\right)+1\left(x-1\right)=0$

Now,

  $\left(x-1\right)\left(5x+1\right)=0$

Hence,

  $x=1$ or $x=-\frac{1}{5}$

So, find value of $y$

  $y=-5x^2+32x+2$ or $y=28x+1$

Now, put $x=1$ or $x=-\frac{1}{5}$

  $y=-5{\left(1\right)}^2+\left(32\times 1\right)+2$ or $y=\left(28\times -\frac{1}{5}\right)+1$

Now,

  $y=-5+32+2$ or $y=-\frac{28}{5}+1$

Hence,

  $y=29$ or $y=-\frac{23}{5}$

Franklin Road and Jefferson Street intersect at $\left(x,y\right)=\left(1,29\right),\left(-\frac{1}{5},-\frac{23}{5}\right)$

(b)

To determine

Solve the equation.

Answer to Problem 21PPS

Franklin Road and Jefferson Street has not possible intersection.

Explanation of Solution

Given Information: $y=-5x^2+32x+2$ and $y=28x+4$

Formula Used: $D=b^2-4ac$

Calculation:

  $D$ equal to negative so $x$

• has no real solution.

$D$ equal to 0 so $x$ has 1 real solution.

$D$ equal to positive so $x$ has 2 real solution.

Franklin Road is modelled $y=-5x^2+32x+2$

and

Jefferson Street is modelled $y=28x+4$

So,

  $y=-5x^2+32x+2$ and $y=28x+4$

Now, equate

  $5x^2-32x-2+28x+4=0$

So, simplify

  $5x^2-4x+2=0$

Now, use formula $D=b^2-4ac$ to check if solution of $x$ exist or not.

As $a=5,b=-4$ and $c=2$

  $D={\left(-4\right)}^2-4\times 5\times 2$

Now,

  $D=16-40$

Hence,

  $D=-24$

As, $D$ equal to negative so $x$ has no real solution.

Franklin Road and Jefferson Street has not possible intersection