Q:

1.) Use the quadratic formula to solve the equation. If necessary, round to the nearest hundredth .x squared minus 21 x equals negative 4 x2.) Which kind of function best models the data in the table? Use differences or ratios.x y0 1.71 6.82 27.23 108.84 435.23.) Solve the system of equations algebraically. Show all of your steps. y equals x squared plus two x; y equals three x plus twentyPLEASE HELP!

Accepted Solution

A:
Part 1
We are given [tex]x^2-21x=-4x[/tex]. This can be rewritten as [tex]x^2-18x=0[/tex].
Therefore, a=1, b=-18, c=0.
Using the quadratic formula
     [tex]x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}=\frac{-\left(-18\right)\pm \sqrt{\left(-18\right)^2-4\left(1\right)\left(0\right)}}{2\left(1\right)}[/tex]
     [tex]x=\frac{18\pm 18}{2}[/tex]

The values of x are
     [tex]x_1=\frac{18-18}{2}=0[/tex]
     [tex]x_2=\frac{18+18}{2}=18[/tex]

Part 2
Since the values of y change drastically for every equal interval of x, the function cannot be linear. Therefore, the kind of function that best suits the given pairs is a quadratic function. 

Part 3.
The first equation is [tex]y=x^2+2[/tex].
The second equation is [tex]y=3x+20[/tex].

We have 
     [tex]x^2+2=3x+20[/tex]
     [tex]x^2-3x-18=0[/tex]
Factoring, we have 
     [tex]\left(x-6\right)\left(x+3\right)=0[/tex]
Equating both factors to zero.
     [tex]x_1-6=0\rightarrow x_1=6[/tex]
     [tex]x_2+3=0\rightarrow x_2=-3[/tex]

When the value of x is 6, the value of y is 
     [tex]y=3\left(6\right)+20=38[/tex]

When the value of x is -3, the value of y is 
     [tex]y=3\left(-3\right)+20=11[/tex]

Therefore, the solutions are (6,38) or (-3,11)