Solve the following equation graphically to two decimal places using a graphing cal..

Sep 2017
10
0
Laredo, Texas
1.2 x^2 - 1.3 x - 5.5

Which values of x, to two decimal places, satisfy the inequality? The answer is -1.67<x<2.75 How do I solve this?

Could I use a graphing calculator so I can solve for these? I must have two answers.
 
Last edited:

skeeter

MHF Helper
Jun 2008
16,217
6,765
North Texas
Re: Solve the following equation graphically to two decimal places using a graphing c

1.2 x^2 - 1.3 x - 5.5 ...

Which values of x, to two decimal places, satisfy the inequality? The answer is -1.67<x<2.75 How do I solve this?

Could I use a graphing calculator so I can solve for these? I must have two answers.
What inequality?. All I see is a quadratic expression ...
 
Sep 2017
10
0
Laredo, Texas
Re: Solve the following equation graphically to two decimal places using a graphing c

I must solve for the two x's.
 
Sep 2017
10
0
Laredo, Texas
Re: Solve the following equation graphically to two decimal places using a graphing c

Never mind. I discovered a method. :0
 
Sep 2017
10
0
Laredo, Texas
Re: Solve the following equation graphically to two decimal places using a graphing c

Don't worry. I discovered this solution myself!
 

HallsofIvy

MHF Helper
Apr 2005
20,249
7,909
Re: Solve the following equation graphically to two decimal places using a graphing c

But we still don't know what the problem was! And we are curious.
 

HallsofIvy

MHF Helper
Apr 2005
20,249
7,909
Re: Solve the following equation graphically to two decimal places using a graphing c

It was probably "find all x for which \(\displaystyle 1.2x^2- 1.3x- 5.5< 0\)". Using a graphing calculator, you graph \(\displaystyle y= 1.2x^2- 1.3x- 5.5\) (I used the one at https://www.desmos.com/calculator). Notice that parabola goes below the x-axis and, if necessary use the zoom feature to determine that the parabola crosses the x-axis at "x= -1.667" (which is probably -5/3) and at "x= 2.75" (11/4). We can check that "probably" by setting x= -5/3 in the inequality- 1.2(25/9)- 1.3(-5/3)- 5.5= 0.

Or you could use the quadratic formula to determine the endpoints of the interval: \(\displaystyle x= \frac{1.3\pm\sqrt{1.3^2- 4(1.2)(5.5)}}{2(1.2)}= \frac{1.3\pm\sqrt{28.09}}{2.4}= \frac{1.3\pm 5.3}{2.4}\).