Problem with the quadratic function

[OdinatorX]

Hello all:

I have run into some difficulties with a question involving the use of the quadratic formula. I have tried using matrices and elimination to solve but have not been getting the correct answer when I line it up into the equation. The answer to this question is not in the back of the book. Here it goes, the quadratic equation is ofcourse S=ap^2+bp+c the first line is S=100 and p=29.95, next line S=70 and p=39.95 and the final line is S=50 and p=49.95. I need to find the value of the a, b and c. Any help given would be appreciated.
thankyou.

 

[Denis]

Well, you've thoroughly confused this Canadian :shock:

Can you post the problem right out of the book?

 

[Unco]

Hello all:

I have run into some difficulties with a question involving the use of the general quadratic equation. I have tried using matrices and elimination to solve but have not been getting the correct answer when I line it up into the equation.
So show us your three equations and subsequent attempt solve, Odi.

The answer to this question is not in the back of the book. Here it goes, the quadratic equation is ofcourse S=ap^2+bp+c the first point is S=100 and p=29.95, next point S=70 and p=39.95 and the final point is S=50 and p=49.95. I need to find the value of the a, b and c. Any help given would be appreciated.
thankyou.

 

[soroban]

Hello, OdinatorX!

Let's see if I understand the problem . . .


We have a quadratic function: \(\displaystyle \:S \:=\:ap^2\,+\,bp\,+\,c\)

And we are told:
. . \(\displaystyle \begin{array}\text{When }p\,=\,29.95, & S\,=\,100 & \;(a) \\
\text{When }p\,=\,39.95, & S\,=\;70 & \;(b) \\
\text{When }p\,=\,49.95, & S\,=\;50 & \;(c)\end{array}\)

\(\displaystyle \begin{array}{ccc}\text{From (a): }a(29.95^2)\,+\,b(29.95)\,+\,c\:=\:100 & \;\Rightarrow\; & 897.0025a\,+\,29.95b\,+\,c\:=\:100 & \;(1)\\
\text{From (b): }a(39.95^2)\,+\,b(39.95)\,+\,c\:=\:70 & \;\Rightarrow\; & 1596.0025a\,+\,39.95b\,+\,c\:=\:70 & \;(2)\\
\text{From (c): }a(49.95^2)\,+\,b(49.95)\,+\,c\:=\:50 & \;\Rightarrow\; & 2495.0025a\,+\,49.95b\,+\,c\:=\:50 & \;(3)
\end{array}\)

Subtract (1) from (2): \(\displaystyle \:699a\,+\,10b\:=\:-30\;\;(4)\)
Subtract (2) from (3): \(\displaystyle \:899a\,+\,10b\:=\:-20\;\;(5)\)

Subtract (4) from (5): \(\displaystyle \:200a \:=\:10\;\;\Rightarrow\;\;\fbox{a\,=\,0.05}\)

Substitute into (4): \(\displaystyle \:699(0.05)\,+\,10b\:=\:=30\;\;\Rightarrow\;\;\fbox{b\,=\,-6.495}\)

Substitute into (1): \(\displaystyle \:897.0225(0.05)\,+\,29.85(-6.495)\,+\,c\:=\:100\;\;\Rightarrow\;\;\fbox{c\,=\,249.675125}\)


Therefore: \(\displaystyle \L\:S\;=\;0.05p^2\,-\,6.495p\,+\,249.675125\)

If you prefer fractions: \(\displaystyle \L\:S\;=\;\frac{1}{20}p^2\,-\,\frac{1299}{200}p\,+\,\frac{1,997,401}{8,000}\)