Solving a Quadratic Equation with Two Variables

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SUMMARY

The forum discussion focuses on solving the quadratic equation \(4x^2 + 4xy - 6y - 9 = 9\) for the variable \(x\). Participants clarify the correct application of the quadratic formula and completing the square, leading to the conclusion that the final expression should be \(2x + y = \pm(y + 3)\). Additionally, the importance of treating \(y\) as a constant during the solution process is emphasized. The discussion also mentions the use of the POLYSMLT tool on a TI-83 calculator for solving polynomial equations.

PREREQUISITES
  • Understanding of quadratic equations and their standard forms
  • Familiarity with the quadratic formula
  • Knowledge of completing the square technique
  • Experience with polynomial solving tools like POLYSMLT on TI-83 calculators
NEXT STEPS
  • Research the quadratic formula and its applications in solving equations
  • Learn how to complete the square for various types of quadratic equations
  • Explore the features and usage of the POLYSMLT tool on TI-83 calculators
  • Study the implications of treating variables as constants in algebraic equations
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Students, educators, and anyone interested in mastering quadratic equations, particularly those involving multiple variables and the application of algebraic techniques.

Amer
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Solve for $x$:

$$4x^2 +4xy - 6y - 9 = 9 $$
 
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I would write the equation as a quadratic in x in standard form:

$\displaystyle 4x^2+4yx-(6(y+3))=0$

Now apply the quadratic formula.
 
Completing the square

$ (2x)^2 + 2(2x)y -6y - 9 = 0 $
$ (2x+y)^2 -y^2 -6y - 9 = 0 $
$(2x+y)^2 - (y+3)^2 = 0 $
$(2x+y)^2 = (y+3)^2 $
$ 2x+y = |y+3| $
 
What happened to the 9 originally on the right side?
 
MarkFL said:
What happened to the 9 originally on the right side?

lol typo
 
I've made a few of those in my time! (Tauri)
 
Amer said:
Completing the square

$ (2x)^2 + 2(2x)y -6y - 9 = 0 $
$ (2x+y)^2 -y^2 -6y - 9 = 0 $
$(2x+y)^2 - (y+3)^2 = 0 $
$(2x+y)^2 = (y+3)^2 $
$ 2x+y = |y+3| $

Last line should be:

\(2x+y = \pm(y+3)\)

CB
 
CaptainBlack said:
Last line should be:

\(2x+y = \pm(y+3)\)

CB

what is wrong in my solution ?
 
Amer said:
Solve for x:

$$4x^2 +4xy - 6y - 9 = 9 $$

Amer said:
what is wrong in my solution ?

Since x is the variable and y is considered to be a constant your solution should read

$$|2x+y| = y+3$$

as CB has pointed out.
 
  • #10
Or you can use POLYSMLT if you have a TI83
 

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