Lesson Completing the Square to Solve General Quadratic Equation

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About josgarithmetic: Academic and job experience with beginning & intermediate Algebra. Tutorial help mostly for Basic Math and up through intermediate algebra.

COMPLETING THE SQUARE TO SOLVE GENERAL QUADRATIC EQUATION

Review the lesson, "What is Completing the Square? With Visual Explanation", before studying this lesson.

A rectangle having side lengths x and x+b has an area, -c.
x is a variable, and b and c are assumed constant.
Additionally, we will use a coefficient on x^2, a constant, a.

The area, -c, could be expressed using ax%5E2%2Bbx=-c

. If all terms are operated to be on one side of the equation, then obtained would be highlight_green%28ax%5E2%2Bbx%2Bc=0%29

.
The left member, the quadratic trinomial, is assumed to not be a perfect square.

The goal is to solve the equation for x. We want to use completing the square, but first a factorization is needed:

a%28x%5E2%2B%28b%2Fa%29x%2Bc%2Fa%29=0

The part of the expression, x%5E2%2B%28b%2Fa%29x

is a representative rectangular area of x%28x%2B%28b%2Fa%29%29

, and it is on this that the completion of the square is used. The missing square term would be %28b%2F%282a%29%29%5E2

, as already learned in the What is Completing The Square lesson.

Continuing,

Add and subtract the square term inside the grouped expression, so that the area remains numerically unchanged in value.

a%28x%5E2%2B%28b%2Fa%29x%2Bc%2Fa%2B%28b%2F%282a%29%29%5E2-%28b%2F%282a%29%29%5E2%29=0

a%28x%5E2%2B%28b%2Fa%29x%2Bc%2Fa%2B%28b%2F%282a%29%29%5E2-%28b%2F%282a%29%29%5E2%29=0

a%28x%5E2%2B%28b%2Fa%29x%2B%28b%2F%282a%29%29%5E2%2Bc%2Fa-%28b%2F%282a%29%29%5E2%29=0

a%28%28x%5E2%2B%28b%2Fa%29x%2B%28b%2F%282a%29%29%5E2%29%2Bc%2Fa-%28b%2F%282a%29%29%5E2%29=0

a%28%28x%2B%28b%2F%282a%29%29%29%5E2%2Bc%2Fa-%28b%5E2%29%2F%284a%5E2%29%29=0

a%28x%2Bb%2F%282a%29%29%5E2%2Bc-%28b%5E2%29%2F%284a%29=0

a%28x%2Bb%2F%282a%29%29%5E2=%28b%5E2%29%2F%284a%29-c

%28x%2Bb%2F%282a%29%29%5E2=%28b%5E2%29%2F%284a%5E2%29-c%2Fa

%28x%2Bb%2F%282a%29%29%5E2=%28b%5E2%29%2F%284a%5E2%29-%284ac%29%2F%284a%5E2%29

%28x%2Bb%2F%282a%29%29%5E2=%28b%5E2-4ac%29%2F%284a%5E2%29

Exponentiate both sides to 1%2F2

power,

x%2Bb%2F%282a%29=0%2B-+sqrt%28b%5E2-4ac%29%2F%282a%29

x=-%28b%2F%282a%29%29%2B-+sqrt%28b%5E2-4ac%29%2F%282a%29

highlight%28x=%28-b%2B-+sqrt%28b%5E2-4ac%29%29%2F%282a%29%29

The General Solution for ax%5E2%2Bbx%2Bc=0

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