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Understanding the General Formula for Solving Quadratic Equations

Introduction to Quadratic Equations

• Daniel Carrión introduces the topic of quadratic equations, emphasizing its importance and relevance.

• A quadratic equation is defined as one where the variable x appears at least squared (e.g., x^2 + 12x + 8 = 0 ).

• The general form of a quadratic equation is presented as ax^2 + bx + c = 0 , with:

• a : quadratic term

• b : linear term

• c : constant term

Identifying Coefficients in Examples

• In the example 3x^2 - 2x + 4 = 0 :

• a = 3

• b = -2

• c = 4

• Another example, 6x^2 + 3x -5 = 0, identifies:

• a = 6

• b = 3

• c = -5

• For the equation x^2 +5x +8 =0:

• Recognizes that if no number accompanies the square term, it defaults to one ( a=1, b=5, c=8).

More Examples and Coefficient Identification

• In the case of 2x^2-x+1=0:

• Identifies coefficients as:

• a =2,

• b=-1,

• and confirms that constant terms are standalone.

The General Formula for Solutions

• The general formula for solving quadratic equations is introduced:

[ x = frac-b ± √b²−4ac2a ]

• Emphasizes that this formula yields two results due to the plus-minus sign indicating two potential solutions.

Step-by-Step Example Solution

• Begins solving an example equation:

[ x² +2x −8 =0]

Identifies coefficients:

• Here,

• a =1,,

• b=2,,

• c=-8.

Substituting Values into the Formula

• Substitutes values into the general formula:

[ x=frac-b±√b²−4ac2a]

Replaces letters with their respective values.

Performing Calculations

• Continues calculations step by step:

• Calculates intermediate steps like squaring and multiplying constants.

• Arrives at simplified expressions leading towards final results.

Final Results from Calculations

• Concludes calculations showing how to derive both possible values for x using both signs from earlier steps.

Verifying Solutions in Original Equation

• Explains how to verify solutions by substituting back into original equations.

• Uses found values of x (e.g., substituting back into original equation).

Solving Quadratic Equations: An Example

Evaluating the First Solution (x1 = 2)

• The equation x^2 + 2x - 8 = 0 is analyzed by substituting x_1 = 2 .

• Calculating 2^2 + 2(2) - 8 :

• 4 + 4 - 8 = 0 , confirming that both sides equal zero.

• This validates that x_1 = 2 is a correct solution since it satisfies the original equation.

Evaluating the Second Solution (x2 = -4)

• The second value, x_2 = -4 , is substituted into the same quadratic equation.

• The expression becomes:

• (-4)^2 + 2(-4) - 8 = 0 , which mirrors the original equation with values replaced.

Performing Calculations for x2

• Squaring negative four results in:

• (-4)(-4) = +16 .

• Continuing with calculations:

• Adding terms gives us +16 + (-8) = +8.

• Thus, we have +16 - 8 = +8, leading to a final evaluation of zero on both sides of the equation.

Conclusion on Solutions