A quadratic equation is a type of equation that can be written in the form:

[
ax^2 + bx + c = 0
]

where:

• ( a ), ( b ), and ( c ) are numbers (with ( a \neq 0 )),
• ( x ) is the variable we are trying to solve for.

Quadratic equations form a parabola when graphed, and they often have two solutions.

Methods to Solve Quadratic Equations

There are several ways to solve quadratic equations:

1. Factoring
2. Using the Quadratic Formula
3. Completing the Square

Let’s explore each method with examples!

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1. Solving by Factoring

Steps:

1. Write the equation in standard form: ( ax^2 + bx + c = 0 ).
2. Factor the quadratic expression on the left side.
3. Set each factor equal to zero.
4. Solve for ( x ).

Example:

Solve ( x^2 – 5x + 6 = 0 ).

Step 1: The equation is already in standard form.

Step 2: Factor the quadratic expression.

We look for two numbers that multiply to ( +6 ) and add up to ( -5 ).

Those numbers are ( -2 ) and ( -3 ).

So,

[
x^2 – 5x + 6 = (x – 2)(x – 3)
]

Step 3: Set each factor equal to zero.

[
\begin{align} x – 2 &= 0 \quad \text{or} \quad x – 3 = 0 \end{align}
]

Step 4: Solve for ( x ).

[
\begin{align} x &= 2 \quad \text{or} \quad x = 3 \end{align}
]

Solution: ( x = 2 ) or ( x = 3 ).

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2. Using the Quadratic Formula

When factoring is difficult or impossible, we can use the quadratic formula:

[
x = \frac{ -b \pm \sqrt{b^2 – 4ac} }{2a}
]

Steps:

1. Identify ( a ), ( b ), and ( c ) from the equation ( ax^2 + bx + c = 0 ).
2. Plug the values into the quadratic formula.
3. Simplify to find the values of ( x ).

Example:

Solve ( 2x^2 + 3x – 2 = 0 ).

Step 1: Identify ( a = 2 ), ( b = 3 ), ( c = -2 ).

Step 2: Plug into the quadratic formula.

[
x = \frac{ -3 \pm \sqrt{(3)^2 – 4 \times 2 \times (-2)} }{2 \times 2}
]

Step 3: Simplify.

First, calculate the discriminant ( D = b^2 – 4ac ):

[
D = (3)^2 – 4 \times 2 \times (-2) = 9 + 16 = 25
]

Now, compute ( x ):

[
x = \frac{ -3 \pm \sqrt{25} }{4}
]

[
x = \frac{ -3 \pm 5 }{4}
]

So we have two solutions:

1. ( x = \frac{ -3 + 5 }{4} = \frac{2}{4} = \frac{1}{2} )
2. ( x = \frac{ -3 – 5 }{4} = \frac{ -8 }{4} = -2 )

Solution: ( x = \frac{1}{2} ) or ( x = -2 ).

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3. Completing the Square

Steps:

1. Write the equation in the form ( x^2 + bx = -c ).
2. Add the square of half of ( b ) to both sides to complete the square.
3. Write the left side as a squared binomial.
4. Solve for ( x ) by taking the square root of both sides.

Example:

Solve ( x^2 + 6x + 5 = 0 ).

Step 1: Move the constant term to the other side.

[
x^2 + 6x = -5
]

Step 2: Calculate ( \left( \frac{b}{2} \right)^2 = \left( \frac{6}{2} \right)^2 = 9 ). Add 9 to both sides.

[
x^2 + 6x + 9 = -5 + 9
]

[
x^2 + 6x + 9 = 4
]

Step 3: Write the left side as a squared binomial.

[
(x + 3)^2 = 4
]

Step 4: Take the square root of both sides.

[
x + 3 = \pm \sqrt{4}
]

[
x + 3 = \pm 2
]

Now, solve for ( x ):

1. ( x + 3 = 2 ) \
   ( x = -1 )
2. ( x + 3 = -2 ) \
   ( x = -5 )

Solution: ( x = -1 ) or ( x = -5 ).

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Summary

• Factoring works when the quadratic expression can be factored into two binomials.
• Quadratic Formula works for all quadratic equations and is especially useful when factoring is hard.
• Completing the Square transforms the equation into a perfect square trinomial, making it easier to solve.

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Remember: Practice makes perfect! Try solving different quadratic equations using these methods to become more comfortable with them.