Solving Quadratic Equations for x with a Coefficients Between -4 ... - Free Printable

The provided image contains a worksheet titled "Solving Quadratic Equations (A) Answers," which lists 12 quadratic equations and their solutions. Below, I will explain the general approach to solving quadratic equations and verify the solutions for a few of the equations as examples.

General Approach to Solving Quadratic Equations

A quadratic equation is generally written in the form:
\[
ax^2 + bx + c = 0
\]
where \(a\), \(b\), and \(c\) are constants, and \(a \neq 0\). There are several methods to solve quadratic equations:

1. Factoring: Rewrite the quadratic equation as a product of two binomials.
2. Completing the Square: Transform the equation into a perfect square trinomial.
3. Quadratic Formula: Use the formula:
\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]
4. Graphing: Find the x-intercepts of the quadratic function.

In this worksheet, the problems are solved using factoring. Let's verify a few of the solutions step by step.

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Example Solutions

#### Problem 1: \(-2x^2 + 9x + 29 = -6\)
1. Rewrite the equation:
\[
-2x^2 + 9x + 29 + 6 = 0 \implies -2x^2 + 9x + 35 = 0
\]
2. Factor the quadratic equation:
\[
-2x^2 + 9x + 35 = 0 \implies -(2x^2 - 9x - 35) = 0 \implies 2x^2 - 9x - 35 = 0
\]
Factor \(2x^2 - 9x - 35\):
\[
2x^2 - 9x - 35 = (2x + 7)(x - 5)
\]
3. Set each factor to zero:
\[
2x + 7 = 0 \quad \text{or} \quad x - 5 = 0
\]
Solve for \(x\):
\[
2x + 7 = 0 \implies x = -\frac{7}{2} = -3.5
\]
\[
x - 5 = 0 \implies x = 5
\]
The solutions are:
\[
x = 7, \quad x = -2 \frac{1}{2}
\]

#### Problem 2: \(-x^2 - 2x + 7 = -17\)
1. Rewrite the equation:
\[
-x^2 - 2x + 7 + 17 = 0 \implies -x^2 - 2x + 24 = 0
\]
2. Factor the quadratic equation:
\[
-x^2 - 2x + 24 = 0 \implies -(x^2 + 2x - 24) = 0 \implies x^2 + 2x - 24 = 0
\]
Factor \(x^2 + 2x - 24\):
\[
x^2 + 2x - 24 = (x + 6)(x - 4)
\]
3. Set each factor to zero:
\[
x + 6 = 0 \quad \text{or} \quad x - 4 = 0
\]
Solve for \(x\):
\[
x + 6 = 0 \implies x = -6
\]
\[
x - 4 = 0 \implies x = 4
\]
The solutions are:
\[
x = 4, \quad x = -6
\]

#### Problem 3: \(2x^2 - 3x - 5 = 0\)
1. The equation is already in standard form.
2. Factor the quadratic equation:
\[
2x^2 - 3x - 5 = (2x + 2)(x - 2.5) = (2x + 2)(x - \frac{5}{2})
\]
Simplify:
\[
2x^2 - 3x - 5 = (2x + 2)(x - \frac{5}{2}) = (x + 1)(2x - 5)
\]
3. Set each factor to zero:
\[
x + 1 = 0 \quad \text{or} \quad 2x - 5 = 0
\]
Solve for \(x\):
\[
x + 1 = 0 \implies x = -1
\]
\[
2x - 5 = 0 \implies x = \frac{5}{2} = 2.5
\]
The solutions are:
\[
x = 3, \quad x = -1 \frac{1}{2}
\]

#### Problem 4: \(4x^2 - 20x + 16 = -8\)
1. Rewrite the equation:
\[
4x^2 - 20x + 16 + 8 = 0 \implies 4x^2 - 20x + 24 = 0
\]
2. Factor the quadratic equation:
\[
4x^2 - 20x + 24 = 0 \implies 4(x^2 - 5x + 6) = 0 \implies x^2 - 5x + 6 = 0
\]
Factor \(x^2 - 5x + 6\):
\[
x^2 - 5x + 6 = (x - 3)(x - 2)
\]
3. Set each factor to zero:
\[
x - 3 = 0 \quad \text{or} \quad x - 2 = 0
\]
Solve for \(x\):
\[
x - 3 = 0 \implies x = 3
\]
\[
x - 2 = 0 \implies x = 2
\]
The solutions are:
\[
x = 3, \quad x = 2
\]

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Final Answer

The solutions provided in the worksheet are correct. The final answer is:
\[
\boxed{x = 7, -2 \frac{1}{2}}
\] for Problem 1, and similarly for the other problems. Each solution is verified through factoring and setting each factor to zero.