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, including:

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: Solve \(-2x^2 + 9x + 29 = -6\)
1. Rewrite the equation:
\[
-2x^2 + 9x + 29 + 6 = 0 \implies -2x^2 + 9x + 35 = 0
\]
2. Factor the quadratic equation:
\[
-2x^2 + 9x + 35 = 0
\]
To factor, we look for two numbers that multiply to \((-2)(35) = -70\) and add to \(9\). These numbers are \(14\) and \(-5\).
\[
-2x^2 + 14x - 5x + 35 = 0
\]
Group the terms:
\[
(-2x^2 + 14x) + (-5x + 35) = 0
\]
Factor out the common factors:
\[
-2x(x - 7) - 5(x - 7) = 0
\]
Factor out the common binomial \((x - 7)\):
\[
(-2x - 5)(x - 7) = 0
\]
3. Set each factor to zero:
\[
-2x - 5 = 0 \quad \text{or} \quad x - 7 = 0
\]
Solve for \(x\):
\[
x = -\frac{5}{-2} = \frac{5}{2} = 2 \frac{1}{2} \quad \text{or} \quad x = 7
\]
So, the solutions are:
\[
x = 7 \quad \text{and} \quad x = -2 \frac{1}{2}
\]

#### Problem 7: Solve \(x^2 - x - 44 = 12\)
1. Rewrite the equation:
\[
x^2 - x - 44 - 12 = 0 \implies x^2 - x - 56 = 0
\]
2. Factor the quadratic equation:
\[
x^2 - x - 56 = 0
\]
To factor, we look for two numbers that multiply to \((-56)\) and add to \((-1)\). These numbers are \(-8\) and \(7\).
\[
x^2 - 8x + 7x - 56 = 0
\]
Group the terms:
\[
(x^2 - 8x) + (7x - 56) = 0
\]
Factor out the common factors:
\[
x(x - 8) + 7(x - 8) = 0
\]
Factor out the common binomial \((x - 8)\):
\[
(x + 7)(x - 8) = 0
\]
3. Set each factor to zero:
\[
x + 7 = 0 \quad \text{or} \quad x - 8 = 0
\]
Solve for \(x\):
\[
x = -7 \quad \text{or} \quad x = 8
\]
So, the solutions are:
\[
x = -7 \quad \text{and} \quad x = 8
\]

#### Problem 12: Solve \(-4x^2 - 26x - 6 = 30\)
1. Rewrite the equation:
\[
-4x^2 - 26x - 6 - 30 = 0 \implies -4x^2 - 26x - 36 = 0
\]
2. Factor the quadratic equation:
\[
-4x^2 - 26x - 36 = 0
\]
First, factor out the greatest common factor \((-2)\):
\[
-2(2x^2 + 13x + 18) = 0
\]
Now, factor \(2x^2 + 13x + 18\). We look for two numbers that multiply to \((2)(18) = 36\) and add to \(13\). These numbers are \(9\) and \(4\).
\[
2x^2 + 9x + 4x + 18 = 0
\]
Group the terms:
\[
(2x^2 + 9x) + (4x + 18) = 0
\]
Factor out the common factors:
\[
x(2x + 9) + 2(2x + 9) = 0
\]
Factor out the common binomial \((2x + 9)\):
\[
(x + 2)(2x + 9) = 0
\]
3. Set each factor to zero:
\[
x + 2 = 0 \quad \text{or} \quad 2x + 9 = 0
\]
Solve for \(x\):
\[
x = -2 \quad \text{or} \quad x = -\frac{9}{2} = -4 \frac{1}{2}
\]
So, the solutions are:
\[
x = -2 \quad \text{and} \quad x = -4 \frac{1}{2}
\]

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Final Answer

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