Problem 37
Question
Solve each equation by factoring. \(4 x^{2}+8 x=-3\)
Step-by-Step Solution
Verified Answer
The solutions are \(x = -\frac{3}{2}\) and \(x = -\frac{1}{2}\).
1Step 1: Move All Terms to One Side
Start by moving all terms to one side of the equation to set it to zero. Add 3 to both sides of the equation: \[4x^2 + 8x + 3 = 0\]
2Step 2: Factor the Quadratic Expression
Factor the quadratic expression \(4x^2 + 8x + 3\) into two binomials. Look for two numbers that multiply to \(4 \times 3 = 12\) and add to \(8\). These numbers are 6 and 2. Rewrite the middle term using these numbers:\[4x^2 + 6x + 2x + 3 = 0\]Now, factor by grouping:\[(4x^2 + 6x) + (2x + 3) = 0\]Factor out the greatest common factors from each group:\[2x(2x + 3) + 1(2x + 3) = 0\]Factor out the common binomial factor:\[(2x + 3)(2x + 1) = 0\]
3Step 3: Solve Each Factor
Set each factor equal to zero and solve for \(x\):1. For \(2x + 3 = 0\): \[2x = -3 \quad \Rightarrow \quad x = -\frac{3}{2}\]2. For \(2x + 1 = 0\): \[2x = -1 \quad \Rightarrow \quad x = -\frac{1}{2}\]
Key Concepts
FactoringZero Product PropertyPolynomial Equations
Factoring
Factoring is a method used to rewrite polynomials as a product of simpler expressions. In the context of quadratic equations, it involves breaking down a polynomial like \(4x^2 + 8x + 3\) into simpler binomials that, when multiplied together, yield the original equation.
The key to factoring is to find two numbers that multiply to the product of the leading coefficient and the constant term, yet add up to the middle coefficient. In our example, we needed two numbers that multiply to \(4 \times 3 = 12\) and add up to \(8\). These numbers are \(6\) and \(2\), allowing us to rewrite \(8x\) as \(6x + 2x\).
After rewriting the polynomial, we proceed by factoring by grouping. This means we group terms so that common factors can be factored out, leading us to the binomials \((2x + 3)(2x + 1)\). It's a critical step as it helps to simplify and solve the polynomial equation.
The key to factoring is to find two numbers that multiply to the product of the leading coefficient and the constant term, yet add up to the middle coefficient. In our example, we needed two numbers that multiply to \(4 \times 3 = 12\) and add up to \(8\). These numbers are \(6\) and \(2\), allowing us to rewrite \(8x\) as \(6x + 2x\).
After rewriting the polynomial, we proceed by factoring by grouping. This means we group terms so that common factors can be factored out, leading us to the binomials \((2x + 3)(2x + 1)\). It's a critical step as it helps to simplify and solve the polynomial equation.
Zero Product Property
The Zero Product Property is a fundamental principle that states if the product of two expressions is zero, then at least one of the expressions must be zero. This property is essential when solving factored equations because it allows us to set each factor to zero separately.
In our exercise, after factoring \(4x^2 + 8x + 3\) into \((2x + 3)(2x + 1) = 0\), we apply the Zero Product Property. This gives us two separate equations to solve:
Thus, the Zero Product Property helps break down complex polynomial equations into simpler linear equations, each of which is easier to solve individually.
In our exercise, after factoring \(4x^2 + 8x + 3\) into \((2x + 3)(2x + 1) = 0\), we apply the Zero Product Property. This gives us two separate equations to solve:
- \(2x + 3 = 0\)
- \(2x + 1 = 0\)
Thus, the Zero Product Property helps break down complex polynomial equations into simpler linear equations, each of which is easier to solve individually.
Polynomial Equations
Polynomial equations are equations involving polynomials, which are expressions made up of variables and coefficients with operations of addition, subtraction, multiplication, and non-negative integer exponents of variables.
In the given exercise, our polynomial equation is quadratic, meaning it involves the second degree of the variable \(x\). Such equations are general expressions like \(ax^2 + bx + c = 0\), where \(a\), \(b\), and \(c\) are constants and \(a eq 0\).
Solving polynomial equations like these involves numerous techniques, with factoring being one of the most straightforward methods for quadratics. The goal is to simplify the equation to be able to apply the Zero Product Property. This method helps us find the values of \(x\), known as roots or solutions, that satisfy the original equation.
Understanding and mastering the solving of polynomial equations opens the door to exploring more complex mathematical concepts and problem-solving techniques.
In the given exercise, our polynomial equation is quadratic, meaning it involves the second degree of the variable \(x\). Such equations are general expressions like \(ax^2 + bx + c = 0\), where \(a\), \(b\), and \(c\) are constants and \(a eq 0\).
Solving polynomial equations like these involves numerous techniques, with factoring being one of the most straightforward methods for quadratics. The goal is to simplify the equation to be able to apply the Zero Product Property. This method helps us find the values of \(x\), known as roots or solutions, that satisfy the original equation.
Understanding and mastering the solving of polynomial equations opens the door to exploring more complex mathematical concepts and problem-solving techniques.
Other exercises in this chapter
Problem 37
Solve each equation by completing the square. \(x^{2}+6 x+13=0\)
View solution Problem 37
Solve each equation by graphing. If exact roots cannot be found, state the consecutive integers between which the roots are located. $$ x^{2}-4 x+2=0 $$
View solution Problem 37
Complete parts a-c for each quadratic function. a. Find the \(y\) -intercept, the equation of the axis of symmetry, and the \(x\) -coordinate of the vertex. b.
View solution Problem 38
Complete parts a–c for each quadratic equation. a. Find the value of the discriminant. b. Describe the number and type of roots. c. Find the exact solutions by
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