Problem 46
Question
Find the value of \(k\) that would make the left side of each equation a perfect square trinomial. $$ x^{2}+k x+\frac{1}{4}=0 $$
Step-by-Step Solution
Verified Answer
The value of \(k\) that would make the left side a perfect square trinomial is \(1\).
1Step 1: Understand a Perfect Square Trinomial
A perfect square trinomial is a quadratic expression of the form \(a^2 + 2ab + b^2\), which factors into \(a + b)^2\). In the given equation, \(\frac{1}{4}\) will be our \(\frac{1}{2}b^2\), since it is the constant term.
2Step 2: Determine the Value of \(b\)
If \(\frac{1}{4}\) is our \(\frac{1}{2}b^2\), then \(b\) would be \(\frac{1}{2}\), because \(\left(\frac{1}{2}\right)^2 = \frac{1}{4}\).
3Step 3: Find the Coefficient of the Middle Term
Given that \(b = \frac{1}{2}\), the middle term in a perfect square trinomial, which is \((2ab)\), will be \(2 \cdot 1 \cdot \frac{1}{2} = k\).
4Step 4: Solve for \(k\)
Therefore, to form a perfect square trinomial, the value of \(k\) must be \(2 \cdot \frac{1}{2} = 1\).
Key Concepts
Understanding Quadratic ExpressionsThe Process of Factoring QuadraticsSolving Algebraic Equations
Understanding Quadratic Expressions
Quadratic expressions are fundamental algebraic expressions which take the form of \(ax^2 + bx + c\), where \(a\), \(b\), and \(c\) are constants, and \(a \eq 0\). They are called 'quadratic' because 'quadra' relates to the square term, \(x^2\), which is the highest degree in these expressions.
Understanding the structure of quadratic expressions is crucial because it underpins much of algebraic problem-solving. These expressions can represent a wide variety of phenomena, from the trajectory of a projectile to the area of a square plot of land, making their study practical and versatile. Grasping the nature of these expressions can also lead to easier factorization and solving, especially when dealing with perfect square trinomials.
Understanding the structure of quadratic expressions is crucial because it underpins much of algebraic problem-solving. These expressions can represent a wide variety of phenomena, from the trajectory of a projectile to the area of a square plot of land, making their study practical and versatile. Grasping the nature of these expressions can also lead to easier factorization and solving, especially when dealing with perfect square trinomials.
The Process of Factoring Quadratics
Perfect Square Trinomials
When factoring quadratics, one often encounters a special case known as the perfect square trinomial. Recognizing when a quadratic is a perfect square trinomial can simplify the factoring process. A perfect square trinomial always factors neatly into \(a+b)^2\) or \(a-b)^2\), essentially 'undoing' the expansion of a binomial squared.The general formula for a perfect square trinomial is \(a^2 + 2ab + b^2\), or its counterpart \(a^2 - 2ab + b^2\). To factor these, you find the square root of the \(a^2\) and \(b^2\) terms, ensuring the middle term is twice the product of your two square roots. If this form fits, you've got a perfect square and can then apply the aforementioned factoring.
Solving Algebraic Equations
Algebraic equations are equalities that include algebraic expressions, typically with one or more variables. Solving these equations is a staple of algebra and involves finding the values of the variables that make the equation true. In the context of quadratics and perfect square trinomials, understanding how to rearrange and solve these equations can often lead to the solutions needed for real-world applications or further algebra challenges.
Solving for a variable, much like finding the value of \(k\) in the exercise involves, at times, recognizing patterns, such as the structure of a perfect square trinomial within the equation. This recognition can be the key to unlocking a simpler path to the solution, such as completing the square or other factoring techniques. Developing a familiarity with these methods boosts mathematical literacy and problem-solving skills.
Solving for a variable, much like finding the value of \(k\) in the exercise involves, at times, recognizing patterns, such as the structure of a perfect square trinomial within the equation. This recognition can be the key to unlocking a simpler path to the solution, such as completing the square or other factoring techniques. Developing a familiarity with these methods boosts mathematical literacy and problem-solving skills.
Other exercises in this chapter
Problem 45
The vertex of a parabola is \((3,2)\) . A second point on the parabola is \((1,7)\) . Which point is also on the parabola? F. \((-1,7)\) G. \((3,7)\) H. \((5,7)
View solution Problem 45
For each function, the vertex of the function's graph is given. Find \(c .\) $$ y=-3 x^{2}+6 x+c ;(1,0) $$
View solution Problem 46
Solve each equation using any method. When necessary, round real solutions to the nearest hundredth. For imaginary solutions, write exact solutions. $$ x^{2}-3
View solution Problem 46
Solve each equation. Check your answers. $$ -5 x^{2}-3=0 $$
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