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Magazine Chapter 11: Problem 45
Solve by completing the square. $$4 a^{2}-7 a+3=0$$
Short Answer
Expert verified
The solutions to the quadratic equation \(4a^2 - 7a + 3 = 0\) by completing the square are \(a = 1\) and \(a = \frac{3}{4}\).
Step by step solution
01
Identify Quadratic Equation
First, identify the given quadratic equation. The equation is:
\(4a^2 - 7a + 3 = 0\)
02
Divide by the Leading Coefficient
If the leading coefficient is not 1, divide the entire equation by the leading coefficient. The leading coefficient in our case is 4:
\(\frac{4a^2 - 7a + 3}{4} = \frac{0}{4}\)
\(a^2 - \frac{7}{4}a + \frac{3}{4} = 0\)
03
Move the Constant to the Other Side
Move the constant term to the other side of the equation, leaving the variable terms on one side:
\(a^2 - \frac{7}{4}a = -\frac{3}{4}\)
04
Find the Term to Complete the Square
Take half of the coefficient of the linear term, square it, and add it to both sides of the equation:
\(\frac{(-\tfrac{7}{4})}{2} = -\frac{7}{8}\)
\((-\tfrac{7}{8})^2 = \frac{49}{64}\)
\(a^2 - \frac{7}{4}a + \frac{49}{64} = -\frac{3}{4} + \frac{49}{64}\)
05
Simplify and Write as a Square
Simplify the equation and rewrite the left side as a perfect square:
\(a^2 - \frac{7}{4}a + \frac{49}{64} = \frac{49}{64} - \frac{48}{64}\)
\((a - \frac{7}{8})^2 = \frac{1}{64}\)
06
Solve for the Variable
Use the square root property to solve for a:
\(a - \frac{7}{8} = \pm \frac{1}{8}\)
For the positive square root:
\(a = \frac{7}{8} + \frac{1}{8} = \frac{8}{8} = 1\)
For the negative square root:
\(a = \frac{7}{8} - \frac{1}{8} = \frac{6}{8} = \frac{3}{4}\)
07
Present the Answer
So the solutions to the quadratic equation are:
\(a = 1 \) and \(a = \frac{3}{4}\)
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Quadratic Equation
A quadratic equation is a polynomial equation of degree 2. This means the highest exponent of the variable is 2. Quadratic equations can be written in the form: \[ ax^2 + bx + c = 0 \] where \( a \), \( b \), and \( c \) are constants, and \( x \) is the variable. The "quadratic" term comes from "quad," which refers to the power of 2. This type of equation can model various real-life situations, such as projectile motion or area problems. **Components of a Quadratic Equation:**
- **Quadratic term**: \( ax^2 \), where \( a eq 0 \).
- **Linear term**: \( bx \).
- **Constant term**: \( c \).
Leading Coefficient
The leading coefficient in a quadratic equation is the number in front of the quadratic term, \( ax^2 \). It plays a crucial role in solving the equation by completing the square. In the equation \( 4a^2 - 7a + 3 = 0 \), the leading coefficient is 4. **Steps Involved:**
- First, check if the leading coefficient is 1. If it is not, you will divide the entire equation by this number. This makes the coefficient of \( x^2 \) equal to 1, simplifying the process of completing the square.
- For our equation, dividing each term by 4 gives: \[ a^2 - \frac{7}{4}a + \frac{3}{4} = 0 \]
Perfect Square
A perfect square is a number or expression that can be expressed as the square of another. In the context of completing the square, we aim to express part of the quadratic equation as a perfect square trinomial. For example, to transform \( a^2 - \frac{7}{4}a \) into a perfect square, we:
- Take half of the linear term's coefficient: \( \frac{-7}{4} \div 2 = \frac{-7}{8} \)
- Square it: \( \left( \frac{-7}{8} \right)^2 = \frac{49}{64} \)
- Add this square to both sides of the equation: \( a^2 - \frac{7}{4}a + \frac{49}{64} = -\frac{3}{4} + \frac{49}{64} \)
Square Root Property
The square root property is a method used to solve equations involving squared terms. It states that if \( x^2 = k \), then \( x = \pm \sqrt{k} \). We apply this principle to solve the equation once it's written in the perfect square form. In our example, after completing the square, the equation becomes: \[ (a - \frac{7}{8})^2 = \frac{1}{64} \] To solve for \( a \), we take the square root of both sides:
- \( a - \frac{7}{8} = \pm \frac{1}{8} \)
- For the positive root: \( a = \frac{7}{8} + \frac{1}{8} = 1 \)
- For the negative root: \( a = \frac{7}{8} - \frac{1}{8} = \frac{3}{4} \)
