Problem 39
Question
Solve the equation by completing the square. $$x^{2}+\frac{3}{5} x-1=0$$
Step-by-Step Solution
Verified Answer
The solutions for the equation are \(-\frac{3 + \sqrt{109}}{10}\) and \(-\frac{3 - \sqrt{109}}{10}\).
1Step 1: Equation Setup
Given equation is \(x^2 + \frac{3}{5}x - 1 = 0\). Separating the constant term from other terms, the equation is rewritten as \(x^2 + \frac{3}{5}x = 1\).
2Step 2: Completing The Square
To complete the square, the square of half of the coefficient of \(x\) is added and subtracted on both sides of equation. So, \((1/2 * 3/5)^2 = (3/10)^2 = 9/100\). Thus, the equation becomes: \(x^2 + \frac{3}{5}x + \frac{9}{100} = 1 + \frac{9}{100}\), which simplifies to \(x^2 + \frac{3}{5}x + \frac{9}{100} = \frac{109}{100}\)
3Step 3: Solving for x
Now, we can rewrite the left hand side as a square of binomial. The equation is simplified to \((x + \frac{3}{10})^2 = \frac{109}{100}\). Taking square root on both sides, we obtain \(x + \frac{3}{10} = \pm\sqrt{\frac{109}{100}}\). Thus, the solutions will be \(x = -\frac{3}{10} \pm \sqrt{\frac{109}{100}}\)
4Step 4: Simplifying The Result
After simplifying, we get \(x = -\frac{3}{10} \pm \frac{\sqrt{109}}{10}\). Therefore, the roots of the equation are \(x1 = -\frac{3 + \sqrt{109}}{10}\) and \(x2 = -\frac{3 - \sqrt{109}}{10}\).
Key Concepts
Quadratic EquationsSquare RootBinomial Square
Quadratic Equations
Quadratic equations are mathematical expressions of the form: \( ax^2 + bx + c = 0 \), where \( a, b, \) and \( c \) are constants, and \( a eq 0 \). These equations are called quadratic because the highest exponent of the variable \( x \) is 2.
Quadratics are prevalent in various mathematical contexts because they model parabolic shapes and appear naturally in physics, engineering, and economics.
To solve a quadratic equation, you can use several methods:
Quadratics are prevalent in various mathematical contexts because they model parabolic shapes and appear naturally in physics, engineering, and economics.
To solve a quadratic equation, you can use several methods:
- Factoring: Expressing the equation as a product of its linear factors.
- Using the Quadratic Formula: Using the formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \).
- Completing the Square: Rewriting the equation so that one side forms a perfect square trinomial.
Square Root
A square root is a value that, when multiplied by itself, gives the original number. In mathematical notation, the square root of a number \( n \) is denoted as \( \sqrt{n} \).
For example, \( \sqrt{9} = 3 \) because \( 3 \times 3 = 9 \).
The square root function is crucial in many areas of mathematics, including solving quadratic equations when using the "completing the square" method.
When solving by completing the square, the equation typically results in a situation where taking a square root is necessary, such as \((x + p)^2 = q\).
To resolve it, you apply the square root to both sides, leading to \( x + p = \pm\sqrt{q}\). The "plus-minus" (\( \pm \)) sign is important because it represents the two possible solutions for \( x \):
For example, \( \sqrt{9} = 3 \) because \( 3 \times 3 = 9 \).
The square root function is crucial in many areas of mathematics, including solving quadratic equations when using the "completing the square" method.
When solving by completing the square, the equation typically results in a situation where taking a square root is necessary, such as \((x + p)^2 = q\).
To resolve it, you apply the square root to both sides, leading to \( x + p = \pm\sqrt{q}\). The "plus-minus" (\( \pm \)) sign is important because it represents the two possible solutions for \( x \):
- The "plus" leads to one value, \( x = -p + \sqrt{q} \).
- The "minus" gives the second value, \( x = -p - \sqrt{q} \).
Binomial Square
The binomial square is a powerful algebraic concept, particularly useful when working with quadratic equations and completing the square method. A binomial is a polynomial with two terms, like \((x + y)\).
When squared, it expands based on the formula: \((x + y)^2 = x^2 + 2xy + y^2\).
This formula enables us to identify or construct a perfect square trinomial from a quadratic expression during the completing the square process.
In completing the square, the key step involves finding the exact third term that makes a quadratic equation a perfect square trinomial:
This transformation achieves a perfect square trinomial, simplifying the process of solving the equation by taking the square root of both sides.
When squared, it expands based on the formula: \((x + y)^2 = x^2 + 2xy + y^2\).
This formula enables us to identify or construct a perfect square trinomial from a quadratic expression during the completing the square process.
In completing the square, the key step involves finding the exact third term that makes a quadratic equation a perfect square trinomial:
- First, take half the coefficient of \( x \).
- Then, square it to form the third term.
This transformation achieves a perfect square trinomial, simplifying the process of solving the equation by taking the square root of both sides.
Other exercises in this chapter
Problem 39
Find the domain of the function. $$y=0.2 \sqrt{x}$$
View solution Problem 39
A field hockey field is a rectangle 60 yards by 100 yards. What is the length of the diagonal from one corner of the field to the opposite corner?
View solution Problem 39
Simplify the expression. $$(2 \sqrt{3}-5)^{2}$$
View solution Problem 40
Solve the equation. Check for extraneous solutions. $$x=\sqrt{11 x-28}$$
View solution