Problem 62 - Elementary Algebra

Question

Solve. $$ 5 x+2 x+1-x x+4=4 $$

Step-by-Step Solution

Answer

\( x = \frac{7 \, \pm \, \\sqrt{53}}{2} \)

Step 1: Understanding the Equation

The equation to solve is: \[ 5x + 2x + 1 - x x + 4 = 4 \] Your goal is to find the value of \( x \) that makes this equation true.

Step 2: Simplifying the Equation

Combine like terms and simplify each side of the equation. Notice that \( 5x + 2x \) can be combined to \( 7x \), and \( -xx \) should be rewritten as \( -x^2 \). The equation then becomes: \[ 7x + 1 - x^2 + 4 = 4 \] Further simplification of the constant terms \( 1 + 4 \) gives \( 5 \), so the equation becomes: \[ -x^2 + 7x + 5 = 4 \]

Step 3: Rearranging the Equation

Rearrange the equation by moving all terms to one side: \[ -x^2 + 7x + 5 - 4 = 0 \] After combining the constant terms, the equation simplifies to: \[ -x^2 + 7x + 1 = 0 \]

Step 4: Changing Signs

Multiply each term by \(-1\) to make the quadratic term positive: \[ x^2 - 7x - 1 = 0 \]

Step 5: Solving the Quadratic Equation

Use the quadratic formula \( x = \frac{-b \, \pm \, \sqrt{b^2 - 4ac}}{2a} \) with \( a = 1 \), \( b = -7 \), and \( c = -1 \). Calculate the discriminant: \( b^2 - 4ac = (-7)^2 - 4(1)(-1) = 49 + 4 = 53 \). The quadratic formula yields: \[ x = \frac{7 \, \pm \, \sqrt{53}}{2} \]

Key Concepts

Algebraic Simplification

In algebra, simplification plays a crucial role in making equations easier to solve. By simplifying expressions, you combine like terms and reduce complexity. In our example, the expression to simplify is:
\[ 5x + 2x + 1 - xx + 4 = 4 \]
First, you notice that terms with the same variable coefficient can be added or subtracted from each other. This is referred to as *combining like terms*. For example, you can add \(5x\) and \(2x\) to get \(7x\). Similarly, \(-xx\) represents \(-x^2\), which can be replaced appropriately in the equation.

• Combine \(5x\) and \(2x\) to get \(7x\).
• Recognize \(-xx\) as \(-x^2\).

Next, don't forget about the constants, which can also be combined. Add \(1\) and \(4\) to simplify further. Once simplification is complete, you'll end up with a neater expression that is easier to manage. Simplification therefore allows you to work with clear forms of an expression, preparing you for subsequent steps like factorization or solving.

Quadratic Formula

The quadratic formula is a pivotal tool in algebra used to find the solutions of quadratic equations. A quadratic equation is typically of the form:
\[ ax^2 + bx + c = 0 \]
In solving the quadratic equation \( x^2 - 7x - 1 = 0 \), you can apply the quadratic formula, which provides solutions for \(x\) by plugging in the coefficients \(a\), \(b\), and \(c\):
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]

• Here, \(a = 1\), \(b = -7\), and \(c = -1\).

The quadratic formula involves finding the *discriminant*, \(b^2 - 4ac\), under the square root. It determines the nature of the roots, whether they are real, imaginary, or rational. For our example:

• \(b^2 = 49\)
• \(-4ac = -4 \times 1 \times -1 = 4\)
• Thus, the discriminant is \(49 + 4 = 53\)

Since the discriminant, \(53\), is positive, the equation has two distinct real solutions. The solutions are calculated as:
\[ x = \frac{7 \pm \sqrt{53}}{2} \]

Solving Equations

Solving equations means finding the value(s) of the unknown variable(s) that satisfy the equation. In our case, we are solving the quadratic equation:
\[ x^2 - 7x - 1 = 0 \]
The goal is to find the values of \(x\) that make this true. By applying *algebraic simplification* as seen in earlier steps, we prepared the equation for easier handling. Applying the quadratic formula confirms the roots, as determined by solving:

• The "+" option of the quadratic formula yields one solution.
• The "−" option gives you another solution.

Quadratic equations, depending on their discriminant, might have:

• Two real and distinct solutions when the discriminant is positive, as shown in this example.
• One real double root when the discriminant is zero.
• Two complex solutions when the discriminant is negative.

Understanding these categories helps in identifying the type of solutions, ultimately enabling you to solve the equations effectively. Solving such equations requires practice in applying algebraic rules consistently and correctly.