Problem 10 - Algebra 1

Question

Write in standard form. Use the quadratic formula to solve the equation. $$2 x^{2}=-x+6$$

Step-by-Step Solution

Answer

The solutions to the quadratic equation \(2x^2 + x - 6 = 0\) are \(x = 1.5\) and \(x = -2\).

Step 1: Rearrange the Equation

Rewrite the equation in standard form, as Ax^2 + Bx + C = 0. This can be done by moving all terms to one side of the equation: \(2x^2 + x - 6 = 0\).

Step 2: Identify a, b, and c Values

Identify the coefficients a, b, and c in the rearranged equation. Here, \(a = 2, b = 1, c = -6\). These will be used in the quadratic formula.

Step 3: Apply the Quadratic Formula

Insert identified a, b, and c values into the quadratic formula \((-b ± sqrt(b^2 - 4ac))/(2a)\). Hence, you will get \((-1 ± sqrt((1)^2 - 4*2*(-6)))/(2*2)\). Calculate the part of the formula inside the square root first, and split the plus-or-minus into two parts, giving two potential solutions for x.

Step 4: Calculate the Results

Continue solving the equation to get the values of x. This gives two solutions: \( x1 = 1.5, x2 = -2 \).

Key Concepts

Standard Form

The concept of standard form is crucial when dealing with quadratic equations. Quadratic equations can appear in various formats, but the standard form is the simplest and most uniform. It is expressed as \(Ax^2 + Bx + C = 0\). This format allows for an easy application of solution methods such as the quadratic formula.

In the exercise, the equation given is \(2x^2 = -x + 6\). To convert this into the standard form, you have to move all terms to one side of the equation. This involves rearranging it to look like \(2x^2 + x - 6 = 0\).

• "A" is the coefficient before \(x^2\), which is 2 in this case.


• "B" is the coefficient before \(x\), which is 1.


• "C" is the constant term, which is -6.

This organization into a standard form is vital for identifying coefficients used in formulas and strategies to find the equation's solutions.

Quadratic Formula

The quadratic formula is a powerful tool used to solve quadratic equations. Once an equation is in standard form, the quadratic formula can be applied. This formula is given by:\[x = \frac{-b ± \sqrt{b^2 - 4ac}}{2a}\]
This formula helps to find the solutions of quadratic equations, no matter how complex they seem. It uses the coefficients from the standard form equation (\(a\), \(b\), and \(c\)).

In the exercise example, after converting to standard form (\(2x^2 + x - 6 = 0\)), you identify:

• \(a = 2\)
• \(b = 1\)
• \(c = -6\)

These values are plugged into the quadratic formula. Calculating further, the equation inside the square root becomes \(1^2 - 4 \times 2 \times (-6)\). Solving this step-by-step is essential to avoid errors and accurately find the solutions for \(x\). The quadratic formula is valuable because it guarantees the solution for any quadratic equation.

Solving Equations

Solving quadratic equations involves finding the values of \(x\) that satisfy the equation. Using the quadratic formula ensures that you consider possible real and imaginary solutions. Once you have plugged your identified \(a\), \(b\), and \(c\) values into the quadratic formula, you'll begin simplifying the equation by performing operations inside the square root.

For our exercise equation, substitute into the quadratic formula: \[x = \frac{-1 ± \sqrt{1^2 - 4 \times 2 \times (-6)}}{2 \times 2}\]
Next, do the math inside the square root: \(1 + 48 = 49\), giving you \(\sqrt{49} = 7\).

• One solution is \(x = \frac{-1 + 7}{4}\), simplifying to \(x = 1.5\).


• The other solution is \(x = \frac{-1 - 7}{4}\), simplifying to \(x = -2\).

Thus, solving the equation gives two potential solutions for \(x\): \(x = 1.5\) and \(x = -2\). Solving intricately by breaking down complex calculations and patiently handling each step ensures accurate results.