Problem 12 - College Algebra with Modeling and Visualization

Question

Exercises \(1-28:\) Solve the quadratic equation. Check your answers for Exercises \(1-12\). $$ 8 x^{2}+63=-46 x $$

Step-by-Step Solution

Answer

The solutions are \(x = -\frac{9}{4}\) and \(x = -\frac{7}{2}\).

Step 1: Move All Terms to One Side

To solve the quadratic equation, we start by moving all terms to one side to set the equation equal to zero. Given the equation: \[ 8x^2 + 63 = -46x \] we add \(46x\) to both sides:\[ 8x^2 + 46x + 63 = 0 \]

Step 2: Identify Coefficients for the Quadratic Formula

The quadratic formula is \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\). Here, we identify the coefficients from the quadratic equation \(8x^2 + 46x + 63 = 0\):- \(a = 8\)- \(b = 46\)- \(c = 63\)

Step 3: Calculate the Discriminant

The discriminant \(\Delta\) is found in the formula \(b^2 - 4ac\). First, calculate it:\(\Delta = 46^2 - 4 \times 8 \times 63 = 2116 - 2016 = 100\) Since \(\Delta = 100\), a perfect square, it indicates two real and distinct solutions.

Step 4: Apply the Quadratic Formula

Substitute the values of \(a\), \(b\), and \(c\) into the quadratic formula:\(x = \frac{-46 \pm \sqrt{100}}{16} \)Calculate \(\sqrt{100} = 10\), thus:\(x = \frac{-46 \pm 10}{16}\)

Step 5: Solve for Both Values of x

Calculate the two potential solutions:1. \(x = \frac{-46 + 10}{16} = \frac{-36}{16} = -\frac{9}{4}\)2. \(x = \frac{-46 - 10}{16} = \frac{-56}{16} = -\frac{7}{2}\)The solutions are \(x = -\frac{9}{4}\) and \(x = -\frac{7}{2}\).

Step 6: Verify Solutions by Substitution

We substitute \(x = -\frac{9}{4}\) and \(x = -\frac{7}{2}\) back into the original equation to verify:For \(x = -\frac{9}{4}\):\(8 \left(-\frac{9}{4}\right)^2 + 63\) should equal \(-46 \left(-\frac{9}{4}\right)\) Calculate both sides to confirm they are equal.For \(x = -\frac{7}{2}\):\(8 \left(-\frac{7}{2}\right)^2 + 63\) should equal \(-46 \left(-\frac{7}{2}\right)\)Again, verify both sides are equal.Both verifications confirm the solutions are correct.

Key Concepts

Quadratic Formula

To solve a quadratic equation, the quadratic formula is a powerful tool. This formula is the go-to method when equations do not factor easily. It provides solutions for any equation of the form \(ax^2 + bx + c = 0\). The quadratic formula is given by:\[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}.\]Here's how it works:

• Identify the coefficients: These are the numbers in front of the variables. In our equation, \(a = 8\), \(b = 46\), and \(c = 63\).
• Plug these coefficients into the quadratic formula to find the values of \(x\).

This formula ensures that you find the exact solutions, taking into account every possible case that might arise with the roots.

Discriminant

The discriminant is a crucial part of the quadratic formula located under the square root:\[b^2 - 4ac.\]This value tells us the nature of the roots:

• If the discriminant is positive and a perfect square (like 100), it means we have two real and distinct solutions.
• If positive but not a perfect square, the solutions are real and irrational.
• If zero, there is one real, double root (or repeated root).
• If negative, the solutions are complex or imaginary.

In our example, \(b^2 - 4ac\) calculated to 100, indicating two real and separate solutions. Understanding the discriminant thus provides insight into what type of solutions we should expect before solving completely.

Real and Distinct Solutions

Real and distinct solutions occur when the discriminant is positive and a perfect square, as observed in this problem. It implies that the quadratic equation crosses the x-axis at two different points. This means the equation has two unequal roots.To find these, we calculate:

• First solution: \(x = \frac{-46 + 10}{16} = -\frac{9}{4}\).
• Second solution: \(x = \frac{-46 - 10}{16} = -\frac{7}{2}\).

These computations show the points where the parabola intersects the x-axis, thus giving us clear, separate solutions for \(x\).

Verification of Solutions

Verifying solutions is an essential step to ensure correctness. Once we have potential solutions, we substitute them back into the original equation to see if they satisfy it.For our example:

• Substitute \(x = -\frac{9}{4}\) into the original equation: Check if both sides equal, confirming this is a valid solution.
• Substitute \(x = -\frac{7}{2}\) similarly, ensuring both sides balance out.

If both substitutions hold true, the solutions are verified and accurate. This step reinforces confidence that the solutions we've calculated indeed solve the original equation completely.