Problem 57 - Algebra and Trigonometry

Question

Solve the equation for the indicated variable. \(A=2 x^{2}+4 x h ; \quad\) for \(x\)

Step-by-Step Solution

Answer

Solve for \(x\) using: \( x = -h \pm \frac{1}{2}\sqrt{4h^2 + 2A} \).

Step 1: Rearrange the equation

Start with the given equation:\[ A = 2x^2 + 4xh \]Move all terms involving \(x\) to the left side:\[ 2x^2 + 4xh - A = 0 \]

Step 2: Identify the correct method

Recognize this as a quadratic equation in terms of \(x\):\[ 2x^2 + 4xh - A = 0 \]Quadratic equations can be solved using the quadratic formula:

Step 3: Apply the quadratic formula

The quadratic formula is:\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]Here, \(a = 2\), \(b = 4h\), and \(c = -A\). Substitute these values into the formula:\[ x = \frac{-(4h) \pm \sqrt{(4h)^2 - 4 \cdot 2 \cdot (-A)}}{2 \cdot 2} \]

Step 4: Simplify the expression under the square root

Calculate the discriminant:\[ b^2 - 4ac = (4h)^2 - 4 \cdot 2 \cdot (-A) = 16h^2 + 8A \]Substitute back into the quadratic formula:

Step 5: Simplify the final expression

Continue with the quadratic formula:\[ x = \frac{-4h \pm \sqrt{16h^2 + 8A}}{4} \]Simplify to get the final solution:\[ x = \frac{-4h \pm \sqrt{16h^2 + 8A}}{4} \] which further breaks down to:\[ x = -h \pm \frac{1}{2}\sqrt{4h^2 + 2A} \]

Step 6: Present the solution

The possible solutions for \(x\) are:\[ x = -h + \frac{1}{2}\sqrt{4h^2 + 2A} \] or\[ x = -h - \frac{1}{2}\sqrt{4h^2 + 2A} \]

Key Concepts

Solving for a variable

Solving for a variable in an equation means we want to find the value of that variable that makes the equation true. In other words, we want to isolate the variable on one side of the equation. When dealing with quadratic equations, this process often involves several steps to reach a solution. In the exercise provided, we're working with a quadratic equation where the variable of interest is \(x\).

The first step in solving for \(x\) is to rearrange the equation so that all terms involving \(x\) are on one side. This creates a standard quadratic form of \(ax^2 + bx + c = 0\). By doing this, we can prepare to solve for \(x\) using various methods like factoring, completing the square, or using the quadratic formula. For this particular equation, the latter is most commonly used. Remember, the goal is to express \(x\) explicitly in terms of the other variables or constants found in the equation.

• Move terms to get \(ax^2 + bx + c = 0\).
• Identify your coefficients: \(a\), \(b\), \(c\).
• Choose appropriate method: factoring, complete square, or quadratic formula.

Quadratic Formula

The quadratic formula is a reliable method for solving any quadratic equation. It's given by the formula:

\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]

This formula works because it considers all components of a quadratic equation's standard form: \(ax^2 + bx + c = 0\). Each term in the formula allows you to compute both potential solutions for \(x\) – called the roots of the equation. The "\(\pm\)" symbol indicates there are generally two solutions, one found by adding the square root and one by subtracting it.

• Substitute \(a\), \(b\), and \(c\) from the quadratic equation.
• Calculate the discriminant (inside the sqrt).
• Determine both potential roots by using the plus and minus operations.

Using the quadratic formula ensures that you find precise solutions, even when factoring is difficult or impossible due to complex numbers.

Discriminant

The discriminant is a part of the quadratic formula that helps determine the nature of the roots without actually calculating them. It is given by:

\(b^2 - 4ac\)

The discriminant's value tells us:

• If it's positive, there are two distinct real roots.
• If it's zero, there is exactly one real root (repeated solution).
• If it's negative, there are no real roots (roots are complex).

In the exercise, the discriminant is computed with \((4h)^2 - 4 \cdot 2 \cdot (-A) = 16h^2 + 8A\). The result determines if \(x\) will have real and distinct, real and repeated, or complex roots. Calculating the discriminant first can save time by predicting which kinds of solutions you should expect, which is especially useful in more complicated scenarios.