Problem 21 - Algebra and Trigonometry

Question

\(5-60\) Find all real solutions of the equation. $$ \frac{x^{2}}{x+100}=50 $$

Step-by-Step Solution

Answer

The real solutions are \( x = 100 \) and \( x = -50 \).

Step 1: Clear the Denominator

Multiply both sides of the equation by \( x + 100 \) to eliminate the fraction. The equation becomes:\[x^2 = 50(x + 100)\]

Step 2: Expand the Right Side

Distribute the 50 on the right-hand side of the equation:\[x^2 = 50x + 5000\]

Step 3: Rearrange into a Standard Quadratic Form

Move all terms to one side of the equation to set it to zero:\[x^2 - 50x - 5000 = 0\]

Step 4: Use the Quadratic Formula

The quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) can be used here for the quadratic \( ax^2 + bx + c = 0 \), where \( a = 1 \), \( b = -50 \), and \( c = -5000 \).Substitute these values into the formula:\[x = \frac{-(-50) \pm \sqrt{(-50)^2 - 4 \times 1 \times (-5000)}}{2 \times 1}\]

Step 5: Simplify Under the Square Root

Calculate the discriminant (value inside the square root):\[(-50)^2 - 4 \times 1 \times (-5000) = 2500 + 20000 = 22500\]

Step 6: Calculate the Square Root

Find the square root of 22500:\[\sqrt{22500} = 150\]

Step 7: Solve for x

Substitute back into the quadratic formula to find the values of \( x \):\[x = \frac{50 \pm 150}{2}\]Solving this gives:\[x = \frac{50 + 150}{2} = 100\]\[x = \frac{50 - 150}{2} = -50\]

Step 8: Check for Valid Solutions

Since the original equation had \( x + 100 \) in the denominator, ensure \( x eq -100 \) to prevent division by zero, which is not an issue for \( x = 100 \) and \( x = -50 \).

Key Concepts

Discriminant

In the world of quadratic equations, the discriminant plays a crucial role in determining the nature of the solutions. When dealing with a quadratic equation of the form \( ax^2 + bx + c = 0 \), the discriminant is given by the formula \( b^2 - 4ac \). It is essentially the part of the quadratic formula under the square root sign.

The discriminant helps us to understand whether the solutions are real or complex. Here's what the discriminant tells us:

• If \( b^2 - 4ac > 0 \), there are two distinct real solutions.
• If \( b^2 - 4ac = 0 \), there is exactly one real solution, also known as a repeated or double root.
• If \( b^2 - 4ac < 0 \), there are no real solutions, only complex solutions.

Knowing this, the discriminant acts like a compass, directing us towards the nature of our solutions. For the given exercise, we calculated the discriminant as 22500, which is greater than zero. Thus, we predict that there are two real solutions. This is confirmed in the final answer with solutions \( x = 100 \) and \( x = -50 \), both are indeed real numbers.

Quadratic Formula

The quadratic formula is an essential tool for solving quadratic equations. It provides a straightforward method to find solutions for any quadratic equation of the standard form \( ax^2 + bx + c = 0 \). The formula itself is:

\( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)

This formula derives from the process of completing the square and is versatile enough to tackle any quadratic equation, no matter how complex it seems.

To apply the quadratic formula, simply plug the coefficients \( a \), \( b \), and \( c \) into it. In our exercise, we had \( a = 1 \), \( b = -50 \), and \( c = -5000 \). Substituting these values into the quadratic formula, we calculated:

\( x = \frac{-(-50) \pm \sqrt{(-50)^2 - 4 \times 1 \times (-5000)}}{2 \times 1} \)

This is the core calculation step that leads to the two real solutions. The formula reveals much about the equation's nature, especially when examined alongside the discriminant, as the square root component is directly derived from it.

Real Solutions

Real solutions in the context of quadratic equations are the values of \( x \) that satisfy the equation and are 'real numbers', as opposed to complex or imaginary numbers.

When the solutions of a quadratic are real, they lie on the number line and can be either positive, negative, or zero.

In our solved exercise, after applying the quadratic formula, we arrived at two potential solutions \( x = 100 \) and \( x = -50 \). Both of these values are real numbers, as opposed to complex numbers with imaginary parts. Additionally, in the context of a physical or practical problem, real solutions often represent feasible or meaningful answers, such as specific measurements or quantities.

It is important to check that these solutions do not violate any conditions set by the original equation. In our example, because the original equation involved division by \( x+100 \), we had to verify that neither solution led to division by zero. Since they did not equal \(-100\), both solutions were valid, underscoring that the discriminant was correct in predicting the presence of real solutions.