Problem 99 - Intermediate Algebra

Question

Solve. \(7-\left(x^{2}-3 x\right)=\sqrt{\left(x^{2}-3 x\right)+5}\)

Step-by-Step Solution

Answer

The solutions are \(x = 4\) and \(x = -1\).

Step 1: Simplify the Left Side of the Equation

We start with the equation \(7 - (x^2 - 3x) = \sqrt{(x^2 - 3x) + 5}\). First, simplify the left side by distributing the negative sign: \(7 - x^2 + 3x\).

Step 2: Set a Substitution

Let \(y = x^2 - 3x\). Thus, the equation becomes \(7 - y = \sqrt{y + 5}\).

Step 3: Isolate the Square Root Term

Rearrange the equation to isolate the square root on one side: \(7 - y = \sqrt{y + 5}\).

Step 4: Square Both Sides

Square both sides of \(7 - y = \sqrt{y + 5}\) to eliminate the square root: \[(7 - y)^2 = y + 5\].

Step 5: Expand and Simplify

Expand the left side: \((7 - y)^2 = 49 - 14y + y^2\),so the equation becomes \[49 - 14y + y^2 = y + 5\].

Step 6: Rearrange into a Quadratic Equation

Rearrange the terms to form a quadratic equation:\[y^2 - 15y + 44 = 0\].

Step 7: Solve the Quadratic Equation

Solve \(y^2 - 15y + 44 = 0\) using the quadratic formula: \[y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\],where \(a = 1, b = -15, c = 44\).

Step 8: Calculate the Discriminant and Solve

Calculate the discriminant: \(b^2 - 4ac = 225 - 176 = 49\). The solutions are:\[y = \frac{15 \pm \sqrt{49}}{2}\]. Thus, \(y = 11\) or \(y = 4\).

Step 9: Substitute Back to Original Variable

Recall that \(y = x^2 - 3x\). Substitute back to find \(x\) from \(x^2 - 3x = 11\) and \(x^2 - 3x = 4\).

Step 10: Solve for \(x\)

For \(x^2 - 3x = 11\):Rearrange to \(x^2 - 3x - 11 = 0\) and solve using the quadratic formula. For \(x^2 - 3x = 4\):Rearrange to \(x^2 - 3x - 4 = 0\) and solve using the quadratic formula. This results in real solutions only for \(x^2 - 3x - 4 = 0\).

Step 11: Verify Solutions

Verify solutions satisfy the original equation. For \(x^2 - 3x = 4\), check roots \((x = 4, x = -1)\) must satisfy \(7 - (x^2 - 3x) = \sqrt{(x^2 - 3x) + 5}\). Only \(x = 4, x = -1\) are valid.

Key Concepts

Substitution Method

In mathematics, the substitution method is a technique used to simplify complex equations, making them easier to solve. It involves replacing a complicated expression with a simpler variable.
For instance, if we have an equation with a common expression, like in the exercise, converting it to a single variable can streamline the process.
This technique is beneficial when handling equations that seem overwhelmingly complicated due to multiple terms.

Here's how it works:

• Identify a repeating element in the equation.
• Replace this element with a single variable, say, 'y'.

Let's take the exercise as an example. We take the expression \(x^2 - 3x\) and substitute it with a new variable, \(y\). This changes the equation from a complex form to a simpler one, \(7 - y = \sqrt{y + 5}\). With this substitution, we can more easily isolate and solve for \(y\). Only after solving it for \(y\), you substitute back to get meaningful solutions for the initial variable \(x\). This method saves time and reduces errors in long tangled equations.

Square Roots

Square roots are a fundamental concept in understanding many mathematical problems and equations. A square root, by definition, is a value that, when multiplied by itself, yields the original number.

Understanding square roots is essential for solving equations where they appear.

• The square root is often denoted by the symbol \(\sqrt{}\).
• To manage square roots algebraically, we square both sides of an equation to eliminate the square root term.

In our exercise, the equation \(7 - y = \sqrt{y + 5}\) includes a square root. To eliminate it, we square both sides, transforming the equation into \((7 - y)^2 = y + 5\). This step is crucial for simplifying and restructuring the equation to make it easier to solve.

Remember that squaring both sides also introduces the possibility of extraneous solutions, so it's always good practice to verify your final answers in the original equation.

Solving Equations

Solving equations, especially quadratic equations, involves finding values for the variables that make the equation true. Quadratic equations are those in the form \(ax^2 + bx + c = 0\) and are common in many mathematical problems.

• The exercise shows how to rearrange an equation into a standard quadratic form.
• Use the quadratic formula for solutions where \(a, b,\) and \(c\) are constants in the equation \(ax^2 + bx + c = 0\).

In our exercise, once the substitution and squaring were done, we ended up with the quadratic equation \(y^2 - 15y + 44 = 0\). Solving this with the quadratic formula, we found \(y = 11\) or \(y = 4\). For each solved \(y\), we equate it back to \(x^2 - 3x\) and solve for the initial variable \(x\) using the same quadratic approach.

Remember, solving equations is not just about arriving at any answer, but verifying it back in the context of the original problem to ensure accuracy and relevance.