Problem 26 - Algebra and Trigonometry with Analytic Geometry

Question

Exer. 1-50: Solve the equation. $$ x-\sqrt{-7 x-24}=-2 $$

Step-by-Step Solution

Answer

The solutions are \(x = -4\) and \(x = -7\).

Step 1: Isolate the Radical Expression

This involves moving the radical expression to one side of the equation. Start with the equation \(x - \sqrt{-7x - 24} = -2\). Add \(\sqrt{-7x - 24}\) to both sides to isolate it. The equation becomes \(x + 2 = \sqrt{-7x - 24}\).

Step 2: Square Both Sides

To eliminate the square root, square both sides of the equation: \((x + 2)^2 = (-7x - 24)\). This yields the equation \(x^2 + 4x + 4 = -7x - 24\).

Step 3: Rearrange the Equation

Next, move all terms to one side to set the equation to zero: \(x^2 + 4x + 4 + 7x + 24 = 0\). Simplify this to \(x^2 + 11x + 28 = 0\).

Step 4: Solve the Quadratic Equation

Now, solve the quadratic equation \(x^2 + 11x + 28 = 0\) using the quadratic formula: \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 1\), \(b = 11\), \(c = 28\). First, calculate the discriminant: \(b^2 - 4ac = 11^2 - 4 \times 1 \times 28 = 121 - 112 = 9\). This gives us two solutions: \(x = \frac{-11 \pm \sqrt{9}}{2}\).

Step 5: Simplify the Solutions

The solutions from the quadratic formula are: \(x = \frac{-11 + 3}{2}\) and \(x = \frac{-11 - 3}{2}\). Simplifying these gives \(x = -4\) and \(x = -7\).

Step 6: Verify the Solutions

Substitute \(x = -4\) back into the original equation: \(-4 - \sqrt{-7(-4) - 24} = -2\). This simplifies to \(-4 - \sqrt{28 - 24} = -2\), which holds true since \(-4 - 2 = -2\). Now, substitute \(x = -7\): \(-7 - \sqrt{-7(-7) - 24} = -2\). This simplifies to \(-7 - \sqrt{49 - 24} = -2\), which is true since \(-7 - 5 = -2\). Both solutions are valid.

Key Concepts

Radical Equations

Radical equations involve variables inside a radical, commonly a square root. To solve radical equations, a common method is to isolate the radical on one side of the equation. This makes it easier to remove the radical by squaring both sides. In our example, the equation \(x - \sqrt{-7x - 24} = -2\) required us to first isolate the radical term \(\sqrt{-7x - 24}\) on one side. We added the radical term to both sides to achieve this, which brought us to \(x + 2 = \sqrt{-7x - 24}\). This step is essential because, without isolating the radical, any operations involving squaring could lead to incorrect results. Once isolated, squaring each side eliminates the radical, allowing us to form a polynomial equation that is easier to handle.

Quadratic Formula

The quadratic formula is a powerful tool used to solve quadratic equations, which are equations of the form \(ax^2 + bx + c = 0\). Given the equation, the formula used is:

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

In the example exercise, after rearranging and simplifying the equation, we arrived at the quadratic equation \(x^2 + 11x + 28 = 0\). Here, \(a = 1\), \(b = 11\), and \(c = 28\). By substituting these values into the quadratic formula, we obtained two potential solutions for \(x\). The plus-minus symbol (±) in the quadratic formula represents the two roots that a quadratic equation might have due to the square root operation, reflecting the symmetry in quadratic functions.

Discriminant

The discriminant is a key component in the quadratic formula, shown as \(b^2 - 4ac\). It helps determine the nature of the roots of a quadratic equation. In our example, the discriminant was calculated by substituting our values for \(b\), \(a\), and \(c\):

• \(b^2 - 4ac = 11^2 - 4 \times 1 \times 28 = 121 - 112 = 9\)

The value of the discriminant reveals:

• If it is positive, there are two distinct real roots. This was the case in our problem.
• If the discriminant is zero, there is exactly one real root (a repeated root).
• If negative, there are no real roots, but two complex roots instead.

Thus, the discriminant not only aids in solving equations but also provides insight into the number and type of solutions possible.

Verification of Solutions

Verifying solutions is a critical step in solving equations, ensuring the solutions satisfy the original equation. After obtaining potential solutions using methods like the quadratic formula, it is important to plug these solutions back into the original equation. This cross-check validates whether they are correct. In the exercise, we found the solutions \(x = -4\) and \(x = -7\). We substituted these back into the original equation \(x - \sqrt{-7x - 24} = -2\) to check their validity:

• For \(x = -4\), substituting gave \(-4 - \sqrt{28 - 24} = -2\), which was true as both sides equated past simplification.
• For \(x = -7\), substituting gave \(-7 - \sqrt{49 - 24} = -2\), which also held true.

Verifying the solutions confirms their correctness and ensures no mistakes were made in previous steps, especially since squaring both sides can introduce extraneous roots.