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Solve. See Examples 1 through \(5 .\). $$ x-\sqrt{19-2 x}-2=0 $$

Short Answer

Expert verified
The solution to the equation is \( x = 5 \).

Step by step solution

01

Isolate the Square Root

Begin by moving all terms except the square root to the other side of the equation. Start with the equation: \[ x - \sqrt{19 - 2x} - 2 = 0 \]Add 2 to both sides to get: \[ x - \sqrt{19 - 2x} = 2 \] Now, isolate the square root by subtracting \(x\) from both sides: \[ -\sqrt{19 - 2x} = 2 - x \] Multiply through by -1 to make the square root positive: \[ \sqrt{19 - 2x} = x - 2 \]
02

Square Both Sides

To eliminate the square root, square both sides of the equation: \[ (\sqrt{19 - 2x})^2 = (x - 2)^2 \]Simplify both sides: \[ 19 - 2x = x^2 - 4x + 4 \]
03

Rearrange the Equation

Move all terms to one side to form a quadratic equation: \[ x^2 - 4x + 4 - 19 + 2x = 0 \] This simplifies to: \[ x^2 - 2x - 15 = 0 \]
04

Solve the Quadratic Equation

Use the quadratic formula, \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 1\), \(b = -2\), and \(c = -15\), to find the roots. Calculate the discriminant: \[b^2 - 4ac = (-2)^2 - 4 \cdot 1 \cdot (-15) = 4 + 60 = 64 \]The roots are: \[x = \frac{-(-2) \pm \sqrt{64}}{2 \cdot 1} = \frac{2 \pm 8}{2} \]This gives two solutions: \[x = \frac{2 + 8}{2} = 5 \] \[x = \frac{2 - 8}{2} = -3 \]
05

Verify the Solutions

Check each solution in the original equation to verify if they are true. For \( x = 5 \): \[5 - \sqrt{19 - 2(5)} - 2 = 5 - \sqrt{9} - 2 = 5 - 3 - 2 = 0 \] For \( x = -3 \): \[-3 - \sqrt{19 - 2(-3)} - 2 = -3 - \sqrt{25} - 2 = -3 - 5 - 2 = -10 \] (not zero, so it's extraneous) Only \( x = 5 \) satisfies the original equation.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Square Root Isolation
One crucial step when solving equations involving square roots is isolating the square root term. This straightforward yet essential process ensures we can effectively apply further operations to solve the equations.
To begin isolating a square root, rearrange the equation so that the term containing the square root stands alone on one side. In our initial equation, we start with \( x - \sqrt{19 - 2x} - 2 = 0 \).
By moving all other terms to the other side, we reach \( \sqrt{19 - 2x} = x - 2 \).
This isolation makes it practical to eliminate the square root through squaring both sides, a crucial step to advance toward finding a solution. Isolating the square root simplifies further computation and lessens potential errors.
Extraneous Solutions
Extraneous solutions often arise when solving equations involving square roots or rational expressions. These are solutions that appear during the algebraic manipulation process but do not satisfy the original equation.
To identify such solutions, it is essential to substitute the potential solutions back into the original equation. In our example, we obtained two potential solutions, \( x = 5 \) and \( x = -3 \).
Verification shows that \( x = 5 \) satisfies the equation, while \( x = -3 \) does not. Substituting \( x = -3 \) back into the equation results in a false statement since:
  • \(-3 - \sqrt{19 - 2(-3)} - 2 = -10\), not zero.
This means \( x = -3 \) is an extraneous solution. Checking solutions not only verifies correctness but also eliminates these spurious results.
Quadratic Formula
The quadratic formula is a method for finding the roots of quadratic equations, which are equations of the form \( ax^2 + bx + c = 0 \).
It is expressed as: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]. In our example, after squaring both sides and simplifying, we derived the quadratic form \( x^2 - 2x - 15 = 0 \).
Here, \( a = 1 \), \( b = -2 \), and \( c = -15 \).
The quadratic formula provides reliable solutions for any quadratic equation, especially when factoring is not an option or seems overly complicated. The key benefit of using this formula is its ability to handle complex numbers when necessary, though this wasn’t needed for this particular problem. By plugging the given values into the formula, we calculate the potential roots, enhancing our solving toolkit.
Discriminant Calculation
The discriminant, denoted by \( b^2 - 4ac \) within the quadratic formula, reveals the nature and number of solutions for a quadratic equation.
In our problem, calculating the discriminant with \( a = 1 \), \( b = -2 \), and \( c = -15 \) gives: \( (-2)^2 - 4 \cdot 1 \cdot (-15) = 64 \).
This positive and perfect square (\( 64 = 8^2 \)) indicates two distinct real solutions exist.
Understanding the discriminant's role is critical as:
  • A positive perfect square means two real, rational roots.
  • A positive non-perfect square means two real, irrational roots.
  • Zero means one real rational root (a repeated or double root).
  • A negative value means two complex roots.
The discriminant's analysis simplifies understanding solutions without solving the equation completely, proving essential in determining the equation's dynamics.

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