Chapter 0: Problem 34
Solve the radical equation for the given variable. $$\sqrt{8-2 x}=2 x-2$$
Short Answer
Expert verified
The solution is \( x = 2 \).
Step by step solution
01
Isolate the Radical
The equation is already set such that the radical \( \sqrt{8-2x} \) is isolated on one side of the equation. The equation is \( \sqrt{8-2x} = 2x-2 \).
02
Square Both Sides
To eliminate the square root, square both sides of the equation: \( (\sqrt{8-2x})^2 = (2x-2)^2 \), which simplifies to \( 8-2x = (2x-2)^2 \).
03
Expand the Quadratic
Expand the right side of the equation \((2x-2)^2\) which gives \((2x-2)(2x-2) = 4x^2 - 8x + 4\).
04
Simplify the Equation
Substitute the expanded form back into the equation: \(8 - 2x = 4x^2 - 8x + 4\). Rearrange all terms to one side: \(0 = 4x^2 - 6x - 4\).
05
Use the Quadratic Formula
Identify coefficients for the quadratic equation \(4x^2 - 6x - 4 = 0\), with \(a = 4\), \(b = -6\), and \(c = -4\). Use the quadratic formula: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] which becomes \[ x = \frac{-(-6) \pm \sqrt{(-6)^2 - 4 \cdot 4 \cdot (-4)}}{2 \cdot 4} \].
06
Calculate the Discriminant
Calculate the discriminant \(b^2 - 4ac = (-6)^2 - 4 \times 4 \times (-4) = 36 + 64 = 100\).
07
Solve for x
Substitute the discriminant into the quadratic formula: \[ x = \frac{6 \pm 10}{8} \]. This results in two possible solutions: \[ x = \frac{16}{8} = 2 \] or \[ x = \frac{-4}{8} = -0.5 \].
08
Verify the Solutions
Substitute \(x = 2\) and \(x = -0.5\) back into the original equation to verify correctness. For \(x = 2\), \(\sqrt{8-2(2)} = 2(2) - 2\) simplifies to \(\sqrt{4} = 2\), which is true. For \(x = -0.5\), \(\sqrt{8-2(-0.5)} = 2(-0.5) - 2\) simplifies to \(\sqrt{9} = -3\), which is false. Therefore, \(x = 2\) is the only solution.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Quadratic Formula
The quadratic formula is a reliable tool used to find solutions of any quadratic equation of the form \( ax^2 + bx + c = 0 \), where \( a \), \( b \), and \( c \) are constants. It is particularly useful when factorizing isn't straightforward. The formula is:
In our example, with coefficients \( a = 4 \), \( b = -6 \), and \( c = -4 \), substituting into the formula helps find possible values of \( x \) that satisfy the original equation.
- \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \)
In our example, with coefficients \( a = 4 \), \( b = -6 \), and \( c = -4 \), substituting into the formula helps find possible values of \( x \) that satisfy the original equation.
Discriminant
The discriminant is a key component of the quadratic formula and plays an essential role in determining the nature and number of solutions to a quadratic equation.
- It is defined as \( b^2 - 4ac \) within the quadratic formula.
- The value of the discriminant reveals important information about the roots of the equation:
- If \( b^2 - 4ac > 0 \), there are two distinct real roots.
- If \( b^2 - 4ac = 0 \), there is exactly one real root (a repeated root).
- If \( b^2 - 4ac < 0 \), no real roots exist (the roots are complex numbers).
Solving Radical Equations
Solving radical equations involves isolating the radical expression and eliminating it through squaring. Let's break it down:
- First, ensure the radical term, such as \( \sqrt{8-2x} \), is isolated. This means the equation should be set so that the square root is the only term on one side.
- Next, to remove the square root, square both sides. Remember, whatever operation you perform on one side of the equation must also be done on the other.
- Squaring eliminates the radical, providing a polynomial equation that can be solved further, often requiring use of the quadratic formula.