Chapter 7: Problem 44
Solve. \(\sqrt{2 x-1}-4=-\sqrt{x-4}\)
Short Answer
Expert verified
The solution is \( x = 5 \).
Step by step solution
01
Isolate the Square Roots
The equation is \( \sqrt{2x - 1} - 4 = -\sqrt{x-4} \). First, let's move one of the square roots to the other side by adding \( \sqrt{x-4} \) to both sides. This gives us: \( \sqrt{2x - 1} = \sqrt{x - 4} + 4 \).
02
Square Both Sides of the Equation
To eliminate the square roots, square both sides of the equation: \( (\sqrt{2x - 1})^2 = (\sqrt{x - 4} + 4)^2 \). Simplifying gives: \( 2x - 1 = (x - 4) + 8 \sqrt{x-4} + 16 \).
03
Simplify the Equation
Combine like terms: \( 2x - 1 = x + 12 + 8 \sqrt{x-4} \). Move \( x \) to the left side to get: \( x - 1 = 12 + 8 \sqrt{x-4} \). Then, isolate the square root term: \( x - 13 = 8 \sqrt{x-4} \).
04
Divide and Square Again
Divide both sides by 8: \( \frac{x - 13}{8} = \sqrt{x - 4} \). Square both sides again to eliminate the square root: \( \left(\frac{x - 13}{8}\right)^2 = x - 4 \). Simplify by expanding and rearranging: \( \frac{(x - 13)^2}{64} = x - 4 \).
05
Cross Multiply and Expand
Multiply through by 64 to eliminate the fraction: \( (x - 13)^2 = 64(x - 4) \). Expand both sides: \( x^2 - 26x + 169 = 64x - 256 \).
06
Rearrange and Solve the Quadratic Equation
Move all terms to one side to form a quadratic equation: \( x^2 - 90x + 425 = 0 \). Solve the quadratic using the quadratic formula: \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \), where \( a = 1, b = -90, c = 425 \).
07
Calculate Discriminant and Roots
Calculate the discriminant: \( b^2 - 4ac = 8100 - 1700 = 6400 \). The roots are \( x = \frac{90 \pm 80}{2} \). Thus, \( x = 85 \) or \( x = 5 \).
08
Verify Roots in Original Equation
Check each solution in the original equation to ensure they work. For \( x = 85 \), \( \sqrt{2(85) - 1} - 4 = -\sqrt{85 - 4} \) simplifies to \( 13 - 4 = -9 \). For \( x = 5 \), \( \sqrt{2(5) - 1} - 4 = -\sqrt{5 - 4} \) simplifies to \( -1 \). Only \( x = 5 \) satisfies both sides of the original equation.
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Square Roots
Square roots are mathematical expressions that represent a value which, when multiplied by itself, gives the original number. They are symbolized by the radical sign \( \sqrt{} \).
For example, the square root of 9 is 3 because \( 3 \times 3 = 9 \). When working with equations that include square roots, one common goal is to remove the square root sign by squaring both sides of the equation.
This process helps in reaching a more manageable form, where conventional solving methods for quadratic equations can be applied.
For example, the square root of 9 is 3 because \( 3 \times 3 = 9 \). When working with equations that include square roots, one common goal is to remove the square root sign by squaring both sides of the equation.
- This often simplifies the equation and leaves a polynomial expression.
- Be cautious, though, because squaring both sides can introduce extraneous solutions.
This process helps in reaching a more manageable form, where conventional solving methods for quadratic equations can be applied.
Quadratic Equation
A quadratic equation is a second-degree polynomial equation in a single variable \( x \) with the general form \( ax^2 + bx + c = 0 \). These equations appear frequently in various algebra problems and have a characteristic curve called a parabola.
- The standard approach to solve quadratic equations is using the quadratic formula: \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \),
- Although sometimes factoring or completing the square might be applicable depending on its complexity.
Discriminant
The discriminant is a key element in understanding the nature of solutions for a quadratic equation. Calculated as \( \Delta = b^2 - 4ac \), it tells us about the number and nature of the roots of the equation:
Using these roots and re-substituting them into the original equation can help verify which, if any, fit the context of the problem. Only roots that satisfy the original given radical equation are valid; hence verification becomes crucial.
- If \( \Delta > 0 \), the quadratic equation has two distinct real roots.
- If \( \Delta = 0 \), there is exactly one real root (it's a perfect square trinomial).
- If \( \Delta < 0 \), the equation has no real roots, but two complex roots.
Using these roots and re-substituting them into the original equation can help verify which, if any, fit the context of the problem. Only roots that satisfy the original given radical equation are valid; hence verification becomes crucial.