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Chapter 3: Problem 12

Find all real solutions to each equation. Check your answers. $$\sqrt{x-1}=x-7$$

Short Answer

Expert verified
The real solution is \(x = 10\).

Step by step solution

01

Isolate the square root

The given equation is \(\sqrt{x-1}=x-7\). The square root is already isolated, so proceed to the next step.
02

Square both sides

Square both sides of the equation to eliminate the square root. This gives \[(\sqrt{x-1})^2 = (x-7)^2\]Simplify to get:\[x-1 = (x-7)^2\]
03

Expand and simplify

Expand the right-hand side and simplify:\[x - 1 = x^2 - 14x + 49\]Now, rearrange the equation to set it to zero:\[x^2 - 15x + 50 = 0\]
04

Solve the quadratic equation

Use the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\) to solve for \(x\). Here, \(a = 1\), \(b = -15\), and \(c = 50\).Calculate the discriminant first:\[b^2 - 4ac = (-15)^2 - 4(1)(50) = 225 - 200 = 25\]Then, find the solutions:\[x = \frac{15 \pm \sqrt{25}}{2} = \frac{15 \pm 5}{2}\]Thus:\[x = 10\]\[x = 5\]
05

Check the solutions

Substitute the solutions back into the original equation to verify.For \(x = 10\):\[\sqrt{10-1} = 10 - 7\]\[3 = 3 \text{ which is true.}\]For \(x = 5\):\[\sqrt{5-1} = 5 - 7\]\[2 = -2 \text{ which is false.}\]Therefore, only \(x = 10\) is a real solution.

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Key Concepts

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

Solving Quadratic Equations
Solving quadratic equations is a fundamental mathematical skill. A quadratic equation takes the form: \[ ax^2 + bx + c = 0 \] To solve for the variable \( x \), follow these steps:
  • First, rearrange the equation so that all terms are on one side. This gives you a standard form quadratic equation.
  • Then, use the quadratic formula: \( x = \frac{-b \, \pm \, \sqrt{b^2 - 4ac}}{2a} \).
    This formula will provide you with the solutions. \[ b^2 - 4ac \] is called the discriminant and tells us about the nature of the roots.
    • If the discriminant is positive, there are two distinct real solutions.
    • If it is zero, there is exactly one real solution.
    • If it is negative, there are no real solutions, only complex ones.

    For example, in our exercise: Given: \( \sqrt{x-1} = x-7 \)
    Rearrange and solve to finally get: \( x^2 - 15x + 50 = 0 \)

    Use the quadratic formula here: \( a = 1 \), \( b = -15 \), and \( c = 50 \), to find the solutions: \( x = 10 \) and \( x = 5 \).
Square Root Isolation
Isolating the square root is an important step in solving some equations. When you have an equation involving a square root, your goal is to get the square root term by itself on one side of the equation:

For example: \( \sqrt{x-1} = x-7 \)

This equation already has the square root isolated. Once isolated, the next step is to get rid of the square root by squaring both sides:
  • Given: \( \sqrt{x-1} = x-7 \)
  • Square both sides: \[ (\sqrt{x-1})^2 = (x-7)^2 \]
  • Simplify: \[ x-1 = x^2 - 14x + 49 \]
Rewriting it in the standard quadratic form: \( x^2 - 15x + 50 = 0 \) allows you to solve it using the quadratic formula.
Checking Solutions
After finding the potential solutions to an equation, the final and crucial step is to verify that these solutions are indeed correct. This ensures they satisfy the original equation.
Let's check the solutions for \( x = 10 \) and \( x = 5 \) in our problem.
For \( x = 10 \) :
  • Substitute back into the original: \( \sqrt{10-1} = 10-7 \)
  • Simplifies to: \( 3 = 3 \)
  • Since this is true, \( x = 10 \) is a valid solution

For \( x = 5 \) :
  • Substitute back into the original: \( \sqrt{5-1} = 5-7 \)
  • Simplifies to: \( 2=-2 \)
  • Since this is not true, \( x = 5 \) is not a valid solution

Therefore, only \( x = 10 \) is the real solution to the equation. Always double check the solutions to avoid any mistakes!

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