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Magazine Chapter 11: Problem 63
Solve each equation. $$ \sqrt{2 x+6}=x-1 $$
Short Answer
Expert verified
x = 5
Step by step solution
01
Isolate the Square Root
The given equation is \ \(\sqrt{2x + 6} = x - 1\). You need to get rid of the square root by isolating it. The square root is already isolated in this equation, so proceed to the next step.
02
Square Both Sides
To remove the square root, square both sides of the equation: \ \[ (\sqrt{2x + 6})^2 = (x - 1)^2 \] \ This simplifies to: \ \[ 2x + 6 = (x - 1)^2 \].
03
Expand the Quadratic Expression
Expand the right-hand side of the equation: \ \[ 2x + 6 = x^2 - 2x + 1 \].
04
Rearrange the Equation
Move all terms to one side of the equation to set it to zero: \ \[ x^2 - 2x + 1 - 2x - 6 = 0 \] \ Simplify: \ \[ x^2 - 4x - 5 = 0 \].
05
Factor the Quadratic Equation
Factor the quadratic equation: \ \[ (x - 5)(x + 1) = 0 \].
06
Solve for x
Set each factor equal to zero and solve for x: \ \[ x - 5 = 0 \quad \text{or} \quad x + 1 = 0 \] \ Thus, \ \[ x = 5 \quad \text{or} \quad x = -1 \].
07
Check Potential Solutions
Substitute each potential solution back into the original equation to verify: \ For \(x = 5\): \ \(\sqrt{2(5) + 6} = 5 - 1\) \ \(\sqrt{10 + 6} = 4\) \ \(\sqrt{16} = 4\) \ True \ For \(x = -1\): \ \(\sqrt{2(-1) + 6} = -1 - 1\) \ \(\sqrt{-2 + 6} = -2\) \ \(\sqrt{4} = -2\) \ False. Thus, -1 is not a valid solution.
08
Conclusion
The valid solution to the equation is \(x = 5\).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Square Root Equations
Square root equations have a variable inside a square root sign. Solving them involves a few key steps to simplify and remove the square root. In our example, the equation given is \(\sqrt{2x + 6} = x - 1\).
To solve this:
To solve this:
- First, ensure the square root expression is isolated. In this case, it already is: \(\sqrt{2x + 6}\) is alone on one side.
- Next, eliminate the square root by squaring both sides. This turns \(\sqrt{2x + 6} = x - 1\) into \((\sqrt{2x + 6})^2 = (x - 1)^2\), simplifying to \(\2x + 6 = (x - 1)^2\).
After squaring both sides, we now have a quadratic equation to solve.
Quadratic Equations
Quadratic equations are equations of the form \(\ax^2 + bx + c = 0\). They involve variables raised to the power of two.
From our earlier steps, squaring both sides of the equation gave us: \(\2x + 6 = (x - 1)^2\).
Next, we expand the right side: \(\2x + 6 = x^2 - 2x + 1\).
To solve for \(\x\), we need to move all terms to one side: \(\x^2 - 2x + 1 - 2x - 6 = 0\).
Combining like terms provides the standard quadratic form: \(\x^2 - 4x - 5 = 0\). Now, it's in a form that we can factor or use the quadratic formula to solve.
From our earlier steps, squaring both sides of the equation gave us: \(\2x + 6 = (x - 1)^2\).
Next, we expand the right side: \(\2x + 6 = x^2 - 2x + 1\).
To solve for \(\x\), we need to move all terms to one side: \(\x^2 - 2x + 1 - 2x - 6 = 0\).
Combining like terms provides the standard quadratic form: \(\x^2 - 4x - 5 = 0\). Now, it's in a form that we can factor or use the quadratic formula to solve.
Factoring
Factoring involves expressing a quadratic equation as a product of its linear factors.
For our quadratic equation \(\x^2 - 4x - 5 = 0\), we look for two numbers that multiply to \(\text{-5}\) and add to \(\text{-4}\).
These numbers are \(\text{-5}\) and \(\text{1}\).
Thus, we can factorize: \(\x^2 - 4x - 5 = (x - 5)(x + 1)\).
Setting each factor to zero provides potential solutions: \(\x - 5 = 0\) or \(\x + 1 = 0\).
Solving these gives \(\x = 5\) or \(\x = -1\).
Lastly, substituting these values back into the original equation to verify, we find only \(\x = 5\) satisfies the original equation, making it the valid solution.
For our quadratic equation \(\x^2 - 4x - 5 = 0\), we look for two numbers that multiply to \(\text{-5}\) and add to \(\text{-4}\).
These numbers are \(\text{-5}\) and \(\text{1}\).
Thus, we can factorize: \(\x^2 - 4x - 5 = (x - 5)(x + 1)\).
Setting each factor to zero provides potential solutions: \(\x - 5 = 0\) or \(\x + 1 = 0\).
Solving these gives \(\x = 5\) or \(\x = -1\).
Lastly, substituting these values back into the original equation to verify, we find only \(\x = 5\) satisfies the original equation, making it the valid solution.
