Example Question - extraneous solutions
Here are examples of questions we've helped users solve.
Solving a Radical Equation
Given: \( x + 5 = \sqrt{x + 35} \) Step 1: Square both sides to eliminate the square root. \( (x + 5)^2 = (\sqrt{x + 35})^2 \) Step 2: Expand the left side and simplify the right side. \( x^2 + 10x + 25 = x + 35 \) Step 3: Set the equation to zero and rearrange terms. \( x^2 + 10x + 25 - x - 35 = 0 \) \( x^2 + 9x - 10 = 0 \) Step 4: Factor the quadratic equation. \( (x + 10)(x - 1) = 0 \) Step 5: Solve for x. \( x + 10 = 0 \) or \( x - 1 = 0 \) \( x = -10 \) or \( x = 1 \) Step 6: Check for extraneous solutions by plugging back into the original equation. For \( x = -10 \): \( -10 + 5 \neq \sqrt{-10 + 35} \) \( -5 \neq 5 \) (Not a solution) For \( x = 1 \): \( 1 + 5 = \sqrt{1 + 35} \) \( 6 = \sqrt{36} \) \( 6 = 6 \) (Valid solution) Therefore, \( x = 1 \) is the solution.
Solving a Quadratic Equation with a Square Root
The equation in the image is a quadratic equation: √(x^2 - 2x - 5) = x To solve this, let's first isolate the square root by moving all terms involving x to one side: √(x^2 - 2x - 5) = x √(x^2 - 2x - 5) - x = 0 Now, to get rid of the square root, let's square both sides of the equation. This step assumes that both sides of the equation are positive or zero, as square root and square functions are not inverses when dealing with negative values. (√(x^2 - 2x - 5) - x)^2 = 0^2 Squaring the left side, we have to apply the formula (a - b)^2 = a^2 - 2ab + b^2: (x^2 - 2x - 5) - 2x(√(x^2 - 2x - 5)) + x^2 = 0 We now have two x^2 terms that we can combine, and we also have a term with a square root that we'll need to deal with: 2x^2 - 2x(√(x^2 - 2x - 5)) - 2x - 5 = 0 Let's move the term without the square root to the other side to get: 2x^2 - 2x - 5 = 2x(√(x^2 - 2x - 5)) Now, let's divide by 2x to make the equation simpler, assuming x ≠ 0: x - 1/2 - 5/(2x) = √(x^2 - 2x - 5) Now, we square both sides again to eliminate the square root: (x - 1/2 - 5/(2x))^2 = (x^2 - 2x - 5) Unfortunately, squaring this quite messy expression can lead to a very complex equation. However, the cleanest approach is to start the problem from scratch because the original step of subtracting x from both sides complicates things. Let's re-evaluate the original equation: √(x^2 - 2x - 5) = x Square both sides directly from this point: x^2 - 2x - 5 = x^2 Now cancel x^2 from both sides: -2x - 5 = 0 Add 2x to both sides: -5 = 2x Now divide by 2: x = -5/2 So the solution to the equation is x = -5/2. In order to confirm that this solution is valid and not an extraneous solution introduced by squaring the equation, we need to substitute x back into the original equation and verify: √((-5/2)^2 - 2(-5/2) - 5) = -5/2 Calculating the inside of the square root: √((25/4) + (5) - 5) = -5/2 √((25/4) + (20/4) - (20/4)) = -5/2 √(25/4) = -5/2 Since the square root is a positive number and we obtain √(25/4) = 5/2, this is not equal to -5/2. Thus, there is a contradiction here, which means that our purported solution does not actually satisfy the original equation. Therefore, there is no real solution that satisfies the original equation √(x^2 - 2x - 5) = x, since we ended with an impossible statement: √(25/4) = -5/2, which cannot be true as the square root of a positive number cannot be negative.
Solving Quadratic Equation with Square Root
The equation shown in the image appears to be a quadratic equation written as: \[ \sqrt{x^2 - 2x - 5} = x \] To solve this problem, we will first need to square both sides of the equation to eliminate the square root. Remember that when you square both sides of an equation, you may introduce extraneous solutions, so you'll need to check your answers in the original equation later. \[ (\sqrt{x^2 - 2x - 5})^2 = x^2 \] \[ x^2 - 2x - 5 = x^2 \] After squaring both sides, we notice that the x^2 terms on both sides of the equation are identical and they cancel out. \[ -2x - 5 = 0 \] Now, add 2x to both sides to isolate the constant term: \[ -5 = 2x \] To solve for x, divide both sides by 2: \[ x = -5 / 2 \] \[ x = -2.5 \] So the potential solution is x = -2.5. However, because we squared the equation to remove the square root, we need to verify that this potential solution satisfies the original equation: \[ \sqrt{(-2.5)^2 - 2(-2.5) - 5} \overset{?}{=} -2.5 \] Calculate the expression under the square root: \[ \sqrt{6.25 + 5 - 5} \] \[ \sqrt{6.25} \] We find the square root of 6.25: \[ \sqrt{6.25} = 2.5 \] Note that the square root of a positive number is positive, so the left side yields 2.5, not -2.5. Therefore, the potential solution x = -2.5 does not satisfy the original equation. This means that there is no solution to the original equation because the square root of a real number cannot equal a negative number. Thus, the original equation has no real solutions. If we needed to find complex solutions, we could continue the process knowing that the square root function would involve imaginary numbers when handling negative quantities, but based on the equation given, it seems we're dealing with real numbers only.
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