Question - Solving a Quadratic Equation with a Square Root
mathsolving quadratic equationisolating square rootextraneous solutionsquadratic equation with square root
Solution:
The equation in the image is a quadratic equation:√(x^2 - 2x - 5) = xTo 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 = 0Now, 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 = 0^2Squaring 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 = 0We 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 = 0Let'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) = xSquare both sides directly from this point:x^2 - 2x - 5 = x^2Now cancel x^2 from both sides:-2x - 5 = 0Add 2x to both sides:-5 = 2xNow divide by 2:x = -5/2So 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) = -5/2Calculating the inside of the square root:√((25/4) + (5) - 5) = -5/2√((25/4) + (20/4) - (20/4)) = -5/2√(25/4) = -5/2Since 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.
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