Question - Solving Quadratic Equation with Square Root
mathothersquare root methodquadratic equation solutionssolving quadratic equationsextraneous solutions
Solution:
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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