Question

Solve: \(\sqrt{3 x+5}-\sqrt{x-2}=\sqrt{x+3}\)

Short answer

The solutions are \( x = \frac{2 + 2\sqrt{31}}{3} \) and \( x = \frac{2 - 2\sqrt{31}}{3} \).

Step-by-step solution

Isolate One Square Root

Move one of the square root terms to the other side of the equation. For instance, write the equation as: \( \sqrt{3x + 5} = \sqrt{x + 3} + \sqrt{x - 2} \).

Square Both Sides

Square both sides of the equation to eliminate the square root on the left side: \[ (\sqrt{3x + 5})^2 = (\sqrt{x + 3} + \sqrt{x - 2})^2 \], which simplifies to \[ 3x + 5 = (x + 3) + (x - 2) + 2\sqrt{(x + 3)(x - 2)} \].

Simplify the Equation

Combine like terms on the right side: \[ 3x + 5 = 2x + 1 + 2\sqrt{(x + 3)(x - 2)} \].

Isolate the Remaining Square Root

Subtract \(2x + 1\) from both sides to isolate the square root term: \[ x + 4 = 2\sqrt{(x + 3)(x - 2)} \].

Square Both Sides Again

Square both sides again to eliminate the remaining square root: \[ (x + 4)^2 = 4(x + 3)(x - 2) \], which simplifies to \[ x^2 + 8x + 16 = 4(x^2 + x - 6) \].

Expand and Simplify

Expand both sides and combine like terms: \[ x^2 + 8x + 16 = 4x^2 + 4x - 24 \], and then move all terms to one side: \[ 0 = 3x^2 - 4x - 40 \].

Solve the Quadratic Equation

Use the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \) where \( a = 3 \), \( b = -4 \), and \( c = -40 \). Calculate the discriminant: \[ b^2 - 4ac = (-4)^2 - 4(3)(-40) = 16 + 480 = 496 \]. Finally, solve for \( x \): \[ x = \frac{4 \pm \sqrt{496}}{6} \].

Simplify the Results

Simplify \( \sqrt{496} = 4\sqrt{31} \), so the solutions are: \( x = \frac{4 + 4\sqrt{31}}{6} \) and \( x = \frac{4 - 4\sqrt{31}}{6} \). Further simplify these to: \( x = \frac{2 + 2\sqrt{31}}{3} \) and \( x = \frac{2 - 2\sqrt{31}}{3} \).

Check for Extraneous Solutions

Verify these solutions in the original equation. If a solution does not satisfy the original equation, it is extraneous and should be discarded.

Key concepts

Quadratic Formula

The quadratic formula helps us solve equations of the form equation: `ax^2 + bx + c = 0`.This formula is very useful because it provides a straightforward method to find the values for `x`, as long as you know the coefficients `a`, `b`, and `c`. The quadratic formula is:\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]Here's how you can use it:

• Identify the coefficients `a`, `b`, and `c` from your quadratic equation.
• Calculate the discriminant: `b^2 - 4ac`.
• Plug the coefficients and discriminant into the formula.
• Solve for `x`, which gives two possible solutions due to the `±` symbol.

For example, in our step-by-step solution to the problem \( \sqrt{3x+5} - \sqrt{x-2} = \sqrt{x+3} \): 1. We re-arranged into a quadratic: `3x^2 - 4x - 40 = 0`, where `a=3`, `b=-4`, `c=-40`.2. Using the formula, we found `x = \frac{4 \pm 4\sqrt{31}}{6}`.Remember, calculating the discriminant is crucial, as it tells us the nature of the roots:* If `b^2 - 4ac > 0`, there are two distinct real roots.* If `b^2 - 4ac = 0`, there is one real root (a repeated root).* If `b^2 - 4ac < 0`, the roots are complex.

Isolating Square Roots

Isolating square roots means making a square root the main subject of the equation. This is important because it helps us effectively eliminate the square roots by squaring both sides of the equation.Follow these steps:

• Move one square root term to the other side of the equation, separating it.
• Make sure only one term with a square root is by itself on one side.

For instance, in our example \( \sqrt{3x+5} - \sqrt{x-2} = \sqrt{x+3} \): 1. We isolate the first square root by rewriting the equation: \( \sqrt{3x + 5} = \sqrt{x + 3} + \sqrt{x - 2} \).Now, we can eliminate the square root by squaring both sides. Remember, squaring both sides can introduce extraneous solutions, so we need to check all solutions at the end.Squaring both sides changes the equation into a form that sometimes may not have a square root, making it easier to solve other parts of the equation.Always double-check by simplifying correctly and make sure no steps create errors in the transformation.

Checking Extraneous Solutions

Extraneous solutions are numbers we find when solving an equation that do not satisfy the original equation. They can appear because squaring both sides of an equation can introduce invalid solutions.To check for extraneous solutions, perform these steps:

• Solve the equation completely and list all tentative solutions.
• Substitute each solution back into the original equation.
• Verify if the left-hand side equals the right-hand side for each solution.

For our equation: \( \sqrt{3x+5} - \sqrt{x-2} = \sqrt{x+3} \).Solutions were found using the quadratic formula as \( x = \frac{2 + 2\sqrt{31}}{3} \)and \( x = \frac{2 - 2\sqrt{31}}{3} \).We need to check both solutions. Substitute `x` back into the left-hand and right-hand sides of the original equation and ensure they balance. If they don't, discard those values as extraneous.This step ensures your final answer is correct and valid.