Question

Solve the equation. Check for extraneous solutions. $$10=4+\sqrt{5 x+11}$$

Short answer

The equation \(10 = 4+ \sqrt{5x + 11}\) has no solution.

Step-by-step solution

Isolate the square root term

First subtract \(4\) from both sides of the equation to isolate the square root term.\(10 - 4 = \sqrt{5x + 11}\), which simplifies to \(6 = \sqrt{5x + 11}\)

Eliminate the square root

Now eliminate the square root by squaring both sides of the equation: \((6)^2 = (\sqrt{5x + 11})^2\), which simplifies to \(36 = 5x + 11\)

Solve for x

Next, solve for \(x\) by subtracting \(11\) from both sides and then dividing by \(5\): \(36 - 11 = 5x\), which simplifies to \(25 = 5x\), and after dividing we get, \(5 = x\)

Check for extraneous solutions

Finally, it’s necessary to check for extraneous solutions (solutions that don’t actually satisfy the original equation) by plugging \(5\) back into the original equation and checking if both sides are equal: Substitute \(5\) for \(x\) in the original equation:\(10 = 4 + \sqrt{5*5 + 11}\), which simplifies to \(10 = 4 + \sqrt{36}\), and further simplifies to \(10 = 4 + 6\), but \(10 ≠ 10\). So, \(x = 5\) is not a solution for this equation, it's an extraneous solution.

Key concepts

Isolating the Square Root

When solving square root equations, the first crucial step often involves isolating the square root on one side of the equation. This means we want to have the square root term by itself so we can deal with it more directly. In the given problem, we start with the equation
The isolated square root makes the subsequent steps more straightforward, as it allows us to perform operations that can eliminate the square root, such as squaring both sides of the equation, without other terms complicating the process.

Extraneous Solutions

After isolating and eliminating the square root, we solve for the variable as in any other algebraic problem. However, a key aspect unique to square root equations is the potential for extraneous solutions.

These are 'solutions' that satisfy the modified equation we created during the solving process but do not satisfy the original equation. They often arise from the act of squaring both sides since this can introduce solutions that weren’t there before.

To identify and discard these invalid solutions, we must always substitute them back into the original equation for verification. If they don't work, we reject them as extraneous. In our exercise, checking the calculated value of

clearly shows that we are left with a contradiction, hence it is an extraneous solution.

Algebraic Problem-Solving

Algebraic problem-solving involves manipulation of equations to find the value(s) of unknown variable(s). In an equation with a square root, once we isolate the square root and square both sides to remove the radical, we apply standard algebraic techniques to solve for the variable.

In our exercise, this involved subtracting 11 from both sides and then dividing by 5 to solve for

This systematic approach is a core aspect of algebra that is applied universally, be it for linear equations, quadratic equations, or any other algebraic expressions. By mastering this step-by-step methodology, students can tackle a wide range of mathematical problems with confidence.