Question

The equation \(\sqrt{x+1}-\sqrt{x-1}=\sqrt{4 x-1}\) has (a) no solution (b) one solution (c) two solutions (d) more than two solutions

Short answer

Based on the given step-by-step solution, we can deduce that the equation has no valid solution, as the solution found does not work in the context of real numbers. Short Answer: The equation has no solution.

Step-by-step solution

Squaring both sides of the equation

To eliminate the square roots from the equation, we will square both sides. This should give us an equation without square roots. So, \((\sqrt{x+1}-\sqrt{x-1})^2=(\sqrt{4x-1})^2\)

Simplifying the equation

Expand the left side of the equation and simplify both sides: \((\sqrt{x+1}-\sqrt{x-1})^2- (\sqrt{4x-1})^2\) = \(0\) \((x + 1 - 2\sqrt{(x+1)(x-1)} + x - 1) - (4x - 1) = 0\) \(2x - 2\sqrt{(x+1)(x-1)} - 4x + 1 = 0\) \(-2x - 2\sqrt{(x+1)(x-1)} + 1 = 0\) Now divide by \(-2\): \(x + \sqrt{(x+1)(x-1)} - \dfrac{1}{2} = 0\) Next, let \(y = \sqrt{(x+1)(x-1)}\) to simplify the equation further: \(x + y - \dfrac{1}{2} = 0\)

Solve the quadratic equation

Now we have an equation in terms of \(x\) and \(y\). To solve for \(x\), we can find the value of \(y\), so we can substitute this expression \(\sqrt{(x+1)(x-1)}\): \(x + \sqrt{(x+1)(x-1)} - \dfrac{1}{2} = 0 \Rightarrow \sqrt{(x+1)(x-1)} = \frac{1}{2} - x\) Squaring both sides of the equation: \((x+1)(x-1) = (\frac{1}{2} - x)^2\) Expanding and simplifying: \(x^2 - 1 = \frac{1}{4} -x + x^2\) Rearranging the terms: \(x = \dfrac{3}{4}\)

Verify the solution with the original equation

We need to make sure if the obtained solution satisfies the original equation with square roots. Substitute the value of \(x\) into the original equation: \(\sqrt{\dfrac{3}{4}+1}-\sqrt{\dfrac{3}{4}-1}=\sqrt{4\cdot\dfrac{3}{4}-1}\) \(\sqrt{\dfrac{7}{4}}-\sqrt{-\dfrac{1}{4}}=\sqrt{3-1}\) Here, we have a square root of a negative number which is not possible for real numbers. Thus, the obtained solution is not valid for the original equation. So, the correct answer is: (a) no solution

Key concepts

Radical Equations

A radical equation is one that contains a variable within a radical, often a square root. Solving these equations often involves isolating the radical on one side and then eliminating it, typically by squaring both sides.
Radical equations can be tricky because squaring both sides can introduce extraneous solutions, which are solutions that don't satisfy the original equation.

• Make sure to isolate the radical before squaring.
• Always check potential solutions by plugging them back into the original equation.

Radical equations might include multiple steps, like isolating more than one radical term. Understanding these basics prepares us for using algebraic techniques to solve them effectively.

Squaring Both Sides

Squaring both sides of an equation is a common technique to remove radicals and simplify the solution process. This method helps to turn a radical equation into a polynomial equation, which is often much simpler to handle.
However, it's essential to be cautious when using this method because it might lead to false solutions.

• After squaring, simplify and arrange the equation into a standard form if possible.
• Be wary of extraneous solutions, which happen because squaring a number hides its original sign.

In our example, squaring both sides transformed the problem into a simpler polynomial. Despite this convenience, we saw that it ended up giving a solution that was invalid once verified.

Solution Verification

Verifying the solution is a crucial final step to ensure that the solution actually satisfies the original equation. With squared equations, outdated or incorrect values might initially seem valid until we recheck.
To verify, substitute the solution back into the original equation and see if both sides are equal. If they are not, the solution is extraneous.

• Check for both real and imaginary numbers when verification is involved.
• If the solution does not check out, there's no valid solution for the equation.

In our case, the solution resulted in a square root of a negative number, revealing the need to dismiss the solution in the real number set.

Complex Numbers

When dealing with radical equations, especially after verification, we may encounter complex numbers if square roots of negative numbers appear. Complex numbers consist of a real part and an imaginary part, denoted most often in the form of \(a + bi \), where \(i = \sqrt{-1}\).
Imaginary numbers arise only if you allow them, as most radical equations deal exclusively with real numbers.

• Living in the realm of real numbers implies any appearance of complex or imaginary solutions cannot be accepted.
• Understand that sometimes the mathematics helps guide whether to look into complex solutions or dismiss them altogether in contexts demanding only real solutions.

In our particular example, the possibility of a negative result under a square root hinted towards complex solutions if further explored, confirming that the original problem had no real solution.