Question

Find the real solutions of the equation $\sqrt{2x+1}+1=x$.

Short answer

The given equation has exactly one real solution $x=4$.

Step-by-step solution

Step 1. By method of quadratic formula.

The roots of the quadratic equation of the form $a{x}^2+bx+c=0$where $a\ne 0$can be evaluated by using the quadratic formula shown below.

$x=\frac{-b\pm \sqrt{b^2-4ac}}{2a}$

Step 2. Find roots of equation by quadratic formula.

The given equation is$\sqrt{2x+1}+1=x$ which can be written as,

role="math" localid="1643875041484" $$\begin{gathered} \sqrt{2x+1}+1=x\mathrm{....}\left(\text{Given}\right) \\ \sqrt{2x+1}=x-1....\left(\text{Rearranged}\right) \\ 2x+1=x^2+1-2x\mathrm{....}\left(\text{Squared}\right) \\ \text{or,}{x}^2-4x=0....\left(1\right) \end{gathered}$$

Evaluate roots of (1) by quadratic formula as shown below.

$$\begin{gathered} x=\frac{-\left(-4\right)\pm \sqrt{{\left(-4\right)}^2-4\left(1\right)\left(0\right)}}{2\left(1\right)} \\ =\frac{4\pm \sqrt{16}}{2} \\ =\frac{4\pm 4}{2} \\ =4,0 \end{gathered}$$

Step 3. Description of steps.

Check the result for$x=0$ for the given equation $\sqrt{2x+1}+1=x$.

role="math" localid="1643875382953" $$\begin{gathered} \sqrt{2x+1}+1=x\mathrm{....}\left(\text{Given}\right) \\ \sqrt{2\left(0\right)+1}+1=\left(0\right)\mathrm{....}\left(\text{For}x=0\right) \\ \sqrt{1}+1=0 \\ 2=0....\left(\text{Whichisfalse}\right) \end{gathered}$$

Similarly for $x=4$

role="math" localid="1643875681655" $$\begin{gathered} \sqrt{2x+1}+1=x\mathrm{....}\left(\text{Given}\right) \\ \sqrt{2\left(4\right)+1}+1=\left(4\right)\mathrm{....}\left(\text{For}x=4\right) \\ \sqrt{9}+1=4 \\ 4=4....\left(\text{Whichistrue}\right) \end{gathered}$$

Since$x=4$ satisfies the given equation$\sqrt{2x+1}+1=x$ and$x=0$ doesn’t satisfy the given equation $\sqrt{2x+1}+1=x$, therefore the given equation has exactly one real solution $x=4$.