Question

Find the real solutions of the equation $2x+\sqrt{x+1}=8$.

Short answer

The given equation has exactly one real solution $x=3$.

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 factorization.

The given equation is$2x+\sqrt{x+1}=8$ which can be written as,

role="math" localid="1643882617927" $$\begin{gathered} 2x+\sqrt{x+1}=8....\left(\text{Given}\right) \\ -\sqrt{x+1}=2x-8....\left(\text{Rearranged}\right) \\ x+1=4x^2+64-32x\mathrm{....}\left(\text{Squared}\right) \\ \text{or,4}{x}^2-33x+63=0....\left(1\right) \end{gathered}$$

Evaluate roots of (1) by factoring the equation as shown below.

role="math" localid="1643882897257" $$\begin{gathered} \text{4}{x}^2-12x-21x+63=0....\left(\text{From}\left(1\right)\right) \\ 4x\left(x-3\right)-21\left(x-3\right)=0....\left(\text{Factorized}\right) \\ \left(x-3\right)\left(4x-21\right)=0 \\ x-3=0\text{or}4x-21=0 \\ x=3\text{or}x=\frac{21}{4} \end{gathered}$$

Step 3. Description of steps.

Check result for$x=3$ for the given equation $2x+\sqrt{x+1}=8$.

role="math" localid="1643883316452" $$\begin{gathered} 2x+\sqrt{x+1}=8....\left(\text{Given}\right) \\ 2\left(3\right)+\sqrt{\left(3\right)+1}=8....\left(\text{For}x=3\right) \\ 6+\sqrt{4}=8 \\ 8=8....\left(\text{Whichistrue}\right) \end{gathered}$$

Similarly for $x=\frac{21}{4}$

role="math" localid="1643883514405" $$\begin{gathered} 2x+\sqrt{x+1}=\mathrm{8....}\left(\text{Given}\right) \\ 2\left(\frac{21}{4}\right)+\sqrt{\left(\frac{21}{4}\right)+1}=8....\left(\text{For}x=\frac{21}{4}\right) \\ \frac{21}{2}+\frac{5}{2}=8 \\ 13=8....\left(\text{Whichisfalse}\right) \end{gathered}$$

Since$x=3$ satisfies the given equation$2x+\sqrt{x+1}=8$ and$x=\frac{21}{4}$ doesn’t satisfy the given equation $2x+\sqrt{x+1}=8$, therefore the given equation has exactly one real solution $x=3$.