Question

Find the real solutions of the equation$\sqrt{5-x}+1=x-2$.

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$\sqrt{5-x}+1=x-2$ which can be written as,

role="math" localid="1643881931756" $$\begin{gathered} \sqrt{5-x}+1=x-2....\left(\text{Given}\right) \\ \sqrt{5-x}=x-3....\left(\text{Rearranged}\right) \\ 5-x=x^2+9-6x\mathrm{....}\left(\text{Squared}\right) \\ \text{or,}{x}^2-5x+4=0....\left(1\right) \end{gathered}$$

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

role="math" localid="1643881102135" $$\begin{gathered} x=\frac{-\left(-5\right)\pm \sqrt{{\left(-5\right)}^2-4\left(1\right)\left(4\right)}}{2\left(1\right)} \\ =\frac{5\pm \sqrt{9}}{2} \\ =\frac{5\pm 3}{2} \\ =4,1 \end{gathered}$$

Step 3. DESCRIPTION OF STEPS.

Check result for$x=1$ for the given equation $\sqrt{5-x}+1=x-2$.

role="math" localid="1643881683299" $$\begin{gathered} \sqrt{5-x}+1=x-2....\left(\text{Given}\right) \\ \sqrt{5-\left(1\right)}+1=\left(1\right)-2....\left(\text{For}x=1\right) \\ \sqrt{4}+1=-1 \\ 3=-1....\left(\text{Whichisfalse}\right) \end{gathered}$$

Similarly for $x=4$

role="math" localid="1643881634379" $$\begin{gathered} \sqrt{5-x}+1=x-2....\left(\text{Given}\right) \\ \sqrt{5-\left(4\right)}+1=\left(4\right)-2....\left(\text{For}x=4\right) \\ \sqrt{1}+1=2 \\ 2=2....\left(\text{Whichistrue}\right) \end{gathered}$$

Sincesatisfies the given equation$\sqrt{5-x}+1=x-2$ and$x=1$ doesn’t satisfy the given equation $\sqrt{5-x}+1=x-2$, therefore the given equation has exactly one real solution $x=4$.