Solve the Algebra Equation Involving Square Root

Find the real values of x from the algebra equation involving square root
Solve the algebra equation involving square root : √(3x + 1) + √(x – 4) = 3
Solution
√(3x + 1) + √(x – 4) = 3
Square both sides to reduce square root terms from the equation
(√(3x + 1) + √(x – 4))² = 3²
Expand the square using, (a + b)² = a² + 2ab + b²
3x + 1 + 2 √(3x + 1) √(x – 4) + x – 4 = 9
Rearrange the algebra equation
4x + 2 √(3x + 1) √(x – 4) = 12
⇒ √(3x + 1) √(x – 4) = 6 – 2x
Square again to remove square roots completely
(3x + 1) (x – 4) = (6 – 2x)²
⇒ 3x² – 12x + x – 4 = 36 – 24x + 4x²
Rearrange the equation
x² – 13x + 40 = 0, this is a quadratic equation
Apply quadratic formula to find the value of x
\begin{aligned} x=&\frac{13\pm\sqrt{(-13)^2-4\times1\times40}}{2\times1} \\ \\ =&\frac{13\pm\sqrt{169-160}}{2} \\ \\ =&\frac{13\pm3}{2} \\ \\ \Rightarrow \ x=&8, \ x= 5 \end{aligned}
We got x = 8 or x = 5
Now check answers. (Plug the solution and make sure they work)
When x = 8
√(3x + 1) + √(x – 4) =? 3
⇒ √(3 × 8 + 1) + √(8 – 4) =? 3
⇒ √25 + √4 =? 3
⇒ 5 + 2 =? 3
Then we get, 7 ≠ 3
So x = 8 is not a solution of the equation
When x = 5
√(3x + 1) + √(x – 4) =? 3
⇒ √(3 × 5 + 1) + √(5 – 4) =? 3
⇒ √16 + √1 =? 3
⇒ 4 + 1 =? 3
Then we get, 5 ≠ 3
So x = 5 is not a solution of the equation
So the equation has not had real solutions