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