Question

Find the real solutions, if any, of each equation. $$ \sqrt{3 x+1}-2 x=-6 $$

Short answer

The solution is x = 5.

Step-by-step solution

Isolate the square root term

Move the linear term involving x to the right side to isolate the square root term. \ \ \ \ \( \sqrt{3x + 1} = -6 + 2x \ \ \)

Eliminate the square root

Square both sides of the equation to remove the square root: \ \ \ \( \left( \sqrt{3x + 1} \right)^2 = \left( -6 + 2x \right)^2 \ \ \) \ This simplifies to: \ \ \ \( 3x + 1 = (-6 + 2x)^2 \)

Expand and simplify

Expand the right side of the equation: \ \ \ \( 3x + 1 = 36 - 24x + 4x^2 \ \ \) \ Combine like terms: \ \ \ \( 4x^2 - 27x + 35 = 0 \)

Solve the quadratic equation

Use the quadratic formula to solve for x: \ \ \ \( x = \frac{ -b \pm \sqrt{b^2 - 4ac} }{ 2a } \) \ Here, \( a = 4 \, \ b = -27 \, \ c = 35 \) \ \ Plugging the values in: \ \ \ \( x = \frac{27 \pm \sqrt{(-27)^2 - 4(4)(35)}}{2(4)} = \frac{27 \pm \sqrt{729 - 560}}{8} = \frac{27 \pm \sqrt{169}}{8} = \frac{27 \pm 13}{8} \)

Find the solutions

Solve for the two values of x: \ \ \ \( x = \frac{27 + 13}{8} = 5 \) \ and \ \( x = \frac{27 - 13}{8} = \frac{14}{8} = 1.75 \)

Verify the solutions

Substitute the values back into the original equation to ensure they are correct: \ \ \ For \( x = 5 \): \ \( \sqrt{3(5) + 1} - 2(5) = \sqrt{16} - 10 = 4 - 10 = -6 \), which is true. \ \ For \( x = 1.75 \): \ \ \ \( \sqrt{3(1.75) + 1} - 2(1.75) = \sqrt{6.25} - 3.5 = 2.5 - 3.5 = -1 \), which is not true. \ So, \( x = 1.75 \) does not satisfy the equation.

Key concepts

quadratic formula

The quadratic formula is a powerful tool for solving quadratic equations. A quadratic equation is any equation of the form

isolating square root

Isolating the square root is essential when solving equations involving a square root term. The goal is to have the square root term by itself on one side of the equation. This makes it easier to eliminate the square root by squaring both sides.

verifying solutions

Verifying solutions is the final important step in solving any equation. Once potential solutions are found, they need to be substituted back into the original equation to ensure they satisfy it. If a solution does not satisfy the equation, it is considered extraneous and should be discarded.