Chapter 6 - Radical Functions and Rational Exponents - 6-5 Solving Square Root and Other Radical Equations - Practice and Problem-Solving Exercises - Page 395: 29

$x = 3$

First, we want to isolate the radical: $\sqrt {11x + 3} = 2x$ Square both sides of the equation to eliminate the radical: $11x + 3 = (2x)^2$ $11x + 3 = 4x^2$ Move all terms to the left side of the equation: $-4x^2 + 11x + 3 = 0$ Divide both sides of the equation by $-1$ so that the $x^2$ term is positive: $4x^2 - 11x - 3 = 0$ We have a quadratic equation, which is in the form $ax^2 + bx + c = 0$. We need to find which factors multiplied together will equal $ac$ but when added together will equal $b$. In this equation, $ac$ is $-12$ and $b$ is $-11$. The factors $-12$ and $1$ will work. Let's rewrite the equation and split the middle term using these two factors: $4x^2 - 12x + x - 3 = 0$ Group the first two and last two terms: $(4x^2 - 12x) + (x - 3) = 0$ Factor out common factors: $4x(x - 3) + (x - 3) = 0$ Factor out $x-3$ $(x - 3)(4x + 1) = 0$ Use the Zero-Product Property by equating each factor to $0$, then solve each equation: First factor: $4x + 1 = 0$ $4x = -1$ Divide each side by $4$: $x = -\frac{1}{4}$ Second factor: $x - 3 = 0$ $x = 3$ To check if we have any extraneous solutions, we substitute each solution into the original equation to see if the two sides equal one another. Let's check the first solution: $\sqrt {11(-\frac{1}{4}) + 3} - 2(-\frac{1}{4}) = 0$ $\sqrt {-\frac{11}{4} + 3} + \frac{2}{4} = 0$ Convert the terms in the radicand into equivalent fractions: $\sqrt {-\frac{11}{4} + \frac{12}{4}} + \frac{2}{4} = 0$ Combine the terms inside the radicand: $\sqrt {\frac{1}{4}} + \frac{2}{4} = 0$ Take the square root: $\frac{1}{2} + \frac{2}{4} = 0$ Convert the fractions into equivalent fractions: $\frac{2}{4} + \frac{2}{4} = 0$ Combine the fractions: $1 = 0$ The sides are not equal; therefore, this is an extraneous solution. Let's try the solution $x = 3$: $\sqrt {11(3) + 3} - 2(3) = 0$ Multiply first: $\sqrt {33 + 3} - 6= 0$ Simplify the radicand: $\sqrt {36} - 6= 0$ Evaluate the radical: $6 - 6 = 0$ Combine like terms: $0 = 0$ Both sides are equal; therefore, this solution is correct.