Video Transcript
Which of the following shows the solution set of the inequality negative square root of three times 𝑥 plus one is less than two times the square root of three times 𝑥?
To sketch the solution set of this inequality, we first want to isolate the variable 𝑥 on the left-hand side of the inequality. We can do this in at least two ways. We can divide the inequality through by the square root of three to get negative 𝑥 plus one is less than two 𝑥. Then, we distribute negative one over the parentheses to get negative 𝑥 minus one is less than two 𝑥.
Next, we recall that we can add any real number to both sides of the inequality to get an equivalent inequality. So we add 𝑥 to both sides to get negative one is less than three 𝑥. Then, we divide through by three to get negative one-third is less than 𝑥, or equivalently 𝑥 is greater than negative one-third.
An alternative method begins with distributing the negative square root of three across the parentheses on the left-hand side of the inequality, resulting in the negative square root of three times 𝑥 minus the square root of three is less than two times the square root of three times 𝑥. Then, we can add the square root of three to both sides to get negative square root of three times 𝑥 is less than two times the square root of three times 𝑥 plus the square root of three. Similarly, we can subtract two square root of three 𝑥 from both sides to get negative three times the square root of three times 𝑥 is less than the square root of three.
Now we can isolate 𝑥 on the left-hand side by dividing both sides of the inequality through by negative three square root of three. We note that whenever we divide or multiply by a negative, we need to switch the direction of the inequality, in this case from less than to greater than. By canceling the square roots of three, we find that 𝑥 is greater than negative one-third.
We can sketch this answer on a number line by drawing an open circle at negative one-third to show that this value is not included and then including all numbers greater than negative one-third. This result matches the diagram given as option (D).
