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Question Video: Solving a Compound Inequality Involving Radical Coefficients Mathematics • Second Year of Preparatory School

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Find the solution set of the inequality π‘₯/√2 < (βˆ’βˆš2)π‘₯ + √2 < 3(√2)π‘₯ in ℝ. Give your answer in interval notation.

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Video Transcript

Find the solution set of the inequality π‘₯ over the square root of two is less than negative square root of two times π‘₯ plus the square root of two is less than three times the square root of two times π‘₯ in the set of real numbers. Give your answer in interval notation.

We begin by recalling that we can solve compound inequalities by treating them as two separate inequalities. This is especially useful when the variable is found in all three parts of the compound inequality. So, for π‘₯ to be a solution of the inequality, it must be a solution of both π‘₯ over the square root of two less than negative square root of two times π‘₯ plus the square root of two and negative square root of two times π‘₯ plus the square root of two less than three times the square root of two times π‘₯.

We can solve each of these inequalities by isolating π‘₯ on either side of the inequality. Let’s start with π‘₯ over the square root of two less than negative square root of two times π‘₯ plus the square root of two. We can simplify the inequality by multiplying through by the square root of two. On the left-hand side, the square roots of two cancel. And on the right-hand side, we distribute the square root of two through the parentheses. This gives us π‘₯ is less than the square root of two times negative square root of two times π‘₯ plus the square root of two times the square root of two.

We know that the square root of two times the square root of two is the square root of four. And because the square root of four equals two, we have π‘₯ is less than negative two π‘₯ plus two. We can then add two π‘₯ to both sides of the inequality. This gives us three π‘₯ is less than two. Now, we divide both sides of the inequality through by three. This yields the inequality π‘₯ is less than two-thirds.

To solve this problem, we also need the second inequality to be satisfied by solutions to the original compound inequality. Our first step will be the same. We will multiply through by the square root of two. On the left-hand side, this means we must distribute the square root of two through the parentheses to get the following inequality. Then, in each set of parentheses, we simplify the product of the square root of two and the square root of two, which is the square root of four. And we know that the square root of four is two, so we have negative two π‘₯ plus two is less than six π‘₯.

Next, we add two π‘₯ to both sides of the inequality. This gives us two is less than eight π‘₯. We can now divide the inequality through by eight to get one-fourth is less than π‘₯. We then combine our two answers to form a compound inequality. That is, one-fourth is less than π‘₯ is less than two-thirds.

We are asked to write our answer in interval notation. Since the endpoints of the interval are not included, the solution set of the given inequality is the open interval from one-fourth to two-thirds. And sometimes, we use outward-facing brackets to notate the open interval.

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