{"id":68801,"date":"2021-01-26T08:25:44","date_gmt":"2021-01-26T14:25:44","guid":{"rendered":"https:\/\/www.mometrix.com\/academy\/?page_id=68801"},"modified":"2026-03-25T10:47:39","modified_gmt":"2026-03-25T15:47:39","slug":"solving-inequalities-involving-square-roots","status":"publish","type":"page","link":"https:\/\/www.mometrix.com\/academy\/solving-inequalities-involving-square-roots\/","title":{"rendered":"Solving Inequalities Involving Square Roots"},"content":{"rendered":"\n\t\t\t<div id=\"mmDeferVideoEncompass_PFemE_zyigs\" style=\"position: relative;\">\n\t\t\t<picture>\n\t\t\t\t<source srcset=\"https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2023\/01\/circle-play-duotone.webp\" type=\"image\/webp\">\n\t\t\t\t<source srcset=\"https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2023\/01\/circle-play-duotone.png\" type=\"image\/jpeg\"> \n\t\t\t\t<img fetchpriority=\"high\" decoding=\"async\" loading=\"eager\" id=\"videoThumbnailImage_PFemE_zyigs\" data-source-videoID=\"PFemE_zyigs\" src=\"https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2023\/01\/circle-play-duotone.png\" alt=\"Solving Inequalities Involving Square Roots Video\" height=\"464\" width=\"825\" class=\"size-full\" data-matomo-title = \"Solving Inequalities Involving Square Roots\">\n\t\t\t<\/picture>\n\t\t\t<\/div>\n\t\t\t<style>img#videoThumbnailImage_PFemE_zyigs:hover {cursor:pointer;} img#videoThumbnailImage_PFemE_zyigs {background-size:contain;background-image:url(\"https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2023\/01\/1711-solving-inequalities-involving-square-roots-copy-1-1.webp\");}<\/style>\n\t\t\t<script defer>\n\t\t\t  jQuery(\"img#videoThumbnailImage_PFemE_zyigs\").click(function() {\n\t\t\t\tlet videoId = jQuery(this).attr(\"data-source-videoID\");\n\t\t\t\tlet helpTag = '<div id=\"mmDeferVideoYTMessage_PFemE_zyigs\" style=\"display: none;position: absolute;top: -24px;width: 100%;text-align: center;\"><span style=\"font-style: italic;font-size: small;border-top: 1px solid #fc0;\">Having trouble? <a href=\"https:\/\/www.youtube.com\/watch?v='+videoId+'\" target=\"_blank\">Click here to watch on YouTube.<\/a><\/span><\/div>';\n\t\t\t\tlet tag = document.createElement(\"iframe\");\n\t\t\t\ttag.id = \"yt\" + videoId;\n\t\t\t\ttag.src = \"https:\/\/www.youtube-nocookie.com\/embed\/\" + videoId + \"?autoplay=1&controls=1&wmode=opaque&rel=0&egm=0&iv_load_policy=3&hd=0&enablejsapi=1\";\n\t\t\t\ttag.frameborder = 0;\n\t\t\t\ttag.allow = \"autoplay; fullscreen\";\n\t\t\t\ttag.width = this.width;\n\t\t\t\ttag.height = this.height;\n\t\t\t\ttag.setAttribute(\"data-matomo-title\",\"Solving Inequalities Involving Square Roots\");\n\t\t\t\tjQuery(\"div#mmDeferVideoEncompass_PFemE_zyigs\").html(tag);\n\t\t\t\tjQuery(\"div#mmDeferVideoEncompass_PFemE_zyigs\").prepend(helpTag);\n\t\t\t\tsetTimeout(function(){jQuery(\"div#mmDeferVideoYTMessage_PFemE_zyigs\").css(\"display\", \"block\");}, 2000);\n\t\t\t  });\n\t\t\t  \n\t\t\t<\/script>\n\t\t\n<p><script>\nfunction mbp_Function() {\n  var x = document.getElementById(\"mbp\");\n  if (x.style.display === \"none\") {\n    x.style.display = \"block\";\n  } else {\n    x.style.display = \"none\";\n  }\n}\n<\/script><\/p>\n<div class=\"moc-toc hide-on-desktop hide-on-tablet\">\n<div><button onclick=\"mbp_Function()\"><img loading=\"lazy\" decoding=\"async\" src=\"https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2024\/12\/toc2.svg\" width=\"16\" height=\"16\" alt=\"show or hide table of contents\"><\/button><\/p>\n<p>On this page<\/p>\n<\/div>\n<nav id=\"mbp\" style=\"display:none;\">\n<ul>\n<li class=\"toc-h2\"><a href=\"#Step_1\" class=\"smooth-scroll\">Step 1<\/a><\/li>\n<li class=\"toc-h2\"><a href=\"#Step_2\" class=\"smooth-scroll\">Step 2<\/a><\/li>\n<li class=\"toc-h2\"><a href=\"#Examples\" class=\"smooth-scroll\">Examples<\/a>\n<ul><\/li>\n<li class=\"toc-h3\"><a href=\"#Example_1\" class=\"smooth-scroll\">Example #1<\/a><\/li>\n<li class=\"toc-h3\"><a href=\"#Example_2\" class=\"smooth-scroll\">Example #2<\/a><\/li>\n<li class=\"toc-h3\"><a href=\"#Example_3\" class=\"smooth-scroll\">Example #3<\/a><\/li>\n<li class=\"toc-h3\"><a href=\"#Example_4\" class=\"smooth-scroll\">Example #4<\/a><\/li>\n<\/ul>\n<\/li>\n<li class=\"toc-h2\"><a href=\"#Solving_Inequalities_with_Square_Roots_Practice_Questions\" class=\"smooth-scroll\">Solving Inequalities with Square Roots Practice Questions<\/a><\/li>\n<\/ul>\n<\/nav>\n<\/div>\n<div class=\"accordion\"><input id=\"transcript\" type=\"checkbox\" class=\"spoiler_button\" \/><label for=\"transcript\">Transcript<\/label><input id=\"PQs\" type=\"checkbox\" class=\"spoiler_button\" \/><label for=\"PQs\">Practice<\/label>\n<div class=\"spoiler\" id=\"transcript-spoiler\">\n<p>Hi, and welcome to this video on solving inequalities with square roots!<\/p>\n<p>When solving inequalities with square roots, we need to follow two simple steps. <\/p>\n<h2><span id=\"Step_1\" class=\"m-toc-anchor\"><\/span>Step 1<\/h2>\n<p>\nThe first step is to solve the inequality for \\(x\\) the same way we would any other inequality. There\u2019s nothing special about this step. <\/p>\n<h2><span id=\"Step_2\" class=\"m-toc-anchor\"><\/span>Step 2<\/h2>\n<p>\nBut the second step is unique to inequalities with even roots. Today, we\u2019ll be looking at square roots specifically. The second step is to set whatever is under the radical symbol greater than or equal to zero and solve for \\(x\\) again. We do this because you cannot have a negative value of \\(x\\) under the square root.<\/p>\n<p>In other words, you can\u2019t take the square root of a negative number. This will then give us two inequalities that we then combine to make a compound inequality if the signs are opposite. If the signs are the same, we use the more restrictive inequality.<\/p>\n<h2><span id=\"Examples\" class=\"m-toc-anchor\"><\/span>Examples<\/h2>\n<p>\nLet\u2019s look at a few examples to see what I\u2019m talking about.<\/p>\n<h3><span id=\"Example_1\" class=\"m-toc-anchor\"><\/span>Example #1<\/h3>\n<div class=\"examplesentence\">\\(\\sqrt{x+3}\\)>\\(4\\)<\/div>\n<p>\n&nbsp;<br \/>\nFirst, we want to solve this inequality for \\(x\\) normally. We do this by following the same steps we would as if we were solving an equation instead of an inequality.<\/p>\n<p>We\u2019re going to start by squaring both sides.<\/p>\n<div class=\"examplesentence\">\\(x+3\\)>\\(16\\)<\/div>\n<p>\n&nbsp;<br \/>\nAnd then we\u2019re gonna subtract 3 from both sides.<\/p>\n<div class=\"examplesentence\">\\(x\\)>\\(13\\)<\/div>\n<p>\n&nbsp;<br \/>\nThis gives us \\(x>13\\).<\/p>\n<p>Now it\u2019s time for our special step. We want to set what is under the radical symbol greater than or equal to 0 and solve for \\(x\\) again.<\/p>\n<p>So \\(x+3\\) is under the radical symbol.<\/p>\n<p>So we&#8217;ll set \\(x+3 \\geq 0\\).<\/p>\n<div class=\"examplesentence\">\\(x+3\\)\u2265\\(0\\)<\/div>\n<p>\n&nbsp;<br \/>\nSo now we&#8217;re going to subtract 3 from both sides.<\/p>\n<div class=\"examplesentence\">\\(x\\)\u2265\\(-3\\)<\/div>\n<p>\n&nbsp;<br \/>\nWhich gives us \\(x \\geq -3\\).<\/p>\n<p>Since our signs are in the same direction for both inequalities, we choose the more restrictive one as our final answer. \\(x>13\\) is more restrictive than \\(x \\geq -3\\) because, according to the second inequality, 0 could be included because it is greater than or equal to -3, but it is not truly in our solution set because it is not greater than 13. Therefore, we only want to consider values of \\(x\\) that are greater than 13.<\/p>\n<h3><span id=\"Example_2\" class=\"m-toc-anchor\"><\/span>Example #2<\/h3>\n<p>\nLet\u2019s try another one.<\/p>\n<p>Solve for \\(x\\).<\/p>\n<div class=\"examplesentence\">\\(\\sqrt{2x-5}+7\\)<\\(12\\)<\/div>\n<p>\n&nbsp;<br \/>\nWe&#8217;re gonna start by subtracting 7 from both sides.<\/p>\n<div class=\"examplesentence\">\\(\\sqrt{2x-5}\\)<\\(5\\)<\/div>\n<p>\n&nbsp;<br \/>\nThen we&#8217;re going to square both sides.<\/p>\n<div class=\"examplesentence\">\\(2x-5\\)<\\(25\\)<\/div>\n<p>\n&nbsp;<br \/>\nAdd 5 to both sides<\/p>\n<div class=\"examplesentence\">\\(2x\\)<\\(30\\)<\/div>\n<p>\n&nbsp;<br \/>\nAnd divide by 2 on both sides.<\/p>\n<div class=\"examplesentence\">\\(x\\)<\\(15\\)<\/div>\n<p>\n&nbsp;<br \/>\nNext, we\u2019re gonna set \\(2x\u20135\\) greater than or equal to 0, and solve for \\(x\\) again.<\/p>\n<div class=\"examplesentence\">\\(2x-5\\)\u2265\\(0\\)<\/div>\n<p>\n&nbsp;<br \/>\nSo we&#8217;re gonna add 5 to both sides.<\/p>\n<div class=\"examplesentence\">\\(2x\\)\u2265\\(5\\)<\/div>\n<p>\n&nbsp;<br \/>\nAnd divide by 2 on both sides.<\/p>\n<div class=\"examplesentence\">\\(x\\)\u2265\\(\\frac{5}{2}\\)<\/div>\n<p>\n&nbsp;<br \/>\nIn this example, our inequality signs face opposite directions, so we want to combine them to make a compound inequality. Our final answer is: \\(\\frac{5}{2}\\)\u2264\\(x\\)<\\(15\\).\n\n\n\n\n\n\n<h3><span id=\"Example_3\" class=\"m-toc-anchor\"><\/span>Example #3<\/h3>\n<p>\nLet\u2019s try one more example together.<\/p>\n<div class=\"examplesentence\">\\(\\sqrt{-3x+1}-9\\)\u2265\\(11\\)<\/div>\n<p>\n&nbsp;<br \/>\nSolve for \\(x\\) normally.<\/p>\n<div class=\"examplesentence\">\\(\\sqrt{-3x+1}\\)\u2265\\(20\\)<br \/>\n\\(-3x+1\\)\u2265\\(400\\)<br \/>\n\\(-3x\\)\u2265\\(399\\)<br \/>\n\\(x\\)\u2264\\(-133\\)<\/div>\n<p>\n&nbsp;<br \/>\nThen solve so what is under the radical is not negative.<\/p>\n<div class=\"examplesentence\">\\(-3x+1\\)\u2265\\(0\\)<br \/>\n\\(-3x\\)\u2265\\(-1\\)<br \/>\n\\(x\\)\u2264\\(\\frac{1}{3}\\)<\/div>\n<p>\n&nbsp;<br \/>\nSince the signs are in the same direction, we choose the more restrictive case as our final answer: \\(x\\)\u2264\\(-133\\).<\/p>\n<h3><span id=\"Example_4\" class=\"m-toc-anchor\"><\/span>Example #4<\/h3>\n<p>\nI want to do one more problem before we go, but this time try it on your own. Once I give you the problem, pause the video and solve the inequality yourself. Then, press play and see if your answer matches up with mine.<\/p>\n<div class=\"examplesentence\">\\(\\sqrt{4x+4}+11\\)<\\(21\\)<\/div>\n<p>\n&nbsp;<br \/>\nThink you\u2019ve got it? Let\u2019s see!<\/p>\n<p>First, we want to solve our inequality normally.<\/p>\n<div class=\"examplesentence\">\\(\\sqrt{4x+4}\\)<\\(10\\)<br \/>\n\\(4x+4\\)<\\(100\\)<br \/>\n\\(4x\\)<\\(96\\)<br \/>\n\\(x\\)<\\(24\\)<\/div>\n<p>\n&nbsp;<br \/>\nThen we set what is under the radical greater than or equal to 0 and solve for <em>x<\/em> again.<\/p>\n<div class=\"examplesentence\">\\(4x+4\\)\u2265\\(0\\)<br \/>\n\\(4x\\)\u2265\\(-4\\)<br \/>\n\\(x\\)\u2265\\(-1\\)<\/div>\n<p>\n&nbsp;<br \/>\nSince our signs are in opposite directions, our final inequality will be a compound inequality: \\(-1\\)\u2264\\(x\\)<\\(24\\).\n\n\nRemember, we cannot take the square root of a negative value, which is why we place the expression under the radical to be greater than or equal to 0. This is true of any even root, like a fourth root or an eighth root. However, it is okay to take an odd root of a negative value. For instance, the cube root of -8 is -2 because \\(-2 \\times -2 \\times -2 = -8\\). \n\n\nIf you are asked to solve an inequality with a different even root, follow the same steps we did here. If it is an odd root, you do not have to worry about the second step and can stop after you initially solve the inequality.<\/p>\n<hr>\n<p>\nI hope this video on solving inequalities with square roots was helpful. Thanks for watching and happy studying!<\/p>\n<\/div>\n<div class=\"spoiler\" id=\"PQs-spoiler\">\n<h2 style=\"text-align:center\"><span id=\"Solving_Inequalities_with_Square_Roots_Practice_Questions\" class=\"m-toc-anchor\"><\/span>Solving Inequalities with Square Roots Practice Questions<\/h2>\n\n\t\t\t\t<div class=\"PQ\">\n\t\t\t\t\t<strong>Question #1:<\/strong>\n\t\t\t\t\t<div style=\"margin-left:10px;\"><p>&nbsp;<br \/>\nSolve this inequality for \\(x\\):<\/p>\n<div class=\"yellow-math-quote\">\\(\\sqrt{x-5} \\lt 1\\)<\/div>\n<\/div>\n\t\t\t\t\t<div class=\"PQ-Choices\"><div class=\"PQ\"  id=\"PQ-1-1\">\\(x \\lt 6\\)<\/div><div class=\"PQ\"  id=\"PQ-1-2\">\\(x \\geq 5\\)<\/div><div class=\"PQ\"  id=\"PQ-1-3\">\\(-5 \\leq x \\lt 6\\)<\/div><div class=\"PQ correct_answer\"  id=\"PQ-1-4\">\\(5 \\leq x \\lt 6\\)<\/div>\n\t\t\t\t\t<\/div>\n\t\t\t\t\t<input id=\"PQ-1\" type=\"checkbox\" class=\"spoiler_button\" \/><label for=\"PQ-1\" style=\"width: 150px;\">Show Answer<\/label>\n\t\t\t\t\t<div class=\"answer\" id=\"PQ-1-spoiler\">\n\t\t\t\t\t\t<strong>Answer:<\/strong><div style=\"margin-left:10px;\"><p>When solving an inequality, we follow the same steps as if we are solving an equation. Because the value under the radical cannot equal zero, we must add an additional step by setting what is under the radical greater than or equal to zero and solve for \\(x\\).<\/p>\n<p style=\"text-align: center; line-height: 40px\">\n\\(\\sqrt{x-5} \\lt 1\\)<br \/>\n\\((\\sqrt{x-5})^2 \\lt 1^2\\)<br \/>\n\\(x-5 \\lt 1\\)<br \/>\n\\(x \\lt 6\\)\n<\/p>\n<p style=\"text-align: center; line-height: 40px\">\n\\(x-5 \\geq 0\\)<br \/>\n\\(x \\geq 5\\)<\/p>\n<p>Since the signs are opposite directions, our answer will be the compound inequality \\(5 \\leq x \\lt 6\\).<\/p>\n<\/div>\n\t\t\t\t\t\t<input id=\"PQ-1-hide\" type=\"checkbox\" class=\"spoiler_button\" \/><label for=\"PQ-1-hide\" style=\"width: 150px;\">Hide Answer<\/label>\n\t\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t<div class=\"PQ\">\n\t\t\t\t\t<strong>Question #2:<\/strong>\n\t\t\t\t\t<div style=\"margin-left:10px;\"><p>&nbsp;<br \/>\nSolve this inequality for \\(x\\):<\/p>\n<div class=\"yellow-math-quote\">\\(\\sqrt{3x+9} \\gt 12\\)<\/div>\n<\/div>\n\t\t\t\t\t<div class=\"PQ-Choices\"><div class=\"PQ\"  id=\"PQ-2-1\">\\(x \\geq -3\\)<\/div><div class=\"PQ correct_answer\"  id=\"PQ-2-2\">\\(x \\gt 45\\)<\/div><div class=\"PQ\"  id=\"PQ-2-3\">\\(-3 \\leq x \\lt 45\\)<\/div><div class=\"PQ\"  id=\"PQ-2-4\">\\(3 \\leq x \\lt 45\\)<\/div>\n\t\t\t\t\t<\/div>\n\t\t\t\t\t<input id=\"PQ-2\" type=\"checkbox\" class=\"spoiler_button\" \/><label for=\"PQ-2\" style=\"width: 150px;\">Show Answer<\/label>\n\t\t\t\t\t<div class=\"answer\" id=\"PQ-2-spoiler\">\n\t\t\t\t\t\t<strong>Answer:<\/strong><div style=\"margin-left:10px;\"><p>When solving an inequality, we use the same steps as when solving an equation. Since we cannot have a zero under the radical, we have an extra step to ensure our solution set does not include any value for \\(x\\) that would make the value under the radical zero.<\/p>\n<p style=\"text-align: center; line-height: 40px\">\n\\(\\sqrt{3x+9} \\gt 12\\)<br \/>\n\\((\\sqrt{3x+9})^2 \\gt 12^2\\)<br \/>\n\\(3x+9 \\gt 144\\)<br \/>\n\\(3x \\gt 135\\)<br \/>\n\\(x \\gt 45\\)\n<\/p>\n<p style=\"text-align: center; line-height: 40px\">\n\\(3x+9 \\geq 0\\)<br \/>\n\\(3x \\geq -9\\)<br \/>\n\\(x \\geq -3\\)\n<\/p>\n<p>Since the signs are in the same direction, we choose the answer that is more restrictive, which is \\(x \\gt 45\\).<\/p>\n<\/div>\n\t\t\t\t\t\t<input id=\"PQ-2-hide\" type=\"checkbox\" class=\"spoiler_button\" \/><label for=\"PQ-2-hide\" style=\"width: 150px;\">Hide Answer<\/label>\n\t\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t<div class=\"PQ\">\n\t\t\t\t\t<strong>Question #3:<\/strong>\n\t\t\t\t\t<div style=\"margin-left:10px;\"><p>&nbsp;<br \/>\nSolve this inequality for \\(x\\).<\/p>\n<div class=\"yellow-math-quote\">\\(\\sqrt{5x-1}-2 \\lt 11\\)<\/div>\n<\/div>\n\t\t\t\t\t<div class=\"PQ-Choices\"><div class=\"PQ\"  id=\"PQ-3-1\">\\(x \\geq \\frac{1}{5}\\)<\/div><div class=\"PQ\"  id=\"PQ-3-2\">\\(x \\lt 34\\)<\/div><div class=\"PQ\"  id=\"PQ-3-3\">\\(-\\frac{1}{5} \\leq x \\lt 34\\)<\/div><div class=\"PQ correct_answer\"  id=\"PQ-3-4\">\\(\\frac{1}{5} \\leq x \\lt 34\\)<\/div>\n\t\t\t\t\t<\/div>\n\t\t\t\t\t<input id=\"PQ-3\" type=\"checkbox\" class=\"spoiler_button\" \/><label for=\"PQ-3\" style=\"width: 150px;\">Show Answer<\/label>\n\t\t\t\t\t<div class=\"answer\" id=\"PQ-3-spoiler\">\n\t\t\t\t\t\t<strong>Answer:<\/strong><div style=\"margin-left:10px;\"><p>We will solve the inequality following the same steps as when solving an equation. Since we cannot have a zero under the radical, we must follow an additional step to ensure our solution set does not include any value that would make the value under the radical zero.<\/p>\n<p style=\"text-align: center; line-height: 40px\">\n\\(\\sqrt{5x-1}-2 \\lt 11\\)<br \/>\n\\(\\sqrt{5x-1} \\lt 13\\)<br \/>\n\\((\\sqrt{5x-1})^2 \\lt 13^2\\)<br \/>\n\\(5x-1 \\lt 169\\)<br \/>\n\\(5x \\lt 170\\)<br \/>\n\\(x \\lt ]34\\)\n<\/p>\n<p style=\"text-align: center; line-height: 40px\">\n\\(5x-1 \\geq 0\\)<br \/>\n\\(5x \\geq 1\\)<br \/>\n\\(x \\geq \\frac{1}{5}\\)\n<\/p>\n<p>Since the signs are in opposite directions, our answer will be in the form of a compound inequality, which is \\(\\frac{1}{5} \\leq x \\lt 34\\).<\/p>\n<\/div>\n\t\t\t\t\t\t<input id=\"PQ-3-hide\" type=\"checkbox\" class=\"spoiler_button\" \/><label for=\"PQ-3-hide\" style=\"width: 150px;\">Hide Answer<\/label>\n\t\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t<div class=\"PQ\">\n\t\t\t\t\t<strong>Question #4:<\/strong>\n\t\t\t\t\t<div style=\"margin-left:10px;\"><p>&nbsp;<br \/>\nSolve this inequality for \\(x\\):<\/p>\n<div class=\"yellow-math-quote\">\\(\\sqrt{6x+10} \\geq 8\\)<\/div>\n<\/div>\n\t\t\t\t\t<div class=\"PQ-Choices\"><div class=\"PQ\"  id=\"PQ-4-1\">\\(x \\geq -\\frac{5}{3}\\)<\/div><div class=\"PQ correct_answer\"  id=\"PQ-4-2\">\\(x \\geq 9\\)<\/div><div class=\"PQ\"  id=\"PQ-4-3\">\\(-\\frac{5}{3} \\leq x \\leq 9\\)<\/div><div class=\"PQ\"  id=\"PQ-4-4\">\\(\\frac{5}{3} \\leq x \\leq 9\\)<\/div>\n\t\t\t\t\t<\/div>\n\t\t\t\t\t<input id=\"PQ-4\" type=\"checkbox\" class=\"spoiler_button\" \/><label for=\"PQ-4\" style=\"width: 150px;\">Show Answer<\/label>\n\t\t\t\t\t<div class=\"answer\" id=\"PQ-4-spoiler\">\n\t\t\t\t\t\t<strong>Answer:<\/strong><div style=\"margin-left:10px;\"><p>When solving an inequality, we follow the same steps as if we are solving an equation. However, since the value under the radical cannot be zero, we follow an additional step to ensure the solution set does not include any value that would make the value under the radical zero.<\/p>\n<p style=\"text-align: center; line-height: 40px\">\n\\((\\sqrt{6x+10}) \\geq 8\\)<br \/>\n\\((\\sqrt{6x+10})^2 \\geq 8^2\\)<br \/>\n\\(6x+10 \\geq 64\\)<br \/>\n\\(6x \\geq 54\\)<br \/>\n\\(x \\geq 9\\)\n<\/p>\n<p style=\"text-align: center; line-height: 40px\">\n\\(6x+10 \\geq 0\\)<br \/>\n\\(6x \\geq -10\\)<br \/>\n\\(x \\geq -\\frac{5}{3}\\)<\/p>\n<p>Since the signs are in the same direction, we choose the answer that is more restrictive and does not include zero, which is \\(x \\geq 9\\).<\/p>\n<\/div>\n\t\t\t\t\t\t<input id=\"PQ-4-hide\" type=\"checkbox\" class=\"spoiler_button\" \/><label for=\"PQ-4-hide\" style=\"width: 150px;\">Hide Answer<\/label>\n\t\t\t\t\t<\/div>\n\t\t\t\t<\/div>\n\t\t\t\t<div class=\"PQ\">\n\t\t\t\t\t<strong>Question #5:<\/strong>\n\t\t\t\t\t<div style=\"margin-left:10px;\"><p>&nbsp;<br \/>\nSolve this inequality for \\(x\\):<\/p>\n<div class=\"yellow-math-quote\">\\(\\sqrt{7x+4}-5 \\geq 7\\)<\/div>\n<\/div>\n\t\t\t\t\t<div class=\"PQ-Choices\"><div class=\"PQ\"  id=\"PQ-5-1\">\\(x \\geq -\\frac{4}{7}\\)<\/div><div class=\"PQ correct_answer\"  id=\"PQ-5-2\">\\(x \\geq 20\\)<\/div><div class=\"PQ\"  id=\"PQ-5-3\">\\(-\\frac{4}{7} \\leq x \\leq 20\\)<\/div><div class=\"PQ\"  id=\"PQ-5-4\">\\(\\frac{4}{7} \\leq x \\leq 20\\)<\/div>\n\t\t\t\t\t<\/div>\n\t\t\t\t\t<input id=\"PQ-5\" type=\"checkbox\" class=\"spoiler_button\" \/><label for=\"PQ-5\" style=\"width: 150px;\">Show Answer<\/label>\n\t\t\t\t\t<div class=\"answer\" id=\"PQ-5-spoiler\">\n\t\t\t\t\t\t<strong>Answer:<\/strong><div style=\"margin-left:10px;\"><p>We will solve the inequality the same way as we would solve an equation. Since this inequality has a radical, we must follow an additional step to ensure our solution set does not include zero. <\/p>\n<p style=\"text-align: center; line-height: 40px\">\n\\(\\sqrt{7x+4}-5 \\geq 7\\)<br \/>\n\\(\\sqrt{7x+4} \\geq 12\\)<br \/>\n\\((\\sqrt{7x+4})^2 \\geq 12^2\\)<br \/>\n\\(7x+4 \\geq 144\\)<br \/>\n\\(7x \\geq 140\\)<br \/>\n\\(x \\geq 20\\)\n<\/p>\n<p style=\"text-align: center; line-height: 40px\">\n\\(7x+4 \\geq 0\\)<br \/>\n\\(7x \\geq -4\\)<br \/>\n\\(x \\geq -\\frac{4}{7}\\)\n<\/p>\n<p>Since the signs are in the same direction, we choose the answer that is more restrictive, which is \\(x \\geq 20\\).<\/p>\n<\/div>\n\t\t\t\t\t\t<input id=\"PQ-5-hide\" type=\"checkbox\" class=\"spoiler_button\" \/><label for=\"PQ-5-hide\" style=\"width: 150px;\">Hide Answer<\/label>\n\t\t\t\t\t<\/div>\n\t\t\t\t<\/div><\/div>\n<\/div>\n\n<div class=\"home-buttons\">\n<p><a href=\"https:\/\/www.mometrix.com\/academy\/algebra-i\/\">Return to Algebra I Videos<\/a><\/p>\n<\/div>\n","protected":false},"excerpt":{"rendered":"<p>Return to Algebra I Videos<\/p>\n","protected":false},"author":1,"featured_media":100699,"parent":0,"menu_order":0,"comment_status":"closed","ping_status":"closed","template":"","meta":{"footnotes":""},"class_list":{"0":"post-68801","1":"page","2":"type-page","3":"status-publish","4":"has-post-thumbnail","6":"page_category-inequalities-videos","7":"page_type-video","8":"content_type-practice-questions","9":"subject_matter-math"},"aioseo_notices":[],"_links":{"self":[{"href":"https:\/\/www.mometrix.com\/academy\/wp-json\/wp\/v2\/pages\/68801","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/www.mometrix.com\/academy\/wp-json\/wp\/v2\/pages"}],"about":[{"href":"https:\/\/www.mometrix.com\/academy\/wp-json\/wp\/v2\/types\/page"}],"author":[{"embeddable":true,"href":"https:\/\/www.mometrix.com\/academy\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/www.mometrix.com\/academy\/wp-json\/wp\/v2\/comments?post=68801"}],"version-history":[{"count":6,"href":"https:\/\/www.mometrix.com\/academy\/wp-json\/wp\/v2\/pages\/68801\/revisions"}],"predecessor-version":[{"id":278938,"href":"https:\/\/www.mometrix.com\/academy\/wp-json\/wp\/v2\/pages\/68801\/revisions\/278938"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/www.mometrix.com\/academy\/wp-json\/wp\/v2\/media\/100699"}],"wp:attachment":[{"href":"https:\/\/www.mometrix.com\/academy\/wp-json\/wp\/v2\/media?parent=68801"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}