{"id":63846,"date":"2020-12-10T17:51:38","date_gmt":"2020-12-10T17:51:38","guid":{"rendered":"https:\/\/www.mometrix.com\/academy\/?page_id=63846"},"modified":"2026-03-28T11:39:57","modified_gmt":"2026-03-28T16:39:57","slug":"solving-equations-involving-roots","status":"publish","type":"page","link":"https:\/\/www.mometrix.com\/academy\/solving-equations-involving-roots\/","title":{"rendered":"Solving Equations Involving Roots"},"content":{"rendered":"\n\t\t\t<div id=\"mmDeferVideoEncompass_3DvUR2DkYEU\" 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_3DvUR2DkYEU\" data-source-videoID=\"3DvUR2DkYEU\" src=\"https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2023\/01\/circle-play-duotone.png\" alt=\"Solving Equations Involving Roots Video\" height=\"464\" width=\"825\" class=\"size-full\" data-matomo-title = \"Solving Equations Involving Roots\">\n\t\t\t<\/picture>\n\t\t\t<\/div>\n\t\t\t<style>img#videoThumbnailImage_3DvUR2DkYEU:hover {cursor:pointer;} img#videoThumbnailImage_3DvUR2DkYEU {background-size:contain;background-image:url(\"https:\/\/www.mometrix.com\/academy\/wp-content\/uploads\/2023\/02\/1709-solving-equations-involving-roots-3.webp\");}<\/style>\n\t\t\t<script defer>\n\t\t\t  jQuery(\"img#videoThumbnailImage_3DvUR2DkYEU\").click(function() {\n\t\t\t\tlet videoId = jQuery(this).attr(\"data-source-videoID\");\n\t\t\t\tlet helpTag = '<div id=\"mmDeferVideoYTMessage_3DvUR2DkYEU\" 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 Equations Involving Roots\");\n\t\t\t\tjQuery(\"div#mmDeferVideoEncompass_3DvUR2DkYEU\").html(tag);\n\t\t\t\tjQuery(\"div#mmDeferVideoEncompass_3DvUR2DkYEU\").prepend(helpTag);\n\t\t\t\tsetTimeout(function(){jQuery(\"div#mmDeferVideoYTMessage_3DvUR2DkYEU\").css(\"display\", \"block\");}, 2000);\n\t\t\t  });\n\t\t\t  \n\t\t\t<\/script>\n\t\t\n<p><p><script>\nfunction Guc_Function() {\n  var x = document.getElementById(\"Guc\");\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=\"Guc_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=\"Guc\" style=\"display:none;\">\n<ul>\n<li class=\"toc-h2\"><a href=\"#What_is_a_Root\" class=\"smooth-scroll\">What is a Root?<\/a><\/li>\n<li class=\"toc-h2\"><a href=\"#Solving_Equations_with_Roots\" class=\"smooth-scroll\">Solving Equations with Roots<\/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<br \/>\n<\/a><\/li>\n<\/ul>\n<\/li>\n<li class=\"toc-h2\"><a href=\"#Solving_Equations_Involving_Roots_Practice_Questions\" class=\"smooth-scroll\">Solving Equations Involving 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 equations involving roots!<\/p>\n<h2><span id=\"What_is_a_Root\" class=\"m-toc-anchor\"><\/span>What is a Root?<\/h2>\n<p>\nBefore we dive in, let\u2019s review <a class=\"ylist\" href=\"https:\/\/www.mometrix.com\/academy\/roots\/\">what a root is<\/a>.<\/p>\n<p>The most common root you will see is a square root. The symbol for a square root is called a radical, and it looks like this: \\(\\sqrt{  }\\). A square root asks the question, what number multiplied by itself (or squared) will give me the number under the radical?<\/p>\n<p>Other roots work in a similar way. When you have a cube root, fourth root, fifth root, etc. it asks the question, what number multiplied by itself three, four, five, times will give me the number under the radical? The way you know how many times to multiply the number by itself is given by a small number put in the hook of the radical symbol. If there is no number there, it is assumed to be a two, or a square root. <\/p>\n<p>This is an example of the fourth root of 16. The fourth root of 16, which you can tell is the fourth root by this little 4 in the hook. Is equal to the fourth root of 2 times 2 times 2 times 2. Since there are four two&#8217;s we can pull those out because we are looking for the fourth root. And our answer is 2.<\/p>\n<p>Now that we\u2019ve gone over what roots are, let\u2019s learn how to solve equations using them by looking at an example:<\/p>\n<h2><span id=\"Solving_Equations_with_Roots\" class=\"m-toc-anchor\"><\/span>Solving Equations with Roots<\/h2>\n<h3><span id=\"Example_1\" class=\"m-toc-anchor\"><\/span>Example #1<\/h3>\n<p>\nSolve the following equation for \\(x\\):<\/p>\n<div class=\"examplesentence\">\\(\\sqrt{x-3}=4\\)<\/div>\n<p>\n&nbsp;<br \/>\nThe first thing we need to do to get x by itself is to get rid of the square root. We do this by doing the opposite operation to both sides of the equation. The opposite of a square root is a square, so we are going to square both sides.<\/p>\n<p>This gives us \\(x-3=16\\).<\/p>\n<p>Now it looks like a regular equation, and we know that all we need to do to get \\(x\\) by itself is add three to both sides.<\/p>\n<p>This gives us our final answer: \\(x=19\\).<\/p>\n<p>That wasn\u2019t too bad! Now let\u2019s look at one with a root other than a two. <\/p>\n<h3><span id=\"Example_2\" class=\"m-toc-anchor\"><\/span>Example #2<\/h3>\n<p>\nLet&#8217;s try this one:<\/p>\n<div class=\"examplesentence\">\\(\\sqrt[5]{x+7}=2\\)<\/div>\n<p>\n&nbsp;<br \/>\nIn order to get rid of our root, we have to do the opposite operation, which is to put both sides to a power. Since we have a fifth root, we want to put both sides to the fifth power.<\/p>\n<p>This gives us \\(x+7=32\\).<\/p>\n<p>Subtracting 7 from both sides gives us our final answer, which is \\(x=25\\). <\/p>\n<h3><span id=\"Example_3\" class=\"m-toc-anchor\"><\/span>Example #3<\/h3>\n<p>\nWhat if the root isn\u2019t the only thing on the left side of our equation? If that\u2019s the case, then we need to follow the <a class=\"ylist\" href=\"https:\/\/www.mometrix.com\/academy\/order-of-operations\/\">order of operations<\/a> backward to get rid of everything until the root is by itself.<\/p>\n<p>Consider the following example:<\/p>\n<div class=\"examplesentence\">\\(4\\sqrt[3]{x-9}+6=18\\)<\/div>\n<p>\n&nbsp;<br \/>\nIf we go by the backward order of operations, we first have to get rid of all our addition and subtraction, so the first thing we do is subtract 6 from both sides.<\/p>\n<div class=\"examplesentence\">\\(4\\sqrt[3]{x-9}=12\\)<\/div>\n<p>\n&nbsp;<br \/>\nThen we want to divide both sides by 4 to get the root by itself.<\/p>\n<div class=\"examplesentence\">\\(\\sqrt[3]{x-9}=3\\)<\/div>\n<p>\n&nbsp;<br \/>\nNow we undo the root by cubing both sides.<\/p>\n<p>This gives us \\(x-9=27\\).<\/p>\n<p>Finally, we add 9 to both sides to get our answer: \\(x=36\\).<\/p>\n<h3>Example #4<\/h4>\n<p>\nI want to do one more example. This time, try to figure out the answer on your own, then check to see if it matches up with mine.<\/p>\n<div class=\"examplesentence\">\\(3\\sqrt{x+8}-12=3\\)<\/div>\n<p>\n&nbsp;<br \/>\nThink you\u2019ve got it? Let\u2019s check!<\/p>\n<p>First, we want to add 12 to both sides.<\/p>\n<div class=\"examplesentence\">\\(3\\sqrt{x+8}=15\\)<\/div>\n<p>\n&nbsp;<br \/>\nThen, we divide both sides by 3.<\/p>\n<div class=\"examplesentence\">\\(\\sqrt{x+8}=15\\)<\/div>\n<p>\n&nbsp;<br \/>\nNow, we are going to square both sides.<\/p>\n<p>This gives us \\(x+8=25\\).<\/p>\n<p>And finally, we are going to subtract 8 from both sides to get our final answer: \\(x=17\\).<\/p>\n<p>I hope this review of solving equations involving 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_Equations_Involving_Roots_Practice_Questions\" class=\"m-toc-anchor\"><\/span>Solving Equations Involving 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 \/>\nFind the value of \\(x\\) in the equation \\(\\sqrt{x+2}=5\\). <\/p>\n<\/div>\n\t\t\t\t\t<div class=\"PQ-Choices\"><div class=\"PQ\"  id=\"PQ-1-1\">\\(x=5\\)<\/div><div class=\"PQ\"  id=\"PQ-1-2\">\\(x=25\\)<\/div><div class=\"PQ\"  id=\"PQ-1-3\">\\(x=3\\)<\/div><div class=\"PQ correct_answer\"  id=\"PQ-1-4\">\\(x=23\\)<\/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>To solve for \\(x\\), perform inverse operations to isolate the variable. Since the opposite of a square root is the exponent 2, raise both sides of the equation to the second power. <\/p>\n<p style=\"text-align: center\">\\((\\sqrt{x+2})^2=5^2\\)<\/p>\n<p>The square root of \\(x+2\\) raised to the second power equals \\(x+2\\), and 5<sup>2<\/sup> is 25. <\/p>\n<p style=\"text-align: center\">\\(x+2=25\\)<\/p>\n<p>Next, isolate the variable by performing the inverse operation of adding 2. Since the opposite of adding 2 is subtracting 2, subtract 2 from both sides of the equation. <\/p>\n<p style=\"text-align: center\">\\(x+2-2=25-2\\)<\/p>\n<p>Since \\(2-2=0\\), we are left with \\(x\\) on the left side of the equation. <\/p>\n<p style=\"text-align: center; line-height: 35px\">\\(25-3=23\\)<br \/>\n\\(x=23\\)<\/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 \/>\nFind the value of \\(x\\) in the equation \\(\\sqrt[\\Large{4}]{x-8}=3\\).<\/p>\n<\/div>\n\t\t\t\t\t<div class=\"PQ-Choices\"><div class=\"PQ\"  id=\"PQ-2-1\">\\(x=3\\)<\/div><div class=\"PQ\"  id=\"PQ-2-2\">\\(x=11\\)<\/div><div class=\"PQ\"  id=\"PQ-2-3\">\\(x=81\\)<\/div><div class=\"PQ correct_answer\"  id=\"PQ-2-4\">\\(x=89\\)<\/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>To solve for \\(x\\), perform inverse operations to isolate the variable. Since the opposite of a fourth root is the exponent 4, raise both sides of the equation to the fourth power.<\/p>\n<p style=\"text-align: center\">\\((\\sqrt[\\Large{4}]{x-8})^4=3^4\\)<\/p>\n<p>The fourth root of \\(x-8\\) raised to the fourth power equals \\(x-8\\). Raising 3 to the fourth power equals \\(3\\times3\\times3\\times3\\), which equals 81. <\/p>\n<p style=\"text-align: center\">\\(x-8=81\\)<\/p>\n<p>Next, isolate the variable by performing the inverse operation of subtracting 8. Since the opposite of subtracting 8 is adding 8, add 8 to both sides of the equation.<\/p>\n<p style=\"text-align: center\">\\(x-8+8=81+8\\)<\/p>\n<p>Since \\(-8 +8 =0\\), we are left with \\(x\\) on the left side of the equation.<\/p>\n<p style=\"text-align: center; line-height:\" 35px\">\\(81 +8 =89\\)<br \/>\n\\(x =89\\)<\/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 \/>\nFind the value of \\(x\\) in the equation \\(2+\\sqrt[\\Large{3}]{8x}=6\\). <\/p>\n<\/div>\n\t\t\t\t\t<div class=\"PQ-Choices\"><div class=\"PQ\"  id=\"PQ-3-1\">\\(x=\\large{\\frac{1}{2}}\\)<\/div><div class=\"PQ correct_answer\"  id=\"PQ-3-2\">\\(x=8\\)<\/div><div class=\"PQ\"  id=\"PQ-3-3\">\\(x=16\\)<\/div><div class=\"PQ\"  id=\"PQ-3-4\">\\(x=64\\)<\/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>To solve for \\(x\\), perform inverse operations to isolate the variable. Since the opposite of adding 2 is subtracting 2, subtract 2 from both sides of the equation.<\/p>\n<p style=\"text-align: center\">\\(2+\\sqrt[\\Large{3}]{8x}-2=6-2\\)<\/p>\n<p>Since \\(2 -2 =0\\), we are left with \\(\\sqrt[\\Large{3}]{8x}\\) on the left side of the equation. Since \\(6 -2 =4\\), write 4 on the right side of the equation.<\/p>\n<p style=\"text-align: center\">\\(\\sqrt[\\Large{3}]{8x}=4\\)<\/p>\n<p>Next, do inverse operations again. Since the opposite of a cube root is the exponent 3, raise both sides of the equation to the third power.<\/p>\n<p style=\"text-align: center\">\\((\\sqrt[\\Large{3}]{8x})^3=4^3\\)<\/p>\n<p>The cube root of \\(8x\\) raised to the third power equals \\(8x\\). Raising 4 to the third power equals \\(4\\times4\\times4\\), which equals 64. <\/p>\n<p style=\"text-align: center\">\\(8x=64\\)<\/p>\n<p>Next, isolate the variable by performing the inverse operation of multiplying by 8. Since the opposite of multiplying by 8 is dividing by 8, divide both sides of the equation by 8. <\/p>\n<p style=\"text-align: center\">\\(\\dfrac{8x}{8}=\\dfrac{64}{8}\\)<\/p>\n<p>We know that \\(8 \\div 8 =1\\), leaving \\(x\\) by itself on the left side of the equation. <\/p>\n<p style=\"text-align: center; line-height: 35px\">\\(64 \\div 8 = 8\\)<br \/>\n\\(x=8\\)<\/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 \/>\nFind the value of \\(x\\) in the equation \\(\\dfrac{(\\sqrt[\\Large{3}]{x+100})-2}{3}=1\\)<\/p>\n<\/div>\n\t\t\t\t\t<div class=\"PQ-Choices\"><div class=\"PQ\"  id=\"PQ-4-1\">\\(x=5\\)<\/div><div class=\"PQ correct_answer\"  id=\"PQ-4-2\">\\(x=25\\)<\/div><div class=\"PQ\"  id=\"PQ-4-3\">\\(x=50\\)<\/div><div class=\"PQ\"  id=\"PQ-4-4\">\\(x=125\\)<\/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>To solve for \\(x\\), perform inverse operations to isolate the variable. Since the opposite of dividing by 3 is multiplying by 3, multiply both sides of the equation by 3.<\/p>\n<p style=\"text-align: center\">\\(\\dfrac{(\\sqrt[\\Large{3}]{x+100})-2}{3}\\times3=1\\times3\\)<\/p>\n<p>Dividing by 3 and multiplying by 3 cancel each other out, so we are left with \\((\\sqrt[\\Large{3}]{x+100})-2\\) on the left side of the equation. Since 1 times 3 equals 3, write 3 on the right side of the equation.<\/p>\n<p style=\"text-align: center\">\\((\\sqrt[\\Large{3}]{x+100})-2=3\\)<\/p>\n<p>Next, do the inverse operation of subtracting 2. Since the opposite of subtracting 2 is adding 2, add 2 to both sides of the equation.<\/p>\n<p style=\"text-align: center\">\\((\\sqrt[\\Large{3}]{x+100})-2+2=3+2\\)<\/p>\n<p>Since \\(-2 +2 =0\\), we are left with \\(\\sqrt[3]{x+100}\\) on the left side of the equation. Since \\(3 +2 =5\\), write 5 on the right side of the equation.<\/p>\n<p style=\"text-align: center\">\\(\\sqrt[\\Large{3}]{x+100}=5\\)<\/p>\n<p>Next, do inverse operations again. Since the opposite of a cube root is the exponent 3, raise both sides of the equation to the third power.<\/p>\n<p style=\"text-align: center\">\\((\\sqrt[\\Large{3}]{x+100})^3=5^3\\)<\/p>\n<p>The cube root of \\(x+100\\) raised to the third power equals \\(x+100\\). Raising 5 to the third power equals \\(5\\times5\\times5\\), which equals 125. <\/p>\n<p style=\"text-align: center\">\\(x+100=125\\)<\/p>\n<p>Next, isolate the variable by performing the inverse operation of adding 100. The opposite of adding 100 is subtracting 100, so subtract 100 from both sides of the equation.<\/p>\n<p style=\"text-align: center\">\\(x+100-100=125-100\\)<\/p>\n<p>We know that \\(100 -100 =0\\), so we are left with \\(x\\) on the left side of the equation.<\/p>\n<p style=\"text-align: center; line-height: 35px\">\\(125 -100 =25\\)<br \/>\n\\(x=25\\)<\/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 \/>\nFind the value of \\(x\\) in the equation \\(6+2\\times \\sqrt[\\Large{4}]{16x}-8=2\\). <\/p>\n<\/div>\n\t\t\t\t\t<div class=\"PQ-Choices\"><div class=\"PQ correct_answer\"  id=\"PQ-5-1\">\\(x=1\\)<\/div><div class=\"PQ\"  id=\"PQ-5-2\">\\(x=2\\)<\/div><div class=\"PQ\"  id=\"PQ-5-3\">\\(x=8\\)<\/div><div class=\"PQ\"  id=\"PQ-5-4\">\\(x=16\\)<\/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>To solve for \\(x\\), perform inverse operations to isolate the variable. Since the opposite of subtracting 8 is adding 8, add 8 to both sides of the equation.<\/p>\n<p style=\"text-align: center\">\\(6+2\\times \\sqrt[\\Large{4}]{16x}-8+8=2+8\\)<\/p>\n<p>Since \\(-8 +8 =0\\), the eights on the left side of the equation cancel out. We know that \\(2 +8 =10\\), so write 10 on the right side of the equation.<\/p>\n<p style=\"text-align: center\">\\(6+2\\times \\sqrt[\\Large{4}]{16x}=10\\)<\/p>\n<p>Next, do the inverse operation of adding 6. Since the opposite of adding 6 is subtracting 6, subtract 6 from both sides of the equation.<\/p>\n<p style=\"text-align: center\">\\(6-6+2\\times \\sqrt[\\Large{4}]{16x}=10-6\\)<\/p>\n<p>Since \\(6 -6 =0\\), the sixes on the left side of the equation cancel out. We know that \\(10 -6 =4\\), so write 4 on the right side of the equation.<\/p>\n<p style=\"text-align: center\">\\(2\\times \\sqrt[\\Large{4}]{16x}=4\\)<\/p>\n<p>Next, do the inverse operation of multiplying by 2. Since the opposite of multiplying by 2 is dividing by 2, divide both sides of the equation by 2.<\/p>\n<p style=\"text-align: center\">\\(\\dfrac{2\\times \\sqrt[\\Large{4}]{16x}}{2}=\\frac{4}{2}\\)<\/p>\n<p>Since \\(2 \\div 2 =1\\), we are left with \\(\\sqrt[\\Large{4}]{16x}\\) on the left side of the equation. We know that \\(4 \\div 2 =2\\), so write 2 on the right side of the equation.<\/p>\n<p style=\"text-align: center\">\\(\\sqrt[\\Large{4}]{16x}=2\\)<\/p>\n<p>Next, do inverse operations again. Since the opposite of a fourth root is the exponent 4, raise both sides of the equation to the fourth power.<\/p>\n<p style=\"text-align: center\">\\((\\sqrt[\\Large{4}]{16x})^4=2^4\\)<\/p>\n<p>The fourth root of \\(16x\\) raised to the fourth power is \\(16x\\). Raising 2 to the fourth power is \\(2\\times2\\times2\\times2\\), which equals 16.<\/p>\n<p style=\"text-align: center\">\\(16x=16\\)<\/p>\n<p>Next, isolate the variable by doing the inverse operation of multiplying by 16. Since the opposite of multiplying by 16 is dividing by 16, divide both sides of the equation by 16.<\/p>\n<p style=\"text-align: center\">\\(\\dfrac{16x}{16}=\\dfrac{16}{16}\\)<\/p>\n<p>We know that \\(16 \\div 16 =1\\), so we are left with \\(x\\) on the left side of the equation. <\/p>\n<p style=\"text-align: center; line-height: 35px\">\\(16 \\div 16 =1\\)<br \/>\n\\(x=1\\)<\/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<br \/>\n&nbsp;<\/p>\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>&nbsp; Return to Algebra I Videos<\/p>\n","protected":false},"author":1,"featured_media":100693,"parent":0,"menu_order":0,"comment_status":"closed","ping_status":"closed","template":"","meta":{"footnotes":""},"class_list":{"0":"post-63846","1":"page","2":"type-page","3":"status-publish","4":"has-post-thumbnail","6":"page_category-algebra-i-videos","7":"page_category-manipulating-expressions-1","8":"page_category-math-advertising-group","9":"page_category-solving-equations-miscellaneous-videos","10":"page_type-video","11":"content_type-practice-questions","12":"subject_matter-math"},"aioseo_notices":[],"_links":{"self":[{"href":"https:\/\/www.mometrix.com\/academy\/wp-json\/wp\/v2\/pages\/63846","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=63846"}],"version-history":[{"count":5,"href":"https:\/\/www.mometrix.com\/academy\/wp-json\/wp\/v2\/pages\/63846\/revisions"}],"predecessor-version":[{"id":285427,"href":"https:\/\/www.mometrix.com\/academy\/wp-json\/wp\/v2\/pages\/63846\/revisions\/285427"}],"wp:featuredmedia":[{"embeddable":true,"href":"https:\/\/www.mometrix.com\/academy\/wp-json\/wp\/v2\/media\/100693"}],"wp:attachment":[{"href":"https:\/\/www.mometrix.com\/academy\/wp-json\/wp\/v2\/media?parent=63846"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}