{"id":3547,"date":"2013-05-08T08:49:52","date_gmt":"2013-05-08T15:49:52","guid":{"rendered":"https:\/\/magoosh.com\/gmat\/?p=3547"},"modified":"2020-01-15T10:49:48","modified_gmt":"2020-01-15T18:49:48","slug":"gmat-math-algebra-equations-with-radicals","status":"publish","type":"post","link":"https:\/\/magoosh.com\/gmat\/gmat-math-algebra-equations-with-radicals\/","title":{"rendered":"GMAT Math: Algebra Equations with Radicals"},"content":{"rendered":"<p>This is a potentially tricky <a href=\"https:\/\/magoosh.com\/gmat\/three-algebra-formulas-essential-for-the-gmat\/\" rel=\"noopener noreferrer\" target=\"_blank\">GMAT algebra<\/a> topic.\u00a0 First, consider these practice questions.<\/p>\n<p><a href=\"https:\/\/magoosh.com\/gmat\/files\/2013\/03\/ewr_img1.png\"><img decoding=\"async\" class=\"alignnone size-full wp-image-3549\" src=\"https:\/\/magoosh.com\/gmat\/files\/2013\/03\/ewr_img1.png\" alt=\"ewr_img1\" width=\"170\" height=\"38\" \/><\/a><\/p>\n<p>1. Which value(s) of x satisfies the equation above?<\/p>\n<p>I. \u20131<\/p>\n<p>II. 4<\/p>\n<p>III. 9<\/p>\n<ul>\n<li style=\"list-style-type: none\">\n<ul>(A) I<\/ul>\n<\/li>\n<\/ul>\n<p>&nbsp;<\/p>\n<ul>\n<li style=\"list-style-type: none\">\n<ul>(B) III<\/ul>\n<\/li>\n<\/ul>\n<p>&nbsp;<\/p>\n<ul>\n<li style=\"list-style-type: none\">\n<ul>(C) I &amp; II<\/ul>\n<\/li>\n<\/ul>\n<p>&nbsp;<\/p>\n<ul>\n<li style=\"list-style-type: none\">\n<ul>(D) I &amp; III<\/ul>\n<\/li>\n<\/ul>\n<p>&nbsp;<\/p>\n<ul>(E) I, II, &amp; III<\/ul>\n<p>2. What is the value of y?<\/p>\n<p><a href=\"https:\/\/magoosh.com\/gmat\/files\/2013\/03\/ewr_img2.png\"><img decoding=\"async\" class=\"alignnone size-full wp-image-3550\" src=\"https:\/\/magoosh.com\/gmat\/files\/2013\/03\/ewr_img2.png\" alt=\"ewr_img2\" width=\"340\" height=\"96\" srcset=\"https:\/\/magoosh.com\/gmat\/files\/2013\/03\/ewr_img2.png 340w, https:\/\/magoosh.com\/gmat\/files\/2013\/03\/ewr_img2-300x84.png 300w\" sizes=\"(max-width: 340px) 100vw, 340px\" \/><\/a><\/p>\n<p>A full discussion of these problems will come at the end of this article.<\/p>\n<p>&nbsp;<\/p>\n<h2>A subtle distinction<\/h2>\n<p>Consider the following, very simple algebra equation<\/p>\n<p><a href=\"https:\/\/magoosh.com\/gmat\/files\/2013\/03\/ewr_img3.png\"><img decoding=\"async\" class=\"alignnone size-full wp-image-3551\" src=\"https:\/\/magoosh.com\/gmat\/files\/2013\/03\/ewr_img3.png\" alt=\"ewr_img3\" width=\"72\" height=\"21\" \/><\/a><\/p>\n<p>Of course, this equation has two solutions, +5 and -5.\u00a0 Of course, we find that by taking the square-root of both sides, remembering this process involves a \u00b1 sign.\u00a0 Now, by contrast, consider this expression:<\/p>\n<p><a href=\"https:\/\/magoosh.com\/gmat\/files\/2013\/03\/ewr_img4.png\"><img decoding=\"async\" class=\"alignnone size-full wp-image-3552\" src=\"https:\/\/magoosh.com\/gmat\/files\/2013\/03\/ewr_img4.png\" alt=\"ewr_img4\" width=\"41\" height=\"30\" \/><\/a><\/p>\n<p>Many folks might think this is, in all respects, the same as the solution in the first process &#8212; i.e. &#8220;taking the square root.&#8221;\u00a0 We need to draw a subtle distinction here, concerning the nature of this sign:<\/p>\n<p><a href=\"https:\/\/magoosh.com\/gmat\/files\/2013\/03\/ewr_img5.png\"><img decoding=\"async\" class=\"alignnone size-full wp-image-3553\" src=\"https:\/\/magoosh.com\/gmat\/files\/2013\/03\/ewr_img5.png\" alt=\"ewr_img5\" width=\"168\" height=\"129\" \/><\/a><\/p>\n<p>The benighted unfortunately will refer to this as a &#8220;square root&#8221; sign, but that is a misleading partial name.\u00a0 The name of this symbol is the &#8220;principal square root&#8221; sign, where the word &#8220;principal&#8221;, in the sense of &#8220;main&#8221; or &#8220;primary&#8221;, here means <b><i>the positive root only<\/i><\/b>.\u00a0 Accordingly, the output of this sign is always positive.<\/p>\n<p>The number 25 has two square roots, one positive and one negative, but it has only one principal square root.\u00a0 A number can have, at most, only one principal square root.\u00a0 The principal square root of 25 is +5 only.<\/p>\n<p>When a squared algebraic expression appears in a problem, and we ourselves, in the process of problem-solving, find a square root, we need to include all roots, positive and negative (a common mistake is to forget the negative roots).\u00a0 BUT, when the symbol above, the principle square root symbol, is printed on the page as part and parcel of the given problem, this means its output will always be positive.<\/p>\n<p>Thus, the paradoxical juxtaposition<\/p>\n<p><a href=\"https:\/\/magoosh.com\/gmat\/files\/2013\/03\/ewr_img6.png\"><img decoding=\"async\" class=\"alignnone size-full wp-image-3554\" src=\"https:\/\/magoosh.com\/gmat\/files\/2013\/03\/ewr_img6.png\" alt=\"ewr_img6\" width=\"244\" height=\"77\" \/><\/a><\/p>\n<p>&nbsp;<\/p>\n<h2>Solving an equation with a radical<\/h2>\n<p>By &#8220;equation with a radical&#8221;, I will be referring to any algebraic equation in which some of the algebra is under a radical sign, a.k.a a principal square root sign.\u00a0 The general strategy for such an equation is (a) isolate the radical (i.e. get it alone by itself on one side of the equation); (b) square both sides, thus eliminating the radical; and (c) solve what remains using algebra (often, this will involve <a href=\"https:\/\/magoosh.com\/gmat\/algebra-on-the-gmat-how-to-factor\/\">factoring a quadratic<\/a>.)\u00a0 All well and good, but there&#8217;s a catch.<\/p>\n<p>You see, when you square both sides of equation, sometimes that creates solutions that weren&#8217;t part of the original equation.\u00a0 Consider the hyper-simple equation x = 5.\u00a0 This &#8220;equation&#8221; has only one solution, positive five.\u00a0 BUT, if we square both sides, then we get the equation we solved in the previous section, with solutions \u00b15.\u00a0 The extra root, x = -5, was not a solution of the original equation, but it <i>became<\/i> a solution once we squared.\u00a0 This is an example of an <b>extraneous root<\/b> &#8212;- a number that is <i><span style=\"text-decoration: underline;\">not<\/span><\/i> a root of the original equation, but which &#8220;becomes&#8221; a root when we square both sides.<\/p>\n<p>We have to square both sides to solve an equation with radicals, but doing so introduces the possibility of an extraneous root.\u00a0 Thus, an essential part of solving any equation with radicals is to check the answers you find, in order to ascertain whether any are extraneous roots.\u00a0 We verify the roots by plugging them into the original equation &#8212;- if the number does not solve the original equation, as given in the problem, then it is not a bonafide solution.<\/p>\n<p>BTW, just as a general point of strategy, regardless of whether radicals are involved, I recommend checking any algebra values you find by plugging them back into the original equation given in the problem, when possible.\u00a0 It&#8217;s just a good habit to check your work.<\/p>\n<p>Extraneous roots play a role in both of the sample problems above.\u00a0 Having read this post, you may want to go back and give them another attempt before reading the solutions below.\u00a0 Questions?\u00a0 Let us know in the comment section at the bottom.<\/p>\n<p>&nbsp;<\/p>\n<h2>Solutions to the sample questions<\/h2>\n<p>1) The radical is already isolated, so square both sides.<\/p>\n<p><a href=\"https:\/\/magoosh.com\/gmat\/files\/2013\/03\/ewr_img7.png\"><img decoding=\"async\" class=\"alignnone size-full wp-image-3555\" src=\"https:\/\/magoosh.com\/gmat\/files\/2013\/03\/ewr_img7.png\" alt=\"ewr_img7\" width=\"188\" height=\"110\" \/><\/a><\/p>\n<p>(x \u2013 9)(x + 1) = 0<\/p>\n<p>preliminary solutions: x = {+9, \u20131}<\/p>\n<p>At this point, an unsuspecting student might be tempted to answer <b>(C)<\/b>, the trap answer.\u00a0 BUT, the problem is: one or both of these answers could be extraneous.\u00a0 We need to check each by plugging back into the original expression.<\/p>\n<p><span style=\"text-decoration: underline;\"><br \/>\nCheck<\/span> x = \u20131<\/p>\n<p><a href=\"https:\/\/magoosh.com\/gmat\/files\/2013\/03\/ewr_img8.png\"><img decoding=\"async\" class=\"alignnone size-full wp-image-3556\" src=\"https:\/\/magoosh.com\/gmat\/files\/2013\/03\/ewr_img8.png\" alt=\"ewr_img8\" width=\"289\" height=\"78\" \/><\/a><\/p>\n<p>This answer does not check &#8212; the left &amp; right sides have different values. \u00a0Thus, x = \u20131 is an extraneous root, not a solution to the problem.<\/p>\n<p>(NB: it&#8217;s often the case that an extraneous root will make the two sides equal to values equal in absolute value and opposite in sign.)<\/p>\n<p><span style=\"text-decoration: underline;\"><br \/>\nCheck<\/span> x = 9<\/p>\n<p><a href=\"https:\/\/magoosh.com\/gmat\/files\/2013\/03\/ewr_img9.png\"><img decoding=\"async\" class=\"alignnone size-full wp-image-3557\" src=\"https:\/\/magoosh.com\/gmat\/files\/2013\/03\/ewr_img9.png\" alt=\"ewr_img9\" width=\"315\" height=\"75\" srcset=\"https:\/\/magoosh.com\/gmat\/files\/2013\/03\/ewr_img9.png 315w, https:\/\/magoosh.com\/gmat\/files\/2013\/03\/ewr_img9-300x71.png 300w\" sizes=\"(max-width: 315px) 100vw, 315px\" \/><\/a><\/p>\n<p>The value x = 9 checks &#8212; it makes the two sides equal, and thus satisfies the original equation.\u00a0 This is the only solution, so only option III contains a root.<\/p>\n<p>Answer = <b>(B)<\/b><\/p>\n<p>2) Using a tried and true <a href=\"https:\/\/magoosh.com\/gmat\/gmat-data-sufficiency-tips\/\">DS strategy<\/a>, start with the easier statement, Statement #2.<\/p>\n<p><a href=\"https:\/\/magoosh.com\/gmat\/files\/2013\/03\/ewr_img10a.png\"><img decoding=\"async\" class=\"alignnone size-full wp-image-3562\" src=\"https:\/\/magoosh.com\/gmat\/files\/2013\/03\/ewr_img10a.png\" alt=\"ewr_img10a\" width=\"268\" height=\"22\" \/><\/a><\/p>\n<p>(y \u2013 4)(y + 2) = 0<\/p>\n<p>y = +4 or y = \u20132<\/p>\n<p>Since there are two values of y, this statement, alone and by itself, is <b>not sufficient<\/b>.<\/p>\n<p><a href=\"https:\/\/magoosh.com\/gmat\/files\/2013\/03\/ewr_img10.png\"><img decoding=\"async\" class=\"alignnone size-full wp-image-3558\" src=\"https:\/\/magoosh.com\/gmat\/files\/2013\/03\/ewr_img10.png\" alt=\"ewr_img10\" width=\"335\" height=\"38\" srcset=\"https:\/\/magoosh.com\/gmat\/files\/2013\/03\/ewr_img10.png 335w, https:\/\/magoosh.com\/gmat\/files\/2013\/03\/ewr_img10-300x34.png 300w\" sizes=\"(max-width: 335px) 100vw, 335px\" \/><\/a><\/p>\n<p>The radical is already isolated, so square both sides.<\/p>\n<p><a href=\"https:\/\/magoosh.com\/gmat\/files\/2013\/03\/ewr_img11.png\"><img decoding=\"async\" class=\"alignnone size-full wp-image-3559\" src=\"https:\/\/magoosh.com\/gmat\/files\/2013\/03\/ewr_img11.png\" alt=\"ewr_img11\" width=\"210\" height=\"114\" \/><\/a><\/p>\n<p>Lo and behold!\u00a0 We have arrived at the same equation we found in Statement #2, with solutions y = +4 or y = \u20132.\u00a0 The na\u00efve conclusion would be &#8212; this statement says exactly the same thing as the other.\u00a0 That&#8217;s incorrect, though, because we don&#8217;t know whether both of these values are valid solutions, or whether one or more is an extraneous root.\u00a0 We need to test this in the original equation.<\/p>\n<p><span style=\"text-decoration: underline;\">Test<\/span> y = +4<\/p>\n<p><a href=\"https:\/\/magoosh.com\/gmat\/files\/2013\/03\/ewr_img12.png\"><img decoding=\"async\" class=\"alignnone size-full wp-image-3560\" src=\"https:\/\/magoosh.com\/gmat\/files\/2013\/03\/ewr_img12.png\" alt=\"ewr_img12\" width=\"353\" height=\"76\" srcset=\"https:\/\/magoosh.com\/gmat\/files\/2013\/03\/ewr_img12.png 353w, https:\/\/magoosh.com\/gmat\/files\/2013\/03\/ewr_img12-300x64.png 300w\" sizes=\"(max-width: 353px) 100vw, 353px\" \/><\/a><\/p>\n<p>This value checks &#8212; y = +4 is a valid solution to the equation<\/p>\n<p><span style=\"text-decoration: underline;\">Test<\/span> y = \u20132<\/p>\n<p><a href=\"https:\/\/magoosh.com\/gmat\/files\/2013\/03\/ewr_img13.png\"><img decoding=\"async\" class=\"alignnone size-full wp-image-3561\" src=\"https:\/\/magoosh.com\/gmat\/files\/2013\/03\/ewr_img13.png\" alt=\"ewr_img13\" width=\"378\" height=\"78\" srcset=\"https:\/\/magoosh.com\/gmat\/files\/2013\/03\/ewr_img13.png 378w, https:\/\/magoosh.com\/gmat\/files\/2013\/03\/ewr_img13-300x61.png 300w\" sizes=\"(max-width: 378px) 100vw, 378px\" \/><\/a><\/p>\n<p>The two sides are not equal, so this does not check!\u00a0 This value, y = \u20132, is an extraneous root.<\/p>\n<p>Thus, the equation given in Statement #1 has only one solution, y = 4, so this equation provides a definitive answer to the prompt question.\u00a0 This statement, alone and by itself, is <b>sufficient<\/b>.<\/p>\n<p>Answer = <b>(A)<\/b><\/p>\n<p>&nbsp;<\/p>\n","protected":false},"excerpt":{"rendered":"<p>This is a potentially tricky GMAT algebra topic.\u00a0 First, consider these practice questions. 1. Which value(s) of x satisfies the equation above? I. \u20131 II. 4 III. 9 (A) I &nbsp; (B) III &nbsp; (C) I &amp; II &nbsp; (D) I &amp; III &nbsp; (E) I, II, &amp; III 2. What is the value of [&hellip;]<\/p>\n","protected":false},"author":26,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[138],"tags":[],"ppma_author":[13209],"class_list":["post-3547","post","type-post","status-publish","format-standard","hentry","category-algebra"],"acf":[],"yoast_head":"<!-- This site is optimized with the Yoast SEO Premium plugin v21.7 (Yoast SEO v21.7) - https:\/\/yoast.com\/wordpress\/plugins\/seo\/ -->\n<title>GMAT Math: Algebra Equations with Radicals | Magoosh Study Resources<\/title>\n<meta name=\"description\" content=\"GMAT expert Mike McGarry discusses algebra equations with radicals on the GMAT.\" \/>\n<meta name=\"robots\" content=\"index, follow, max-snippet:-1, max-image-preview:large, max-video-preview:-1\" \/>\n<link rel=\"canonical\" href=\"https:\/\/magoosh.com\/gmat\/gmat-math-algebra-equations-with-radicals\/\" \/>\n<meta property=\"og:locale\" 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He was also featured as \"member of the month\" for over two years at <a href=\"https:\/\/gmatclub.com\/blog\/2012\/09\/mike-mcgarrys-gmat-experience\/\" rel=\"noopener noreferrer\">GMAT Club<\/a>. Mike holds an A.B. in Physics (graduating <em>magna cum laude<\/em>) and an M.T.S. in Religions of the World, both from Harvard. Beyond standardized testing, Mike has over 20 years of both private and public high school teaching experience specializing in math and physics. In his free time, Mike likes smashing foosballs into orbit, and despite having no obvious cranial deficiency, he insists on rooting for the NY Mets. 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