{"id":24710,"date":"2025-06-18T17:29:28","date_gmt":"2025-06-18T17:29:28","guid":{"rendered":"https:\/\/gaviki.com\/blog\/?p=24710"},"modified":"2025-06-18T17:29:30","modified_gmt":"2025-06-18T17:29:30","slug":"show-all-work-to-multiply-quantity-2-plus-the-square-root-of-negative-25-end-quantity-times-quantity-4-minus-the-square-root-of-negative-100-end-quantity","status":"publish","type":"post","link":"https:\/\/gaviki.com\/blog\/show-all-work-to-multiply-quantity-2-plus-the-square-root-of-negative-25-end-quantity-times-quantity-4-minus-the-square-root-of-negative-100-end-quantity\/","title":{"rendered":"Show all work to multiply quantity 2 plus the square root of negative 25 end quantity times quantity 4 minus the square root of negative 100 end quantity"},"content":{"rendered":"\n<p>Show all work to multiply quantity 2 plus the square root of negative 25 end quantity times quantity 4 minus the square root of negative 100 end quantity<\/p>\n\n\n\n<p><strong><mark style=\"background-color:rgba(0, 0, 0, 0)\" class=\"has-inline-color has-vivid-green-cyan-color\">The Correct Answer and Explanation is:<\/mark><\/strong><\/p>\n\n\n\n<p>We are asked to multiply:(2+\u221225)(4\u2212\u2212100)(2 + \\sqrt{-25})(4 &#8211; \\sqrt{-100})(2+\u221225\u200b)(4\u2212\u2212100\u200b)<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\"><strong>Step 1: Simplify the square roots of negative numbers<\/strong><\/h3>\n\n\n\n<p>We use the imaginary unit <strong>i<\/strong>, where i=\u22121i = \\sqrt{-1}i=\u22121\u200b.<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>\u221225=25\u22c5\u22121=5i\\sqrt{-25} = \\sqrt{25} \\cdot \\sqrt{-1} = 5i\u221225\u200b=25\u200b\u22c5\u22121\u200b=5i<\/li>\n\n\n\n<li>\u2212100=100\u22c5\u22121=10i\\sqrt{-100} = \\sqrt{100} \\cdot \\sqrt{-1} = 10i\u2212100\u200b=100\u200b\u22c5\u22121\u200b=10i<\/li>\n<\/ul>\n\n\n\n<p>So the expression becomes:(2+5i)(4\u221210i)(2 + 5i)(4 &#8211; 10i)(2+5i)(4\u221210i)<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\"><strong>Step 2: Use the distributive property (FOIL)<\/strong><\/h3>\n\n\n\n<p>We apply the FOIL method:<\/p>\n\n\n\n<p><strong>First:<\/strong><br>2\u22c54=82 \\cdot 4 = 82\u22c54=8<\/p>\n\n\n\n<p><strong>Outer:<\/strong><br>2\u22c5(\u221210i)=\u221220i2 \\cdot (-10i) = -20i2\u22c5(\u221210i)=\u221220i<\/p>\n\n\n\n<p><strong>Inner:<\/strong><br>5i\u22c54=20i5i \\cdot 4 = 20i5i\u22c54=20i<\/p>\n\n\n\n<p><strong>Last:<\/strong><br>5i\u22c5(\u221210i)=\u221250i25i \\cdot (-10i) = -50i^25i\u22c5(\u221210i)=\u221250i2<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\"><strong>Step 3: Combine the results<\/strong><\/h3>\n\n\n\n<p>Now add them:8\u221220i+20i\u221250i28 &#8211; 20i + 20i &#8211; 50i^28\u221220i+20i\u221250i2<\/p>\n\n\n\n<p>Note that the imaginary terms cancel:\u221220i+20i=0-20i + 20i = 0\u221220i+20i=0<\/p>\n\n\n\n<p>We are left with:8\u221250i28 &#8211; 50i^28\u221250i2<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\"><strong>Step 4: Simplify i2i^2i2<\/strong><\/h3>\n\n\n\n<p>We know:i2=\u22121i^2 = -1i2=\u22121<\/p>\n\n\n\n<p>So:\u221250i2=\u221250(\u22121)=50-50i^2 = -50(-1) = 50\u221250i2=\u221250(\u22121)=50<\/p>\n\n\n\n<p>Now add:8+50=588 + 50 = 588+50=58<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\"><strong>Final Answer:<\/strong><\/h3>\n\n\n\n<p>58\\boxed{58}58\u200b<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\"><strong>Explanation <\/strong><\/h3>\n\n\n\n<p>In this problem, we are multiplying two complex-number-like expressions that involve square roots of negative numbers. Square roots of negative numbers do not exist in the real number system, but they are defined in the complex number system using the imaginary unit <strong>i<\/strong>, where i=\u22121i = \\sqrt{-1}i=\u22121\u200b.<\/p>\n\n\n\n<p>The first step is to simplify the square roots. For example, \u221225\\sqrt{-25}\u221225\u200b becomes 5i5i5i because 25=5\\sqrt{25} = 525\u200b=5 and \u22121=i\\sqrt{-1} = i\u22121\u200b=i. Similarly, \u2212100=10i\\sqrt{-100} = 10i\u2212100\u200b=10i. Replacing these in the original expression transforms it into (2+5i)(4\u221210i)(2 + 5i)(4 &#8211; 10i)(2+5i)(4\u221210i).<\/p>\n\n\n\n<p>To multiply two binomials, we use the distributive property, commonly known as the FOIL method. This method stands for First, Outer, Inner, and Last terms. Multiplying each of these term pairs gives us the components that make up the expanded expression.<\/p>\n\n\n\n<p>Adding the results, we notice that the imaginary parts cancel out. This leaves us with two real number components: 8 and \u221250i2-50i^2\u221250i2. Remembering that i2=\u22121i^2 = -1i2=\u22121, we substitute and find that \u221250i2-50i^2\u221250i2 becomes 50. Adding 8 and 50 gives the final result of 58.<\/p>\n\n\n\n<p>This shows how imaginary numbers can interact algebraically to produce real results. The key is understanding how to handle iii and applying basic algebra rules carefully.<\/p>\n\n\n\n<figure class=\"wp-block-image size-full\"><img loading=\"lazy\" decoding=\"async\" width=\"852\" height=\"1024\" src=\"https:\/\/gaviki.com\/blog\/wp-content\/uploads\/2025\/06\/learnexams-banner8-11.jpeg\" alt=\"\" class=\"wp-image-24715\" srcset=\"https:\/\/gaviki.com\/blog\/wp-content\/uploads\/2025\/06\/learnexams-banner8-11.jpeg 852w, https:\/\/gaviki.com\/blog\/wp-content\/uploads\/2025\/06\/learnexams-banner8-11-250x300.jpeg 250w, https:\/\/gaviki.com\/blog\/wp-content\/uploads\/2025\/06\/learnexams-banner8-11-768x923.jpeg 768w\" sizes=\"auto, (max-width: 852px) 100vw, 852px\" \/><\/figure>\n","protected":false},"excerpt":{"rendered":"<p>Show all work to multiply quantity 2 plus the square root of negative 25 end quantity times quantity 4 minus the square root of negative 100 end quantity The Correct Answer and Explanation is: We are asked to multiply:(2+\u221225)(4\u2212\u2212100)(2 + \\sqrt{-25})(4 &#8211; \\sqrt{-100})(2+\u221225\u200b)(4\u2212\u2212100\u200b) Step 1: Simplify the square roots of negative numbers We use the [&hellip;]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"open","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[1],"tags":[],"class_list":["post-24710","post","type-post","status-publish","format-standard","hentry","category-quiz-questions"],"_links":{"self":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/24710","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/comments?post=24710"}],"version-history":[{"count":1,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/24710\/revisions"}],"predecessor-version":[{"id":24716,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/24710\/revisions\/24716"}],"wp:attachment":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/media?parent=24710"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/categories?post=24710"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/tags?post=24710"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}