{"id":37165,"date":"2025-06-25T14:03:46","date_gmt":"2025-06-25T14:03:46","guid":{"rendered":"https:\/\/gaviki.com\/blog\/?p=37165"},"modified":"2025-06-25T14:03:48","modified_gmt":"2025-06-25T14:03:48","slug":"if-the-positive-square-root-of-sqrt90-sqrt80-is-multiplied-by-sqrt2-1-and-the-product-is-raised-to-the-power-of-four-the-result-would-be-1-100-2-1600-3-11520000-4-10","status":"publish","type":"post","link":"https:\/\/gaviki.com\/blog\/if-the-positive-square-root-of-sqrt90-sqrt80-is-multiplied-by-sqrt2-1-and-the-product-is-raised-to-the-power-of-four-the-result-would-be-1-100-2-1600-3-11520000-4-10\/","title":{"rendered":"If the positive square root of (sqrt(90) + sqrt(80)) is multiplied by (sqrt(2) &#8211; 1) and the product is raised to the power of four, the result would be: 1. 100 2. 1600 3. 11520000 4. 10"},"content":{"rendered":"\n<p>If the positive square root of (sqrt(90) + sqrt(80)) is multiplied by (sqrt(2) &#8211; 1) and the product is raised to the power of four, the result would be: 1. 100 2. 1600 3. 11520000 4. 10<\/p>\n\n\n\n<p><strong><mark style=\"background-color:rgba(0, 0, 0, 0)\" class=\"has-inline-color has-vivid-cyan-blue-color\">The Correct Answer and Explanation is:<\/mark><\/strong><\/p>\n\n\n\n<p>We are given the expression:<\/p>\n\n\n\n<blockquote class=\"wp-block-quote is-layout-flow wp-block-quote-is-layout-flow\">\n<p>(90+80\u22c5(2\u22121))4(\\sqrt{\\sqrt{90} + \\sqrt{80}} \\cdot (\\sqrt{2} &#8211; 1))^4(90\u200b+80\u200b\u200b\u22c5(2\u200b\u22121))4<\/p>\n<\/blockquote>\n\n\n\n<p>We will simplify this step by step.<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">Step 1: Simplify the inner square roots<\/h3>\n\n\n\n<p>We begin with:90+80\\sqrt{90} + \\sqrt{80}90\u200b+80\u200b<\/p>\n\n\n\n<p>Break these into their simplified radical form:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>90=9\u22c510=310\\sqrt{90} = \\sqrt{9 \\cdot 10} = 3\\sqrt{10}90\u200b=9\u22c510\u200b=310\u200b<\/li>\n\n\n\n<li>80=16\u22c55=45\\sqrt{80} = \\sqrt{16 \\cdot 5} = 4\\sqrt{5}80\u200b=16\u22c55\u200b=45\u200b<\/li>\n<\/ul>\n\n\n\n<p>So we get:90+80=310+45\\sqrt{90} + \\sqrt{80} = 3\\sqrt{10} + 4\\sqrt{5}90\u200b+80\u200b=310\u200b+45\u200b<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">Step 2: Take the square root of the sum<\/h3>\n\n\n\n<p>We now compute:310+45\\sqrt{3\\sqrt{10} + 4\\sqrt{5}}310\u200b+45\u200b\u200b<\/p>\n\n\n\n<p>This is complicated to simplify exactly, so let\u2019s approximate using decimals:<\/p>\n\n\n\n<ul class=\"wp-block-list\">\n<li>10\u22483.162\\sqrt{10} \\approx 3.16210\u200b\u22483.162, so 310\u22483\u22c53.162=9.4863\\sqrt{10} \\approx 3 \\cdot 3.162 = 9.486310\u200b\u22483\u22c53.162=9.486<\/li>\n\n\n\n<li>5\u22482.236\\sqrt{5} \\approx 2.2365\u200b\u22482.236, so 45\u22484\u22c52.236=8.9444\\sqrt{5} \\approx 4 \\cdot 2.236 = 8.94445\u200b\u22484\u22c52.236=8.944<\/li>\n<\/ul>\n\n\n\n<p>Now add:310+45\u22489.486+8.944=18.433\\sqrt{10} + 4\\sqrt{5} \\approx 9.486 + 8.944 = 18.43310\u200b+45\u200b\u22489.486+8.944=18.43<\/p>\n\n\n\n<p>Then take the square root:18.43\u22484.29\\sqrt{18.43} \\approx 4.2918.43\u200b\u22484.29<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">Step 3: Multiply by 2\u22121\\sqrt{2} &#8211; 12\u200b\u22121<\/h3>\n\n\n\n<p>2\u22481.414\\sqrt{2} \\approx 1.4142\u200b\u22481.414, so 2\u22121\u22480.414\\sqrt{2} &#8211; 1 \\approx 0.4142\u200b\u22121\u22480.414<\/p>\n\n\n\n<p>Now multiply:4.29\u22c50.414\u22481.7764.29 \\cdot 0.414 \\approx 1.7764.29\u22c50.414\u22481.776<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">Step 4: Raise the result to the power of 4<\/h3>\n\n\n\n<p>(1.776)4\u2248(1.7762)2=(3.154)2\u22489.95(1.776)^4 \\approx (1.776^2)^2 = (3.154)^2 \\approx 9.95(1.776)4\u2248(1.7762)2=(3.154)2\u22489.95<\/p>\n\n\n\n<p>Rounded to the nearest whole number, the answer is:10\\boxed{10}10\u200b<\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">Final Answer:<\/h3>\n\n\n\n<p><strong>Option 4: 10<\/strong><\/p>\n\n\n\n<hr class=\"wp-block-separator has-alpha-channel-opacity\"\/>\n\n\n\n<h3 class=\"wp-block-heading\">Explanation:<\/h3>\n\n\n\n<p>The problem requires simplifying a compound expression involving nested radicals and exponentiation. First, we simplify the square roots of 90 and 80 using their prime factors. These simplify to 3103\\sqrt{10}310\u200b and 454\\sqrt{5}45\u200b respectively. Their sum does not simplify further using exact radicals, so we approximate to a decimal. Taking the square root of this sum, we get an estimate around 4.29. Multiplying by 2\u22121\\sqrt{2} &#8211; 12\u200b\u22121 (approximately 0.414) gives roughly 1.776. Finally, raising this value to the 4th power gives close to 10. This shows how estimation using known square root approximations can help evaluate seemingly complicated expressions.<\/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-banner8-863.jpeg\" alt=\"\" class=\"wp-image-37226\" srcset=\"https:\/\/gaviki.com\/blog\/wp-content\/uploads\/2025\/06\/learnexams-banner8-863.jpeg 852w, https:\/\/gaviki.com\/blog\/wp-content\/uploads\/2025\/06\/learnexams-banner8-863-250x300.jpeg 250w, https:\/\/gaviki.com\/blog\/wp-content\/uploads\/2025\/06\/learnexams-banner8-863-768x923.jpeg 768w\" sizes=\"auto, (max-width: 852px) 100vw, 852px\" \/><\/figure>\n","protected":false},"excerpt":{"rendered":"<p>If the positive square root of (sqrt(90) + sqrt(80)) is multiplied by (sqrt(2) &#8211; 1) and the product is raised to the power of four, the result would be: 1. 100 2. 1600 3. 11520000 4. 10 The Correct Answer and Explanation is: We are given the expression: (90+80\u22c5(2\u22121))4(\\sqrt{\\sqrt{90} + \\sqrt{80}} \\cdot (\\sqrt{2} &#8211; 1))^4(90\u200b+80\u200b\u200b\u22c5(2\u200b\u22121))4 [&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-37165","post","type-post","status-publish","format-standard","hentry","category-quiz-questions"],"_links":{"self":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/37165","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=37165"}],"version-history":[{"count":1,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/37165\/revisions"}],"predecessor-version":[{"id":37230,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/posts\/37165\/revisions\/37230"}],"wp:attachment":[{"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/media?parent=37165"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/categories?post=37165"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/gaviki.com\/blog\/wp-json\/wp\/v2\/tags?post=37165"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}