What is the following product? (5sqrt(2) – 4sqrt(3))(5sqrt(2) – 4sqrt(3)) <\/p>\n\n\n\n
-2 <\/p>\n\n\n\n
2 <\/p>\n\n\n\n
22 – 40sqrt(6) <\/p>\n\n\n\n
98 – 40sqrt(6)<\/p>\n\n\n\n
The Correct Answer and Explanation is :<\/strong><\/mark><\/p>\n\n\n\n The correct answer is: 98 – 40\\sqrt{6}<\/strong><\/mark><\/p>\n\n\n\n Let’s simplify the expression step by step. The product you’re trying to find is:<\/p>\n\n\n\n [ This is of the form ((a – b)(a – b)), which is a special case of the expansion formula:<\/p>\n\n\n\n [ Here, (a = 5\\sqrt{2}) and (b = 4\\sqrt{3}). Applying the formula:<\/p>\n\n\n\n [ Now, let’s calculate each part:<\/p>\n\n\n\n Now, putting everything together:<\/p>\n\n\n\n [ Simplify the constants:<\/p>\n\n\n\n [ So, the expression becomes:<\/p>\n\n\n\n [ Thus, the correct answer is:<\/p>\n\n\n\n [ This type of problem involves expanding binomials with square roots. The key is recognizing the pattern ((a – b)(a – b)), which can be expanded using the difference of squares formula (a^2 – 2ab + b^2). After calculating each term, we combine the constant parts and the radical parts separately. The result is a mixed number with both a constant term (98) and a radical term involving the square root of 6. This technique is common in algebra when simplifying products of square roots or binomials.<\/p>\n","protected":false},"excerpt":{"rendered":" What is the following product? (5sqrt(2) – 4sqrt(3))(5sqrt(2) – 4sqrt(3)) -2 2 22 – 40sqrt(6) 98 – 40sqrt(6) The Correct Answer and Explanation is : The correct answer is: 98 – 40\\sqrt{6} Let’s simplify the expression step by step. The product you’re trying to find is: [(5\\sqrt{2} – 4\\sqrt{3})(5\\sqrt{2} – 4\\sqrt{3})] This is of the […]<\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"closed","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"site-sidebar-layout":"default","site-content-layout":"","ast-site-content-layout":"default","site-content-style":"default","site-sidebar-style":"default","ast-global-header-display":"","ast-banner-title-visibility":"","ast-main-header-display":"","ast-hfb-above-header-display":"","ast-hfb-below-header-display":"","ast-hfb-mobile-header-display":"","site-post-title":"","ast-breadcrumbs-content":"","ast-featured-img":"","footer-sml-layout":"","ast-disable-related-posts":"","theme-transparent-header-meta":"","adv-header-id-meta":"","stick-header-meta":"","header-above-stick-meta":"","header-main-stick-meta":"","header-below-stick-meta":"","astra-migrate-meta-layouts":"default","ast-page-background-enabled":"default","ast-page-background-meta":{"desktop":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"tablet":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"mobile":{"background-color":"","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""}},"ast-content-background-meta":{"desktop":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"tablet":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""},"mobile":{"background-color":"var(--ast-global-color-5)","background-image":"","background-repeat":"repeat","background-position":"center center","background-size":"auto","background-attachment":"scroll","background-type":"","background-media":"","overlay-type":"","overlay-color":"","overlay-opacity":"","overlay-gradient":""}},"footnotes":""},"categories":[25],"tags":[],"class_list":["post-152315","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/152315","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/comments?post=152315"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/152315\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=152315"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=152315"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=152315"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}
(5\\sqrt{2} – 4\\sqrt{3})(5\\sqrt{2} – 4\\sqrt{3})
]<\/p>\n\n\n\n
(a – b)(a – b) = a^2 – 2ab + b^2
]<\/p>\n\n\n\n
a^2 – 2ab + b^2
]<\/p>\n\n\n\n\n
[
(5\\sqrt{2})^2 = 5^2 \\times (\\sqrt{2})^2 = 25 \\times 2 = 50
]<\/li>\n\n\n\n
[
(4\\sqrt{3})^2 = 4^2 \\times (\\sqrt{3})^2 = 16 \\times 3 = 48
]<\/li>\n\n\n\n
[
-2 \\times (5\\sqrt{2}) \\times (4\\sqrt{3}) = -2 \\times 5 \\times 4 \\times \\sqrt{2} \\times \\sqrt{3} = -40 \\times \\sqrt{6}
]<\/li>\n<\/ol>\n\n\n\n
a^2 – 2ab + b^2 = 50 – 40\\sqrt{6} + 48
]<\/p>\n\n\n\n
50 + 48 = 98
]<\/p>\n\n\n\n
98 – 40\\sqrt{6}
]<\/p>\n\n\n\n
\\boxed{98 – 40\\sqrt{6}}
]<\/p>\n\n\n\nExplanation:<\/h3>\n\n\n\n