The solution to x2 \u2013 10x = 24 is
\u2714 12 or \u20132<\/p>\n\n\n\n
The solution to 2×2 \u2013 11 = 87 is
\u2714 7 or \u20137<\/p>\n\n\n\n
The solution to 3×2 \u2013 12x + 24 = 0 is
\u2714 2 + 2i or 2 \u2013 2i<\/p>\n\n\n\n
The Correct Answer and Explanation is:<\/mark><\/strong><\/p>\n\n\n\n Let’s solve the equations one by one and provide the detailed explanation.<\/p>\n\n\n\n Rearrange the equation to standard quadratic form: Substitute these values into the quadratic formula: Rearrange the equation: We can solve this quadratic equation using the quadratic formula<\/strong>: Substitute these values into the quadratic formula: These solutions are derived through solving each quadratic equation either by factoring, completing the square, or using the quadratic formula.<\/p>\n","protected":false},"excerpt":{"rendered":" The solution to x2 \u2013 10x = 24 is\u2714 12 or \u20132 The solution to 2×2 \u2013 11 = 87 is\u2714 7 or \u20137 The solution to 3×2 \u2013 12x + 24 = 0 is\u2714 2 + 2i or 2 \u2013 2i The Correct Answer and Explanation is: Let’s solve the equations one by one […]<\/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-164533","post","type-post","status-publish","format-standard","hentry","category-exams-certification"],"_links":{"self":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/164533","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=164533"}],"version-history":[{"count":0,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/posts\/164533\/revisions"}],"wp:attachment":[{"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/media?parent=164533"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/categories?post=164533"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/www.learnexams.com\/blog\/wp-json\/wp\/v2\/tags?post=164533"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}
\n\n\n\n1. Solution to (x^2 – 10x = 24)<\/h3>\n\n\n\n
[
x^2 – 10x – 24 = 0
]
Now, we can solve this quadratic equation using the quadratic formula<\/strong>:
[
x = \\frac{-b \\pm \\sqrt{b^2 – 4ac}}{2a}
]
For the equation (x^2 – 10x – 24 = 0), the coefficients are (a = 1), (b = -10), and (c = -24).<\/p>\n\n\n\n
[
x = \\frac{-(-10) \\pm \\sqrt{(-10)^2 – 4(1)(-24)}}{2(1)}
]
[
x = \\frac{10 \\pm \\sqrt{100 + 96}}{2}
]
[
x = \\frac{10 \\pm \\sqrt{196}}{2}
]
[
x = \\frac{10 \\pm 14}{2}
]
Thus, the two possible solutions are:
[
x = \\frac{10 + 14}{2} = \\frac{24}{2} = 12
]
and
[
x = \\frac{10 – 14}{2} = \\frac{-4}{2} = -2
]
So, the solutions are 12<\/strong> and -2<\/strong>.<\/p>\n\n\n\n
\n\n\n\n2. Solution to (2x^2 – 11 = 87)<\/h3>\n\n\n\n
[
2x^2 – 11 = 87 \\quad \\Rightarrow \\quad 2x^2 = 87 + 11 \\quad \\Rightarrow \\quad 2x^2 = 98
]
Now, divide both sides by 2:
[
x^2 = \\frac{98}{2} = 49
]
Taking the square root of both sides:
[
x = \\pm \\sqrt{49} = \\pm 7
]
So, the solutions are 7<\/strong> and -7<\/strong>.<\/p>\n\n\n\n
\n\n\n\n3. Solution to (3x^2 – 12x + 24 = 0)<\/h3>\n\n\n\n
[
x = \\frac{-b \\pm \\sqrt{b^2 – 4ac}}{2a}
]
For the equation (3x^2 – 12x + 24 = 0), the coefficients are (a = 3), (b = -12), and (c = 24).<\/p>\n\n\n\n
[
x = \\frac{-(-12) \\pm \\sqrt{(-12)^2 – 4(3)(24)}}{2(3)}
]
[
x = \\frac{12 \\pm \\sqrt{144 – 288}}{6}
]
[
x = \\frac{12 \\pm \\sqrt{-144}}{6}
]
Since the discriminant ((\\sqrt{-144})) is negative, the solutions will be complex numbers. We can express the square root of (-144) as:
[
\\sqrt{-144} = 12i
]
Thus, the solutions are:
[
x = \\frac{12 \\pm 12i}{6}
]
Simplifying:
[
x = \\frac{12}{6} \\pm \\frac{12i}{6}
]
[
x = 2 \\pm 2i
]
So, the solutions are (2 + 2i)<\/strong> and (2 – 2i)<\/strong>.<\/p>\n\n\n\n
\n\n\n\nSummary of Solutions:<\/h3>\n\n\n\n
\n