{"id":6679,"date":"2021-10-19T19:19:15","date_gmt":"2021-10-19T13:49:15","guid":{"rendered":"https:\/\/mathemerize.com\/?p=6679"},"modified":"2021-11-27T23:26:48","modified_gmt":"2021-11-27T17:56:48","slug":"solve-quadratic-equation-using-quadratic-formula","status":"publish","type":"post","link":"https:\/\/mathemerize.com\/solve-quadratic-equation-using-quadratic-formula\/","title":{"rendered":"Solve Quadratic Equation Using Quadratic Formula"},"content":{"rendered":"

Here you will learn how to solve quadratic equation using quadratic formula or sridharacharya formula with examples.<\/p>\n

Let’s begin – <\/p>\n

Solve Quadratic Equation Using Quadratic Formula (Sridharacharya Formula)<\/h2>\n

For the equation \\(ax^2 + bx + c\\) = 0, if \\(b^2 – 4ac\\) \\(\\ge\\) 0, then<\/p>\n

\n

x = (\\(-b + \\sqrt{b^2 -4ac}\\over 2a\\))  and  x = (\\(-b – \\sqrt{b^2 -4ac}\\over 2a\\))<\/p>\n<\/blockquote>\n

This formula was given by indian mathematician sridharacharya. Therefore , it is also called sridharacharya formula<\/strong>.<\/p>\n

If \\(b^2 – 4ac\\) < 0, i.e. negative, then \\(\\sqrt{b^2 -4ac}\\) is not real and therefore, the equation does not have any real roots.<\/p>\n

Therefore, if \\(b^2 – 4ac\\) \\(\\ge\\) 0, then the quadratic equation \\(ax^2 + bx + c\\) = 0 has two roots \\(\\alpha\\) and \\(\\beta\\), given by <\/p>\n

\n

\\(\\alpha\\) = (\\(-b + \\sqrt{b^2 -4ac}\\over 2a\\))  and  \\(\\beta\\) = (\\(-b – \\sqrt{b^2 -4ac}\\over 2a\\))<\/p>\n<\/blockquote>\n

Example<\/span><\/strong> : Solve the following quadratic equation using quadratic formula.<\/p>\n

(i) \\(6x^2 + 7x – 10\\) = 0<\/p>\n

(ii) \\(15x^2 – 28\\) = x<\/p>\n

Solution<\/span><\/strong> : <\/p>\n

(i) We have, \\(6x^2 + 7x – 10\\) = 0<\/p>\n

Here, a = 6, b = 7, c = -10<\/p>\n

\\(\\therefore\\)    D = \\(b^2 – 4ac\\) = \\((7)^2 – 4\\times 6 \\times (-10)\\)<\/p>\n

D = 49 + 240 = 289 > 0<\/p>\n

So, the given equation has real roots, given by<\/p>\n

x =  \\(-b + \\sqrt{b^2 -4ac}\\over 2a\\) = \\(-7 + \\sqrt{289}\\over 12\\) = \\(10\\over 12\\) = \\(5\\over 6\\)<\/p>\n

or,   x = \\(-b – \\sqrt{b^2 -4ac}\\over 2a\\) = \\(-7 – \\sqrt{289}\\over 12\\) = \\(24\\over 12\\) = -2<\/p>\n

Therefore, the roots of the given equations are \\(5\\over 6\\) and -2.<\/p>\n

(ii) We have, \\(15x^2 – x – 28\\) = 0<\/p>\n

Here, a = 15, b = -1, c = -28<\/p>\n

\\(\\therefore\\)    D = \\(b^2 – 4ac\\) = \\((-1)^2 – 4\\times 15 \\times (-28)\\)<\/p>\n

D = 1 + 1680 = 1681 > 0<\/p>\n

So, the given equation has real roots, given by<\/p>\n

x =  \\(-b + \\sqrt{b^2 -4ac}\\over 2a\\) = \\(-(-1) + \\sqrt{1681}\\over 30\\) = \\(42\\over 30\\) = \\(7\\over 5\\)<\/p>\n

or,   x = \\(-b – \\sqrt{b^2 -4ac}\\over 2a\\) = \\(-(-1) – \\sqrt{1681}\\over 30\\) = \\(40\\over 30\\) = -\\(4\\over 3\\)<\/p>\n

Therefore, the roots of the given equations are \\(7\\over 5\\) and -\\(4\\over 3\\).<\/p>\n\n\n

\n
Next – Nature of Roots of Quadratic Equation<\/a><\/div>\n<\/div>\n\n\n\n
\n
Previous – Solve Quadratic Equation by Completing the Square<\/a><\/div>\n<\/div>\n","protected":false},"excerpt":{"rendered":"

Here you will learn how to solve quadratic equation using quadratic formula or sridharacharya formula with examples. Let’s begin –  Solve Quadratic Equation Using Quadratic Formula (Sridharacharya Formula) For the equation \\(ax^2 + bx + c\\) = 0, if \\(b^2 – 4ac\\) \\(\\ge\\) 0, then x = (\\(-b + \\sqrt{b^2 -4ac}\\over 2a\\))  and  x = …<\/p>\n

Solve Quadratic Equation Using Quadratic Formula<\/span> Read More »<\/a><\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"site-sidebar-layout":"default","site-content-layout":"default","ast-global-header-display":"","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":"","theme-transparent-header-meta":"default","adv-header-id-meta":"","stick-header-meta":"","header-above-stick-meta":"","header-main-stick-meta":"","header-below-stick-meta":""},"categories":[42],"tags":[560,563,564,565],"yoast_head":"\nSolve Quadratic Equation Using Quadratic Formula - Mathemerize<\/title>\n<meta name=\"description\" content=\"In this post you will learn how to solve quadratic equation using quadratic formula or sridharacharya formula with examples.\" \/>\n<meta name=\"robots\" content=\"index, follow, max-snippet:-1, max-image-preview:large, max-video-preview:-1\" \/>\n<link rel=\"canonical\" href=\"https:\/\/mathemerize.com\/solve-quadratic-equation-using-quadratic-formula\/\" \/>\n<meta property=\"og:locale\" content=\"en_US\" \/>\n<meta property=\"og:type\" content=\"article\" \/>\n<meta property=\"og:title\" content=\"Solve Quadratic Equation Using Quadratic Formula - 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