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