{"id":6673,"date":"2021-10-19T18:08:35","date_gmt":"2021-10-19T12:38:35","guid":{"rendered":"https:\/\/mathemerize.com\/?p=6673"},"modified":"2021-11-27T23:32:19","modified_gmt":"2021-11-27T18:02:19","slug":"solution-of-quadratic-equation-of-class-10","status":"publish","type":"post","link":"https:\/\/mathemerize.com\/solution-of-quadratic-equation-of-class-10\/","title":{"rendered":"Solution of Quadratic Equation of Class 10"},"content":{"rendered":"
Here you will learn how to find solution of quadratic equation of class 10 and different methods used to solve quadratic equation.<\/p>\n
Let’s begin – <\/p>\n
(a) The general form of quadratic equation is \\(ax^2 + bx + c\\) = 0, a \\(\\ne\\) 0.<\/p>\n
The roots or solution of quadratic equation can be found in following manner.<\/p>\n
\n\\(a(x^2 + {b\\over a}x + {c\\over a})\\) = 0 \\(\\implies\\) \\((x + {b\\over 2a})^2\\) + \\(c\\over a\\) – \\(b^2\\over 4a^2\\) = 0<\/p>\n
\\((x + {b\\over 2a})^2\\) = \\(b^2\\over 4a^2\\) – \\(c\\over a\\) <\/p>\n
\\(\\implies\\) x = \\(-b \\pm \\sqrt{b^2 – 4ac}\\over 2a\\)<\/p>\n<\/blockquote>\n
This \\(-b \\pm \\sqrt{b^2 – 4ac}\\over 2a\\) expression can be directly used to find the two roots of a quadratic equation. This formula is known as sridharacharya formula.<\/p>\n
(b) This expression \\(b^2 – 4a\\) = D<\/strong> is called the discriminant of the quadratic equation.<\/p>\n
Example<\/strong><\/span> : Find the roots of the quadratic equation \\(x^2 + 3x + 2\\) = 0. Also find sum of roots and product of roots.<\/p>\n
Solution<\/span><\/strong> : We have, \\(x^2 + 3x + 2\\) = 0<\/p>\n
By using the formula described above,<\/p>\n
x = \\(-3 \\pm \\sqrt{9 – 8}\\over 2\\) = \\(-3 \\pm 1\\over 2\\)<\/p>\n
\\(\\implies\\) x = -1 and -2<\/p>\n
Let \\(\\alpha\\) = -1 and \\(\\beta\\) = -2<\/p>\n
Sum of Roots = \\(\\alpha + \\beta\\) = -3<\/p>\n
Product of Roots = \\(\\alpha \\beta\\) = 2<\/p>\n
Different Methods for Solving Quadratic Equation<\/h2>\n
Following three methods are used to solve quadratic equation. These methods are discussed in detail in next sections.<\/p>\n
(1) By the method of factorisation<\/a>.<\/p>\n