{"id":6675,"date":"2021-10-19T18:11:02","date_gmt":"2021-10-19T12:41:02","guid":{"rendered":"https:\/\/mathemerize.com\/?p=6675"},"modified":"2021-11-27T23:30:46","modified_gmt":"2021-11-27T18:00:46","slug":"solve-quadratic-equation-by-factorisation","status":"publish","type":"post","link":"https:\/\/mathemerize.com\/solve-quadratic-equation-by-factorisation\/","title":{"rendered":"Solve Quadratic Equation by Factorisation"},"content":{"rendered":"
Here you will learn how to solve quadratic equation by factorisation with examples.<\/p>\n
Let’s begin –<\/p>\n
\nStep 1<\/strong>. Splitting of middle term :<\/p>\n
(i) If the product of a and c = +ac<\/p>\n
then we have to choose two factors ac whose sum is equal to b.<\/p>\n
(ii) If the product of a and c = -ac<\/p>\n
then we have to choose two factors of ac whose difference is equal to b.<\/p>\n
Step 2<\/strong> : Let the factors of \\(ax^2 + bx + c\\) be (dx + e) and (fx + g)<\/p>\n
\\(\\implies\\) (dx + e) (fx + g) = 0<\/p>\n
Either\u00a0 dx + e =\u00a0 or\u00a0 fx + g = 0\u00a0<\/p>\n
\\(\\implies\\)\u00a0 x = -\\(e\\over d\\)\u00a0 \u00a0or\u00a0 \u00a0x = -\\(g\\over f\\)<\/p>\n<\/blockquote>\n
Example<\/strong><\/span> : Solve the equation : \\(2x^2 – 11x + 12\\) = 0.<\/p>\n
Solution<\/span><\/strong> : We have, \\(2x^2 – 11x + 12\\) = 0<\/p>\n
\\(2x^2 – 8x – 3x + 12\\) = 0<\/p>\n
2x (x – 4) – 3 (x – 4) = 0<\/p>\n
(x – 4) (2x – 3) = 0<\/p>\n
\\(\\implies\\)\u00a0 x – 4 = 0\u00a0 \u00a0or\u00a0 \u00a02x – 4 = 0<\/p>\n
\\(\\implies\\)\u00a0 x = 4\u00a0 \u00a0or\u00a0 \u00a0x\u00a0 = \\(3\\over 2\\)<\/p>\n
Hence, x = 4 and x = \\(3\\over 2\\)\u00a0 are the roots of the given equation.<\/p>\n
Example<\/strong><\/span> : Solve the equation : \\(3x^2 – 14x – 5\\) = 0.<\/p>\n
Solution<\/span><\/strong> : We have, \\(3x^2 – 14x – 5\\) = 0<\/p>\n
\\(3x^2 – 15x + x – 5\\) = 0<\/p>\n
3x (x – 5) + 1 (x – 5) = 0<\/p>\n
(x – 5) (3x + 1) = 0<\/p>\n
\\(\\implies\\)\u00a0 x – 5 = 0\u00a0 \u00a0or\u00a0 \u00a03x + 1 = 0<\/p>\n
\\(\\implies\\)\u00a0 x = 5\u00a0 \u00a0or\u00a0 \u00a0x\u00a0 = -\\(1\\over 3\\)<\/p>\n
Hence, x = 5 and x = -\\(1\\over 3\\)\u00a0 are the roots of the given equation.<\/p>\n\n\n