{"id":6677,"date":"2021-10-19T18:32:48","date_gmt":"2021-10-19T13:02:48","guid":{"rendered":"https:\/\/mathemerize.com\/?p=6677"},"modified":"2021-11-27T23:29:33","modified_gmt":"2021-11-27T17:59:33","slug":"solve-quadratic-equation-by-completing-the-square","status":"publish","type":"post","link":"https:\/\/mathemerize.com\/solve-quadratic-equation-by-completing-the-square\/","title":{"rendered":"Solve Quadratic Equation by Completing the Square"},"content":{"rendered":"
Here, you will learn completion of square method and how to solve quadratic equation by completing the square with examples.<\/p>\n
Let’s begin –\u00a0<\/p>\n
Let us consider the equation \\(x^2 + 8x + 4\\) = 0<\/p>\n
If we want to factorize the left hand side of the equation using the method of splitting the middle term, we must determine two integer factors of 4 whose sum is 8.<\/p>\n
But the factors of 4 are 1, 4; -1, -4; -2, -2; and 2, 2. In these cases the sum is not 8.<\/p>\n
Therefore, using factorization, we cannot solve the given equation \\(x^2 + 8x + 4\\) = 0.<\/p>\n
Here, we shall discuss a method known as completing the square to solve such quadratic equations.<\/p>\n
\nIn the method completion of square we simply add and subtract \\(({1\\over 2} coefficient of x)^2\\) in LHS.\u00a0<\/p>\n<\/blockquote>\n
Let’s understand the concept of completing the square by taking an example.<\/p>\n
Example<\/span><\/strong> : Solve the given quadratic equation \\(x^2 + 8x + 4\\) = 0 by using completion of square method.<\/p>\n
Solution<\/strong><\/span> : We have, \\(x^2 + 8x + 4\\) = 0<\/p>\n
We add and subtract \\(({1\\over 2} coefficient of x)^2\\) in LHS and get<\/p>\n
\\(x^2 + 8x + ({1\\over 2}\\times 8)^2 – ({1\\over 2}\\times 8)^2\\) + 4 = 0<\/p>\n
\\(x^2 + 8x + 16 – 16 + 4\\) = 0<\/p>\n
\\(x^2 + 2(4x) + (4)^2 – 12\\) = 0<\/p>\n
\\((x + 4)^2\\) – \\((\\sqrt{12})^2\\) = 0<\/p>\n
\\((x + 4)^2\\) = \\((\\sqrt{12})^2\\)<\/p>\n
\\(\\implies\\) x + 4 = \\(\\pm \\sqrt{12}\\)<\/p>\n
\\(\\implies\\) x = -4 \\(\\pm \\sqrt{12}\\)<\/p>\n
This gives\u00a0 x = -4 + \\(\\sqrt{12}\\)\u00a0 \u00a0or\u00a0 \u00a0 x = -4 – \\(\\sqrt{12}\\)<\/p>\n\n\n