{"id":6671,"date":"2021-10-19T16:27:02","date_gmt":"2021-10-19T10:57:02","guid":{"rendered":"https:\/\/mathemerize.com\/?p=6671"},"modified":"2021-11-27T23:33:25","modified_gmt":"2021-11-27T18:03:25","slug":"roots-of-quadratic-equation","status":"publish","type":"post","link":"https:\/\/mathemerize.com\/roots-of-quadratic-equation\/","title":{"rendered":"Roots of Quadratic Equation"},"content":{"rendered":"
Here you will learn what are the roots of quadratic equation with examples.<\/p>\n
Let’s begin – <\/p>\n
If p(x) = 0 is a quadratic equation, then the zeroes of the poynomial p(x) are called the roots of the quadratic equation p(x) = 0.<\/p>\n
Thus, x = 1 is a root of p(x) = \\(x^2 + x – 2\\) = 0<\/p>\n
Since p(1) = \\((1)^2\\) + 1 – 2 = 0<\/p>\n
and, x = -2 is also a root of \\(x^2 + x – 2\\) = 0.<\/p>\n
Since p(-2) = \\((-2)^2\\) + (-2) – 2 = 4 – 2 – 2 = 0.<\/p>\n
Note<\/strong> : <\/p>\n 1). x = \\(\\alpha\\) is a root of p(x) = 0, iff \\(p(\\alpha)\\) = 0.<\/p>\n 2). Every quadratic equation can have atmost two real roots.<\/p>\n 3). Finding the zeroes of the quadratic equations is known as solving the quadratic equation.<\/p>\n Example<\/strong><\/span> : In each of the following quadratic equations the values of x are given. Determine whether these values are roots of the equations or not.<\/p>\n (i) \\(2x^2 + 3x + 1\\) = 0; x = -1<\/p>\n (ii) \\(4x^2 – 5x – 6\\) = 0; x = 2<\/p>\n Solution<\/span><\/strong> : <\/p>\n (i) Putting x = -1, on the left side of the given equation, we get<\/p>\n L.H.S = \\(2\\times (-1)^2 + 3\\times (-1) + 1\\) = 2 – 3 + 1 = 0 = R.H.S<\/p>\n So, x = -1 is a root of the given equation.<\/p>\n (ii) Putting x = 2, on the left side of the given equation, we get<\/p>\n L.H.S = \\(4\\times (2)^2 – 5\\times (2) – 6\\) = 16 – 10 – 6 = 0 = R.H.S<\/p>\n So, x = 2 is a root of the given equation.<\/p>\n\n\n