{"id":6681,"date":"2021-10-19T19:41:51","date_gmt":"2021-10-19T14:11:51","guid":{"rendered":"https:\/\/mathemerize.com\/?p=6681"},"modified":"2021-11-27T23:25:22","modified_gmt":"2021-11-27T17:55:22","slug":"nature-of-roots-of-quadratic-equation","status":"publish","type":"post","link":"https:\/\/mathemerize.com\/nature-of-roots-of-quadratic-equation\/","title":{"rendered":"Nature of Roots of Quadratic Equation"},"content":{"rendered":"
Here you will learn how to find the nature of roots of quadratic equation using discriminant with examples.<\/p>\n
Let’s begin –<\/p>\n
(a) Consider the quadratic equation \\(ax^2 + bx + c\\) = 0 where a, b, c \\(\\in\\) R & a \\(\\ne\\) 0 , then<\/p>\n
x = \\(-b \\pm \\sqrt{D}\\over 2a\\) where D = \\(b^2 – 4ac\\)<\/p>\n
So, a quadratic equation \\(ax^2 + bx + c\\) = 0 <\/p>\n
\n(i) has no real roots if D < 0.<\/p>\n
(ii) has two equal real roots if D = 0.<\/p>\n
(iii) has two distinct real roots if D > 0.<\/p>\n
(iv) has real roots if D \\(\\ge\\) 0.<\/p>\n
(v) If p + iq is one of the root of a quadratic equation, then the other root must be the conjugate p – iq & vice versa. (p, q \\(\\in\\) R & i = \\(\\sqrt{-1}\\)).<\/p>\n<\/blockquote>\n
(b) Consider the quadratic equation \\(ax^2 + bx + c\\) = 0 where a, b, c \\(\\in\\) Q & a \\(\\in\\) 0, then :<\/p>\n
\n(i) If D is a perfect square, then roots are rational.<\/p>\n
(ii) if \\(\\alpha\\) = p + \\(\\sqrt{q}\\) is one root in this case, (where p is rational & \\(\\sqrt{q}\\) is a surd) then other root will be p – \\(\\sqrt{q}\\).<\/p>\n<\/blockquote>\n
Example<\/strong><\/span> : Examine, whether the following equations have real roots :<\/p>\n
(i) \\(x^2 + x + 1\\) = 0<\/p>\n
(ii) \\(3x^2 + 2x – 1\\) = 0<\/p>\n
Solution<\/span><\/strong> : <\/p>\n
(i) We have, \\(x^2 + x + 1\\) = 0<\/p>\n
Here, a = 1, b = 1, c = 1<\/p>\n
Therefore, Discriminant D = \\(b^2 – 4ac\\) = \\((1)^2 – 4(1)(1)\\) = -3 < 0.<\/p>\n
Since, D < 0, the equation has no real roots.<\/p>\n
(ii) We have, \\(3x^2 + 2x – 1\\) = 0<\/p>\n
Here, a = 3, b = 2, c = -1<\/p>\n
Therefore, Discriminant D = \\(b^2 – 4ac\\) = \\((2)^2 – 4(3)(-1)\\) = 16 > 0.<\/p>\n
Since, D > 0, the equation has real roots.<\/p>\n\n\n