- Nature of Roots Based on Discriminant
- two distinct real roots, two equal real roots, no real roots
- Solutions of Quadratic Equations by Using Quadratic Formula and Nature of Roots
Two distinct real roots if b2 – 4ac > 0
Two equal real roots if b2 – 4ac = 0
No real roots if b2 – 4ac < 0
have seen that the roots of the equation ax2 + bx + c = 0 are given by
If b2 – 4ac > 0, we get two distinct real roots, `-b/(2a)+(b^2-4ac)/(2a)` and `-d/(2a)-(sqrt(b^2-4ac))/(2a)`
If b2 – 4ac = 0,then x= `-b/(2a)+-0` i.e., `x=-b/(2a) or -b/(2a)`
So, the roots of the equation ax2 + bx + c = 0 are both `-b/(2a)`
Therefore, we say that the quadratic equation ax2 + bx + c = 0 has two equal real roots in this case.
If b2 – 4ac < 0, then there is no real number whose square is b2 – 4ac. Therefore, there are no real roots for the given quadratic equation in this case.
Since b2 – 4ac determines whether the quadratic equation ax2 + bx + c = 0 has real roots or not, b2 – 4ac is called the discriminant of this quadratic equation.
So, a quadratic equation ax2 + bx + c = 0 has
(i) two distinct real roots, if b2 – 4ac > 0,
(ii) two equal real roots, if b2 – 4ac = 0,
(iii) no real roots, if b2 – 4ac < 0.
Let us consider one examples.
Example : Find the discriminant of the quadratic equation 2x2 – 4x + 3 = 0, and hence find the nature of its roots.
Solution : The given equation is of the form ax2 + bx + c = 0, where a = 2, b = – 4 and c = 3.
Therefore, the discriminant
b2 – 4ac = (– 4)2 – (4 × 2 × 3) = 16 – 24 = – 8 < 0
So, the given equation has no real roots.
The 4th term of an A.P. is 22 and the 15th term is 66. Find the first terns and the common
difference. Hence find the sum of the series to 8 terms.
If -5 is a root of the quadratic equation 2x2 + px – 15 = 0 and the quadratic equation p(x2 + x)k = 0 has equal roots, find the value of k.
Solve for x using the quadratic formula. Write your answer corrected to two significant figures. (x - 1)2 - 3x + 4 = 0