Class 10 Nature of Roots

### Topic to be coverd

♦ Nature of Roots .

### Nature of Roots

In the previous section, you have seen that the roots of the equation ax^2 + bx + c = 0 are given by

" " color(blue)(x = (-b pm sqrt(b^2-4ac))/(2a))

color(red)("Case 1 : ") If color(red)(b^2 – 4ac > 0), we get two distinct real roots color(blue)(- b/(2a) + sqrt(b^2-4ac)/(2a)) and color(blue)(-b/(2a) - sqrt(b^2-4ac)/(2a))

color(red)("Case 2 : ") If color(red)(b^2 – 4ac = 0), then color(blue)(x = -b/(2a) pm 0) i.e. x = -b/(2a) or -b/(2a)

So, the roots of the equation ax^2 + bx + c = 0 are both color(blue)((-b)/(2a))

Therefore, we say that the quadratic equation ax^2 + bx + c = 0 has two equal real roots in this case.

color(red)("Case 3 : ") If color(red)(b^2 – 4ac < 0,) then there is no real number whose square is b^2 – 4ac.

Therefore, there are no real roots for the given quadratic equation in this case.

color(green)(ul★"Discriminant of this Quadratic Equation")

Since color(red)(b^2 – 4ac) determines whether the quadratic equation ax^2 + bx + c = 0 has real roots or not, b^2 – 4ac is called color(blue)("the discriminant of this quadratic equation.")

So, a quadratic equation color(red)(ax^2 + bx + c = 0) has

color(red)((i)) two distinct real roots, if color(blue)(b^2 – 4ac > 0),

color(red)((ii)) two equal real roots, if color(blue)(b^2 – 4ac = 0),

color(red)((iii)) no real roots, if color(blue)(b^2 – 4ac < 0).
Q 3119178910

Find the discriminant of the quadratic equation 2x^2 – 4x + 3 = 0, and hence find the nature of its roots.
Class 10 Chapter 4 Example 16
Solution:

The given equation is of the form ax^2 + bx + c = 0, where a = 2, b = – 4 and c = 3. Therefore, the discriminant
b^2 – 4ac = (– 4)^2 – (4 × 2 × 3) = 16 – 24 = – 8 < 0
So, the given equation has no real roots.
Q 3139178912

A pole has to be erected at a point on the boundary of a circular park of diameter 13 metres in such a way that the differences of its distances from two diametrically opposite fixed gates A and B on the boundary is 7 metres. Is it possible to do so? If yes, at what distances from the two gates should the pole be erected?
Class 10 Chapter 4 Example 17
Solution:

Let us first draw the diagram (see Fig).

Let P be the required location of the pole. Let the distance of the pole from the gate B be x m, i.e., BP = x m. Now the difference of the distances of the pole from the two gates = AP – BP (or, BP – AP) = 7 m. Therefore, AP = (x + 7) m.

Now, AB = 13m, and since AB is a diameter,
∠APB = 90° (Why?)
Therefore, AP^2 + PB^2 = AB^2 (By Pythagoras theorem)
i.e., (x + 7)^2 + x^2 = 132
i.e., x^2 + 14x + 49 + x^2 = 169
i.e., 2x^2 + 14x – 120 = 0
So, the distance ‘x’ of the pole from gate B satisfies the equation
x^2 + 7x – 60 = 0
So, it would be possible to place the pole if this equation has real roots. To see if this
is so or not, let us consider its discriminant. The discriminant is b^2 – 4ac = 72 – 4 × 1 × (– 60) = 289 > 0.
So, the given quadratic equation has two real roots, and it is possible to erect the
pole on the boundary of the park.
Solving the quadratic equation x^2 + 7x – 60 = 0, by the quadratic formula, we get

x = (-7 pm sqrt(289))/2 = (-7 pm 17)/2

Therefore, x = 5 or – 12.
Since x is the distance between the pole and the gate B, it must be positive.
Therefore, x = – 12 will have to be ignored. So, x = 5.
Thus, the pole has to be erected on the boundary of the park at a distance of 5m
from the gate B and 12m from the gate A.
Q 3109178918

Find the discriminant of the equation 3x^2 -2x +1/3 = 0 and hence find the nature of its roots. Find them, if they are real.

Class 10 Chapter 4 Example 18
Solution:

Here a = 3, b = – 2 and c = 1/3.

Therefore, discriminant b^2 - 4ac = (-2)^2 - 4xx3 = 4-4 = 0

Hence, the given quadratic equation has two equal real roots.

The roots are  (-b)/(2a) , (-b)/(2a)

i.e. 2/6 , 2/6

i.e. 1/3 , 1/3