Class 10 Nature of Roots

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Class 10 Chapter 4 - QUADRATIC EQUATIONS

Nature of Roots

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`♦` Nature of Roots .

`♦` 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).`

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`

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