# Nature of Roots

(1) Equation ax^{2 }+ bx + c = 0; a Â¹0 will have real roots only when D = b^{2 }- 4ac > 0.

(2) When D = b^{2 }- 4ac<0, then the equation ax^{2 }+ bx + c = 0 will have no real roots.

(3) When D = b^{2 }- 4ac = 0, then the roots will be real and equal, each being equal to .

**Sum and Product of the Roots of a Quadratic Equation**

Let there be a quadratic equation ax^{2}+bx+c=0, whose discriminant, D=b^{2}-4ac â‰¥ 0.

Then its two real roots Î± and Î² will be written as

Î±= and Î² =

Therefore, the sum of the roots of the equation,

Î± + Î² =

=

and the product of the roots will be written as,

Î± Î² =

=

Therefore, the sum of the roots = and product of the roots = .

Find the discriminant of the following equations:

(i) 2x^{2 }+ 5x - 6 = 0 (ii) 7x^{2 }- 2x + 6 = 0

(iii) (iv) 3x^{2 }- 5x + 2 = 0

(v) 6x^{2 }- 7cx + 8 = 0

(i) The given equation is 2x^{2 }+ 5x - 6 = 0

Here a = 2, b = 5 and c = -6

Therefore, discriminant D = b^{2 }- 4ac

=5^{2 }- 4(2)(-6) = 73

(ii) The given equation is

7x^{2}-2x+6=0 in which

a=7, b=-2, c=6

D=b^{2}-4ac=(-2)^{2}-(4)(7)(6)=-164

(iii)

a=

D = b^{2}-4ac =

Substitute âˆš3 = 1.732

=192+76.2 = 268.2

(iv) 3x^{2 }- 5x + 2 = 0

Here a = 3, b = -5, c = 2

D = b^{2 }- 4ac = (-5)^{2 }- (4)(3)(2) = 1

(v) 6x^{2 }- 7cx + 8 = 0

Here a = 6, b = -7c, c = 8

D = b^{2 }- 4ac = (-7c)^{2 }- (4)(6)(8)

= 49c^{2 }- 192.

Comment on the nature of the roots of the following equations:

(i) 3x^{2 }+ 5x + 1 = 0 (ii) 4x^{2 }- 2x + 3 = 0

(iii) 7x^{2 }- 13x + 5 = 0 (iv)

(v) x^{2 }+ 6x + 9 = 0

(i) The given equation is

3x^{2 }+ 5x + 1 = 0

Here a = 3, b = 5, c = 1

Discriminant D = b^{2 }- 4ac

= (5)^{2 }- (4)(3)(1) = 13

Here, D = 13>0. Hence the roots will be real and unequal.

(ii) For the equation 4x^{2 }- 2x + 3 = 0

a = 4, b = -2, c = 3

D = b^{2 }- 4ac = (-2)^{2 }- (4)(4)(3) = -44

Here D = -44 < 0

Therefore there are no real roots.

(iii) 7x^{2 }- 13x + 5 = 0

a = 7, b = -13, c = 5

D = b^{2 }- 4ac = (13)^{2 }- (4)(7)(5) = 29

Here D>0. Therefore unequal and real roots will exist.

(iv)

a =

D = b^{2 }- 4ac = (-

=7 + 36 = 43>0

.^{.}. Unequal and real roots exists.

(v) x^{2 }+ 6x + 9 = 0

a = 1,b = 6,c = 9

D = b^{2 }- 4ac = (6)^{2 }-(4)(1)(9) = 0

Therefore real and equal roots.

Find the value of p, for which the following equations have two equal roots.

(i)

(ii) 4x^{2 }- 8x + p = 0

(i) The given equation is

Here,

For real and equal roots, D = 0

âˆ´

where p is any real number.

(ii) 4x^{2 }- 8x + p = 0

a = 4, b = -8, c = p

D = b^{2 }- 4ac, i.e. D = (-8)^{2 }- 4.4. p

=64 - 16p

For real and equal roots, D = 0

âˆ´ 64 - 16p = 0 or p = 4.

where p is any real number.

In the following, determine whether the given quadratic equations have real roots and if so, find the roots.

(i) x^{2 }+ 7x + 10 = 0 (ii) 4x^{2 }- 3x - 10 = 0

(iii) 8x^{2 }+ 8x + 1 = 0 (iv) 2x^{2 }+ 4x + 1 = 0

(v)

(i) The given equation is x^{2 }+ 7x + 10 = 0

Here, a = 1, b = 7, c = 10

D = b^{2 }- 4ac=7^{2 }-(4)(1)(10) = 9 >0

As D>0, therefore real roots are possible.

Again, x^{2 }+ 7x + 10 = 0

x^{2} + 5x + 2x +10 = 0

x(x + 5) + 2(x + 5) = 0

(x + 5)(x + 2) = 0

Therefore, x = -5 and -2.

(ii) The given equation is

4x^{2 }- 3x - 10 = 0

and a = 4, b = -3, c = -10

D = b^{2 }-4ac =(-3)^{2 }- (4)(4)(-10)

= 9 + 160 =169

As D>0, therefore real roots will exist. The two real roots of the equation will be written as,

Î± = Î² =

Î± = Î² =

Î± = 2 Î² =

Therefore, the roots are 2 and

(iii) 8x^{2 }+ 8x + 1=0

a = 8, b = 8, c = 1

D = b^{2 }- 4ac = (8)^{2 }- (4)(8)(1) = 32

As D>0, therefore the equation will have real roots given by,

Î± =

Î² =

(iv) 2x^{2 }+ 4x + 1 = 0

a = 2, b = 4, c = 1

D = b^{2 }- 4ac = (4)^{2 }- (4)(2)(1) = 8.

As D > 0, so the equation will have real roots given by,

Î±=

Î²=

(v)

a = , b = -3, c = +

D = (-3)^{2 }- 4. . = 9 - 20 = -11

As D < 0, no real roots are possible.

Prove that, both the roots of the equation (x - p) (x - q) + (x - q) (x - r)+ (x - r) (x - p) = 0 are real, but they are equal only when p = q = r.

The given equation is

(x - p) (x - q) + (x - q) (x - r) + ( x - r) (x - p) = 0

3x^{2 }- 2x(p + q + r) + (pq + qr + rp)=0

D = b^{2 }- 4ac

= [-2(p + q + r)]^{2 }- [4 Ã— 3 (pq + qr + rp)]

= 4 (p + q + r)^{2 }- 12(pq + qr + rp)

= 4 (p^{2} + q^{2} + r^{2} - pq - qr - rp)

= 2 (2p^{2} + 2q^{2} + 2r^{2} - 2pq - 2qr - 2rp)

= 2(p^{2 }- 2pq + q^{2 }+ q^{2 }- 2qr + r^{2} + r^{2 }- 2pr + p^{2})

= 2 [(p - q)^{2} + (q - r)^{2} + (r - p)^{2}] â‰¥ 0

Therefore, the roots of the equation are real. For equal roots of a quadratic equation D must be equal to zero.

Hence, 2[(p - q)^{2} + (q - r)^{2} + (r - p)^{2}] = 0

This is possible only when,

p - q = 0 or p = q

and q - r = 0 or q = r

and r - p = 0 or r = p

i.e. p = q = r

Therefore, the roots of the given equation are equal, only when p = q = r.