# FIND THE MISSING VALUE OF QUADRATIC EQUATION USING NATURE OF ROOTS

Nature of the roots of the Equation ax2 + bx + c = 0

The two roots of the Quadratic equation ax2 + bx + c = 0 are:

b2 – 4ac > 0

## Nature

is a perfect square, the roots are real, rational and unequal.

is not a perfect square, then roots are real, irrational and unequal.

b2 – 4ac = 0

then roots will be real, equal and rational.

b2 – 4ac < 0

then roots will be imaginary and unequal.

Example 1 :

For what value of K the roots of equation 2x2 + 5x + k = 0 will be rational and equal.

Solution :

Since the roots are real and equal,

b2 - 4ac = 0

a = 2, b = 5 and c = k

52 - 4(2) (k) = 0

25 - 8k = 0

8k = 25

k = 25/8

For what value of K the roots of the given equations are equal.

Example 2 :

x2 + 3(k + 1)x + 4k + 5 = 0

Solution :

a = 1, b = 3(k + 1) and c = 4k + 5

b2 - 4ac = 0

[3(k + 1)]2 - 4(1) (4k + 5) = 0

9(k + 1)2 - 4(4k + 5) = 0

9(k2 + 2k + 1) - 16k - 20 = 0

9k2 + 18k + 9 - 16k - 20 = 0

9k2 + 2k - 11 = 0

(k - 1) (9k + 11) = 0

k - 1 = 0 and 9k + 11 = 0

k = 1 and 9k = -11

k = -11/9

So, the value of k are 1 and -11/9.

Example 3 :

x2 + 2(k – 2)x – 8k = 0

Solution :

a = 1, b = 2(k - 2) and c = -8k

b2 - 4ac = 0

[2(k - 2)]2 - 4(1) (-8k) = 0

4(k - 2)2 + 32k = 0

4(k2 - 4k + 4) + 32k = 0

4k2 - 16k + 16 + 32k = 0

4k2 + 16k + 16 = 0

k2 + 4k + 4 = 0

(k + 2) (k + 2) = 0

k = -2 and k = -2

So, the value of k is -2.

Example 4 :

(3k + 6)x2 + 6x + k = 0

Solution :

a = 3k + 6, b = 6 and c = k

b2 - 4ac = 0

62 - 4(3k + 6) (k) = 0

36 - 4k(3k + 6) = 0

36 - 12k2 - 24k = 0

Dividing by -12

k2 + 2k - 3 = 0

(k + 3)(k - 1) = 0

k = -3 and k = 1

So, the values of k are -3 and 1.

Example 5 :

(k + 2)x2 – 2kx + k – 1 = 0

Solution :

a = k + 2, b = -2k and c = k - 1

b2 - 4ac = 0

(-2k)2 - 4(k + 2) (k - 1) = 0

4k2 - 4(k2 + k - 2) = 0

4k2 - 4k2 - 4k + 8 = 0

-4k + 8 = 0

k = 2

Example 6 :

For what value of k the equation (4–k) x2 + 2(k+2) x + 8k + 1 = 0 will be a perfect square.

(Hint : The equation will be perfect square if Disc. b2 – 4ac = 0 )

Solution :

a = 4 - k, b = 2(k + 2) and c = 8k + 1

b2 - 4ac = 0

[2(k + 2)]2 - 4(4 - k) (8k + 1) = 0

4(k + 2)2 - 4(32k + 4 - 8k2 - k) = 0

4(k2 + 4k + 4) - 4(31k - 8k2 + 4) = 0

4k2 + 16k + 16 - 124k + 32k2 - 16 = 0

36k2 - 108k = 0

36k(k - 3) = 0

k  = 0 and k = 3

So, the value of k are 0 and 3.

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