Nature of the roots of the Equation ax^{2} + bx + c = 0
The two roots of the Quadratic equation ax^{2} + bx + c = 0 are:
Conditionb^{2} – 4ac > 0 |
Natureis a perfect square, the roots are real, rational and unequal. is not a perfect square, then roots are real, irrational and unequal. |
b^{2} – 4ac = 0 |
then roots will be real, equal and rational. |
b^{2} – 4ac < 0 |
then roots will be imaginary and unequal. |
Example 1 :
For what value of K the roots of equation 2x^{2} + 5x + k = 0 will be rational and equal.
Solution :
Since the roots are real and equal,
b^{2} - 4ac = 0
a = 2, b = 5 and c = k
5^{2} - 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 :
x^{2} + 3(k + 1)x + 4k + 5 = 0
Solution :
a = 1, b = 3(k + 1) and c = 4k + 5
b^{2} - 4ac = 0
[3(k + 1)]^{2} - 4(1) (4k + 5) = 0
9(k + 1)^{2} - 4(4k + 5) = 0
9(k^{2} + 2k + 1) - 16k - 20 = 0
9k^{2} + 18k + 9 - 16k - 20 = 0
9k^{2} + 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 :
x^{2} + 2(k – 2)x – 8k = 0
Solution :
a = 1, b = 2(k - 2) and c = -8k
b^{2} - 4ac = 0
[2(k - 2)]^{2} - 4(1) (-8k) = 0
4(k - 2)^{2} + 32k = 0
4(k^{2} - 4k + 4) + 32k = 0
4k^{2} - 16k + 16 + 32k = 0
4k^{2} + 16k + 16 = 0
k^{2} + 4k + 4 = 0
(k + 2) (k + 2) = 0
k = -2 and k = -2
So, the value of k is -2.
Example 4 :
(3k + 6)x^{2} + 6x + k = 0
Solution :
a = 3k + 6, b = 6 and c = k
b^{2} - 4ac = 0
6^{2} - 4(3k + 6) (k) = 0
36 - 4k(3k + 6) = 0
36 - 12k^{2} - 24k = 0
Dividing by -12
k^{2} + 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)x^{2} – 2kx + k – 1 = 0
Solution :
a = k + 2, b = -2k and c = k - 1
b^{2} - 4ac = 0
(-2k)^{2} - 4(k + 2) (k - 1) = 0
4k^{2} - 4(k^{2} + k - 2) = 0
4k^{2} - 4k^{2} - 4k + 8 = 0
-4k + 8 = 0
k = 2
Example 6 :
For what value of k the equation (4–k) x^{2} + 2(k+2) x + 8k + 1 = 0 will be a perfect square.
(Hint : The equation will be perfect square if Disc. b^{2} – 4ac = 0 )
Solution :
a = 4 - k, b = 2(k + 2) and c = 8k + 1
b^{2} - 4ac = 0
[2(k + 2)]^{2} - 4(4 - k) (8k + 1) = 0
4(k + 2)^{2} - 4(32k + 4 - 8k^{2} - k) = 0
4(k^{2} + 4k + 4) - 4(31k - 8k^{2} + 4) = 0
4k^{2} + 16k + 16 - 124k + 32k^{2} - 16 = 0
36k^{2 }- 108k = 0
36k(k - 3) = 0
k = 0 and k = 3
So, the value of k are 0 and 3.
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