To find missing coefficient in a quadratic equation using the concept of discriminant, we have to be aware of nature of roots of a quadratic equation.
Without solving the equation completely, by using the formula b^{2} - 4ac, we can easily figure out how the roots will be.
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. |
Problem 1 :
Find all the values of a such that
ax^{2} + 3x + 5 = 0
has two real roots.
Solution :
Since the given quadratic equation has two real roots,
b^{2} - 4ac > 0
By comparing the given equation with ax^{2} + bx + c = 0, we get
a = a, b = 3 and c = 5
3^{2} - 4(a)(5) > 0
9 - 20a > 0
-20a > -9
20a < 9
a < 9/20
Problem 2 :
Find all values of a such that ax^{2} + 48x + 64 = 0 has one real root (a double root)
Solution :
Since the given quadratic equation has one real and equal roots.
b^{2} - 4ac = 0
By comparing the given equation with ax^{2} + bx + c = 0, we get
a = a, b = 48 and c = 64
48^{2} - 4(a)(64) = 0
2304 - 256a = 0
-256a = -2304
a = 2304/256
a = 9
Problem 3 :
Find all the values of a such that
ax^{2} + 3x - 6 = 0
has two imaginary roots.
Solution :
Since the equation has two imaginary roots, we use the condition.
b^{2} - 4ac < 0
a = a, b = 3 and c = -6
(-3)^{2} - 4a(-6) < 0
9 + 24a < 0
24a < -9
a < -9/24
a < -3/8
Problem 4 :
Find all the values of such that
2x^{2} - 6x + c = 0
has two imaginary roots.
Solution :
Since it has two imaginary roots,
b^{2} - 4ac < 0
a = 2, b = -6 and c = c
(-6)^{2} - 4(2)c < 0
36 - 8a < 0
-8a < -36
a > 36/8
a > 9/2
Problem 5 :
Find all values of c such that -4x^{2} + 8x + c = 0 has two real roots.
Solution :
Since it has two imaginary roots,
b^{2} - 4ac > 0
a = -4, b = 8 and c = c
8^{2} - 4(-4) c > 0
64 + 16c > 0
16c > -64
Dividing by -16 on both sides.
c > -64/16
c > -4
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