In this section, we will examine the roots of a quadratic equation.
That is, we will analyze whether the roots of a quadratic equation are equal or unequal, real or imaginary and rational or irrational.
To examine the roots of a quadratic equation, let us consider the general form a quadratic equation.
ax2 + bx + c = 0
(Here a, b and c are real and rational numbers)
To know the nature of the roots of a quadratic-equation, we will be using the discriminant b2 - 4ac.
Because b2 - 4ac discriminates the nature of the roots.
Let us see how this discriminant b2 - 4ac can be used to know the nature of the roots of a quadratic-equation.
Examples 1-5 : Examine the nature of the roots of the quadratic equation.
Example 1 :
3x2 + 8x + 4 = 0
Solution :
The given quadratic equation is in general form
ax2 + bx + c = 0
Then, we have a = 3, b = 8 and c = 4.
Find the value of the discriminant b2 - 4ac.
b2 - 4ac = 82 - 4(3)(4)
b2 - 4ac = 64 - 48
b2 - 4ac = 16
Here, b2 - 4ac > 0 and also a perfect square.
So, the roots are real, unequal and rational.
Example 2 :
2x2 - 3x - 1 = 0
Solution :
The given quadratic equation is in general form
ax2 + bx + c = 0
Then, we have a = 2, b = -3 and c = -1.
Find the value of the discriminant b2 - 4ac.
b2 - 4ac = (-3)2 - 4(2)(-1)
b2 - 4ac = 9 + 8
b2 - 4ac = 17
Here, b2 - 4ac > 0, but not a perfect square.
So, the roots are real, unequal and irrational.
Example 3 :
x2 - 16x + 64 = 0
Solution :
The given quadratic equation is in general form
ax2 + bx + c = 0
Then, we have a = 1, b = -16 and c = 64.
Find the value of the discriminant b2 - 4ac.
b2 - 4ac = (-16)2 - 4(1)(64)
b2 - 4ac = 256 - 256
b2 - 4ac = 0
So, the roots are real, equal and rational.
Example 4 :
5x2 - 4x + 2 = 0
Solution :
The given quadratic equation is in general form
ax2 + bx + c = 0
Then, we have a = 5, b = -4 and c = 2.
Find the value of the discriminant b2 - 4ac.
b2 - 4ac = (-4)2 - 4(5)(2)
b2 - 4ac = 16 - 40
b2 - 4ac = -24
Here, b2 - 4ac < 0.
So, the roots are imaginary.
Example 5 :
1/(x + 1) + 2/(x - 4) = 2
Solution :
The given quadratic equation is not in general form.
First, write the given quadratic equation in general form.
1/(x + 1) + 2/(x - 4) = 2
Add the two fractions on the right side of the equation using cross multiplication.
[(x - 4) + 2(x + 1)]/[(x + 1)(x - 4)] = 2
(x - 4 + 2x + 2)/(x2 - 3x - 4) = 2
(3x - 2)/(x2 - 3x - 4) = 2
Multiply each side by (x2 - 3x - 4).
3x - 2 = 2(x2 - 3x - 4)
3x - 2 = 2x2 - 6x - 8
2x2 - 9x - 6 = 0
Now, the quadratic equation is general form
ax2 + bx + c = 0
Then, we have a = 2, b = -9 and c = -6.
Find the value of the discriminant b2 - 4ac.
b2 - 4ac = (-9)2 - 4(2)(-6)
b2 - 4ac = 81 + 48
b2 - 4ac = 129
Here, b2 - 4ac > 0, but not a perfect square.
So, the roots are real, unequal and irrational.
Example 6 :
x = 1 - 2y
4y = x2 + 3
How many ordered pairs (x, y) in the xy-plane are solutions to the system of equations above?
A) 0
B) 1
C) 2
D) Infinitely Many
Solution :
x = 1 - 2y ----(1)
4y = x2 + 3 ----(2)
(1) :
x = 1 - 2y
Add 2y to both sides.
x + 2y = 1
Subtract x from both sides.
2y = 1 - x
Multiply both sides by 2.
4y = 2 - 2x
Substitute 4y = 2 - 2x in (2).
2 - 2x = x2 + 3
x2 + 2x + 1 = 0
Comparing ax2 + bx + c = 0 and x2 + 2x + 1 = 0,
a = 1, b = 2 and c = 1
Find the value of the discriminant b2 - 4ac.
b2 - 4ac = 22 - 4(1)(1)
b2 - 4ac = 4 - 4
b2 - 4ac = 0
Here, b2 - 4ac = 0.
So, the roots are real and equal. It means, there is only one real value for x. Hence, we will have only one real value for y.
Therefore, only one ordered pair (x, y) in the xy-plane is solution to the given system of equations.
The correct answer choice is (B).
Example 7 :
2x2 + 8x - m3 = 0
If the roots of the quadratic equation above are equal, then find the value of m.
Solution :
Comparing ax2 + bx + c = 0 and 2x2 + 8x - m3 = 0,
a = 2, b = 8 and c = -m3
Because the roots of the given equation are equal,
b2 - 4ac = 0
82 - 4(2)(-m3) = 0
64 + 8m3 = 0
Subtract 64 from both sides.
8m3 = -64
Divide both sides by 8.
m3 = -8
m3 = (-2)3
m = -2
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