Problem 1 :
Examine the nature of the roots of the following quadratic equation.
x2 + 5x + 6 = 0
Problem 2 :
Examine the nature of the roots of the following quadratic equation.
2x2 - 3x - 1 = 0
Problem 3 :
Examine the nature of the roots of the following quadratic equation.
x2 - 16x + 64 = 0
Problem 4 :
Examine the nature of the roots of the following quadratic equation.
3x2 + 5x + 8 = 0
Problem 5 :
If the roots of the equation 2x2 + 8x - m³ = 0 are equal, then find the value of m.
Problem 6 :
If the roots of the equation x2 - (p + 4)x + 2p + 5 = 0 are equal, then find the value of p.
Problem 7 :
If the roots of the equation x2 + (2s - 1)x + s2 = 0 are real, then find the value of a.
Problem 8 :
If the roots of the equation x2 - 16x + k = 0 are real and equal, then find the value of k.
Problem 9 :
Examine the nature of the roots of the following quadratic equation.
x2 - 5x = 2(5x + 1)
Problem 10 :
Examine the nature of the roots of the following quadratic equation.
1/(x+1) + 2/(x-4) = 2
Problem 1 :
Examine the nature of the roots of the following quadratic equation.
x2 + 5x + 6 = 0
Solution :
The given quadratic equation is in the general form
ax2 + bx + c = 0
Then, we have a = 1, b = 5 and c = 6.
Find the value of the discriminant b2 - 4ac.
b2 - 4ac = 52 - 4(1)(6)
b2 - 4ac = 25 - 24
b2 - 4ac = 1
Here, b2 - 4ac > 0 and also a perfect square.
So, the roots are real, unequal and rational.
Problem 2 :
Examine the nature of the roots of the following quadratic equation.
2x2 - 3x - 1 = 0
Solution :
The given quadratic equation is in the 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.
Problem 3 :
Examine the nature of the roots of the following quadratic equation.
x2 - 16x + 64 = 0
Solution :
The given quadratic equation is in the 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.
Problem 4 :
Examine the nature of the roots of the following quadratic equation.
3x2 + 5x + 8 = 0
Solution :
The given quadratic equation is in the general form
ax2 + bx + c = 0
Then, we have a = 3, b = 5 and c = 8.
Find the value of the discriminant b2 - 4ac.
b2 - 4ac = 52 - 4(3)(8)
b2 - 4ac = 25 + 96
b2 - 4ac = 121
Here, b2 - 4ac > 0 and also a perfect square.
So, the roots are real, unequal and rational.
Problem 5 :
If the roots of the equation 2x2 + 8x - m³ = 0 are equal, then find the value of m.
Solution :
The given quadratic equation is in the general form
ax2 + bx + c = 0
Then, we have 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 each side.
8m3 = -64
Divide each side by 8.
m3 = -8
m3 = (-2)3
m = -2.
So, the value of m is -2.
Problem 6 :
If the roots of the equation x2 - (p + 4)x + 2p + 5 = 0 are equal, then find the value of p.
Solution :
The given quadratic equation is in the general form
ax2 + bx + c = 0
Then, we have a = 1, b = -(p + 4) and c = (2p + 5).
Because the roots of the given equation are equal,
b2 - 4ac = 0
[-(p + 4)]2 - 4(1)(2p + 5) = 0
Simplify.
(p + 4)2 - 4(2p + 5) = 0
p2 + 8p + 16 -8p -20 = 0
p2 - 4 = 0
p2 - 22 = 0
(p + 2)(p - 2) = 0
p + 2 = 0 or p - 2 = 0
p = -2 or p = 2
So, the value of p is ±2.
Problem 7 :
If the roots of the equation x2 + (2s - 1)x + s2 = 0 are real, then find the value of a.
Solution :
The given quadratic equation is in the general form
ax2 + bx + c = 0
Then, we have a = 1, b = (2s - 1) and c = s2.
Because the roots of the given equation are equal,
b2 - 4ac ≥ 0
(2s - 1)2 - 4(1)(s2) ≥ 0
Simplify.
4s2 - 4s + 1 - 4s2 ≥ 0
-4s + 1 ≥ 0
-4s ≥ -1
Divide each side by -4.
s ≤ 1/4
So, the value of s is less than or equal to 1/4.
Note :
Whenever we multiply or divide both sides of an inequality by a negative number, we have to flip the inequality sign.
Problem 8 :
If the roots of the equation x2 - 16x + k = 0 are real and equal, then find the value of k.
Solution :
The given quadratic equation is in the general form
ax2 + bx + c = 0
Then, we have a = 1, b = -16 and c = k.
Because the roots of the given equation are equal,
b2 - 4ac = 0
(-16)2 - 4(1)(k) = 0
256 - 4k = 0
Subtract 256 from each side.
-4k = -256
Divide each side by -4.
k = 64
So, the value of k is 64.
Problem 9 :
Examine the nature of the roots of the following quadratic equation.
x2 - 5x = 2(5x + 1)
Solution :
The given quadratic equation is not in the general form.
First, write the given quadratic equation in the general form.
x2 - 5x = 2(5x + 1)
x2 - 5x = 10x + 2
x2 - 15x - 2 = 0
Now, the quadratic equation is the general form
ax2 + bx + c = 0
Then, we have a = 1, b = -15 and c = -2.
Find the value of the discriminant b2 - 4ac.
b2 - 4ac = (-15)2 - 4(1)(-2)
b2 - 4ac = 225 + 8
b2 - 4ac = 233
Here, b2 - 4ac > 0, but not a perfect square.
So, the roots are real, unequal and irrational.
Problem 10 :
Examine the nature of the roots of the following quadratic equation.
1/(x+1) + 2/(x-4) = 2
Solution :
The given quadratic equation is not in the general form.
First, write the given quadratic equation in the 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 the 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.
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