Problems 1-5 : Examine the nature of the roots of the quadratic equation.
Problem 1 :
x^{2} + 5x + 6 = 0
Problem 2 :
2x^{2} - 3x - 1 = 0
Problem 3 :
x^{2} - 16x + 64 = 0
Problem 4 :
x^{2} - 5x = 2(5x + 1)
Problem 5 :
1/(x + 1) + 2/(x - 4) = 2
Problem 6 :
If the roots of the equation 2x^{2} + 8x - m^{3} = 0 are equal, then find the value of m.
Problem 7 :
If the roots of the equation x^{2} - (p + 4)x + 2p + 5 = 0 are equal, then find the value of p.
Problem 8 :
If the roots of the equation x^{2} + (2s - 1)x + s^{2 }= 0 are real, then find the value of a.
Problem 9 :
If the roots of the equation x^{2} - 16x + k = 0 are real and equal, then find the value of k.
Problem 10 :
x = 1 - 2y
4y = x^{2} + 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
1. Answer :
x^{2} + 5x + 6 = 0
The given quadratic equation is in general form
ax^{2} + bx + c = 0
Then, we have a = 1, b = 5 and c = 6.
Find the value of the discriminant b^{2} - 4ac.
b^{2} - 4ac = 5^{2} - 4(1)(6)
b^{2} - 4ac = 25 - 24
b^{2} - 4ac = 1
Here, b^{2} - 4ac > 0 and also a perfect square.
So, the roots are real, unequal and rational.
2. Answer :
2x^{2} - 3x - 1 = 0
The given quadratic equation is in general form
ax^{2} + bx + c = 0
Then, we have a = 2, b = -3 and c = -1.
Find the value of the discriminant b^{2} - 4ac.
b^{2} - 4ac = (-3)^{2} - 4(2)(-1)
b^{2} - 4ac = 9 + 8
b^{2} - 4ac = 17
Here, b^{2} - 4ac > 0, but not a perfect square.
So, the roots are real, unequal and irrational.
3. Answer :
x^{2} - 16x + 64 = 0
The given quadratic equation is in general form
ax^{2} + bx + c = 0
Then, we have a = 1, b = -16 and c = 64.
Find the value of the discriminant b^{2} - 4ac.
b^{2} - 4ac = (-16)^{2} - 4(1)(64)
b^{2} - 4ac = 256 - 256
b^{2} - 4ac = 0
So, the roots are real, equal and rational.
4. Answer :
x^{2} - 5x = 2(5x + 1)
The given quadratic equation is not in general form.
First, write the given quadratic equation in the general form.
x^{2} - 5x = 2(5x + 1)
x^{2} - 5x = 10x + 2
x^{2} - 15x - 2 = 0
Now, the quadratic equation is in general form
ax^{2} + bx + c = 0
Then, we have a = 1, b = -15 and c = -2.
Find the value of the discriminant b^{2} - 4ac.
b^{2} - 4ac = (-15)^{2} - 4(1)(-2)
b^{2} - 4ac = 225 + 8
b^{2} - 4ac = 233
Here, b^{2} - 4ac > 0, but not a perfect square.
So, the roots are real, unequal and irrational.
5. Answer :
1/(x + 1) + 2/(x - 4) = 2
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)/(x^{2} - 3x - 4) = 2
(3x - 2)/(x^{2} - 3x - 4) = 2
Multiply each side by (x^{2} - 3x - 4).
3x - 2 = 2(x^{2} - 3x - 4)
3x - 2 = 2x^{2} - 6x - 8
2x^{2} - 9x - 6 = 0
Now, the quadratic equation is in general form
ax^{2} + bx + c = 0
Then, we have a = 2, b = -9 and c = -6.
Find the value of the discriminant b^{2} - 4ac.
b^{2} - 4ac = (-9)^{2} - 4(2)(-6)
b^{2} - 4ac = 81 + 48
b^{2} - 4ac = 129
Here, b^{2} - 4ac > 0, but not a perfect square.
So, the roots are real, unequal and irrational.
6. Answer :
2x^{2} + 8x - m^{3} = 0
The given quadratic equation is in general form
ax^{2} + bx + c = 0
Then, we have a = 2, b = 8 and c = -m^{3}.
Because the roots of the given equation are equal,
b^{2} - 4ac = 0
8^{2} - 4(2)(-m^{3}) = 0
64 + 8m^{3} = 0
Subtract 64 from each side.
8m^{3} = -64
Divide each side by 8.
m^{3} = -8
m^{3} = (-2)^{3}
m = -2.
So, the value of m is -2.
7. Answer :
x^{2} - (p + 4)x + 2p + 5 = 0
The given quadratic equation is in general form
ax^{2} + bx + c = 0
Then, we have a = 1, b = -(p + 4) and c = (2p + 5).
Because the roots of the given equation are equal,
b^{2} - 4ac = 0
[-(p + 4)]^{2} - 4(1)(2p + 5) = 0
Simplify.
(p + 4)^{2} - 4(2p + 5) = 0
p^{2} + 8p + 16 -8p -20 = 0
p^{2} - 4 = 0
p^{2} - 2^{2} = 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.
8. Answer :
x^{2} + (2s - 1)x + s^{2 }= 0
The given quadratic equation is in general form
ax^{2} + bx + c = 0
Then, we have a = 1, b = (2s - 1) and c = s^{2}.
Because the roots of the given equation are equal,
b^{2} - 4ac ≥ 0
(2s - 1)^{2} - 4(1)(s^{2}) ≥ 0
Simplify.
4s^{2} - 4s + 1 - 4s^{2} ≥ 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.
9. Answer :
x^{2} - 16x + k = 0
The given quadratic equation is in general form
ax^{2} + bx + c = 0
Then, we have a = 1, b = -16 and c = k.
Because the roots of the given equation are equal,
b^{2} - 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
10. Answer :
x = 1 - 2y ----(1)
4y = x^{2} + 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 = x^{2} + 3
x^{2} + 2x + 1 = 0
Comparing ax^{2} + bx + c = 0 and x^{2} + 2x + 1 = 0,
a = 1, b = 2 and c = 1
Find the value of the discriminant b^{2} - 4ac.
b^{2} - 4ac = 2^{2} - 4(1)(1)
b^{2} - 4ac = 4 - 4
b^{2} - 4ac = 0
Here, b^{2} - 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).
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