ANSWERS
A. ULLUSTRATIONS OF QUADRATIC QUATIONS
ACTIVITY 1 ACTIVITY 2 ACTIVITY 3
1. Quadratic 1. 2x² - 4 – 2x = 0 1. x²-3x + 2 = 0
2. Not Quadratic 2. x² + 11x - 12 = 0 2. x²-9x + 6 = 0
3. Not Quadratic 3. −6x² +15x+36=0 3. x²-12x + 35 = 0
4. Quadratic 4. x² +4x +81 = 0 4. 10x²+ 3x - 100 = 0
5. x²+2x + 8 = 0 5. 5t2 - 8t -21 = 0
5. Not Quadratic
6. Not Quadratic 6. 6 x² −30x−36=0
7. Quadratic 7. + 15x + 36 = -6x²
8. x² + 1x - 6 = 0
8. Quadratic
9. 2x2 + 2x - 24 = 0
9. Quadratic
10. x 2−27 x +50=0
10. Quadratic
B. SOLVING QUADRATIC EQUATIONS
B. I EXTRACTING SQUARE ROOT
Activity No 1. Solve by factoring Activity No 2. Solve by extracting the roots.
and then solve by extracting roots.
1. ±9 11. (x± 2) 21. ±7
2. ±1 12. ± 2 √ 6 22. ± 5 √ 37
1. 4, - 4 3. ± √19 13. ± i 23. ± 2 √2
2. 6, - 6 14. ± 10i
4. ± 2 √29 24. ± 3 √2
−1 1
, 5. ± 2 √3 1 25. ± 2 √2 i
3.
4.
3 3
−5 5
,
6. ± 3 √2
3
15. ±
5
5
√ 26. ± 5 √5 i
2
5.
2 2
1,3
7. ±
8. ±
4
5
16. ±
6
√
√2
27. ±
√
5
1
6.
7.
-3 ,1
1 7
,
2
1
17. ±i
4
28. ±
√
3
8.
2 2
−1 5
3
,−
3
9. ±
√ 2
12
18. ±i √
± 2i
15
6
29. -5, -9
30. -15, -15
9.
10.
10, 0
4, 0
10. ±
√ 3
19.
20. ±i
B.II FACTORING
Activity No. 1. SOLVE THE QUADRATIC EQUATION BELOW
USING FACTORING METHOD
1) 6, 3 11) 7, 4
2) -1, -4 12) -7, -8
3) ±8 −7
13) , -3
4) 0, -5 3
−1 −3 −5
5) , 14) 6,
5 7 8
2 −2 15) -7
6) ,
5 3 16) 2, 0
−4 −7
7) 6, 17) 2,
7 2
7 −7
8) , -7 18) ,-7
3 3
9) 7, -1 3
19) 2,
10) 4, 3 5
−4
20) , -4
3
B.III COMPLETING THE SQUARE
ACTIVITY 1
1) −7 ± √ 87
2) −3 ± 2 √ 17
3) 3 ,−17
4) 11, 1
5) −2 ,−4
6) 3 ,−1
7) 1 ,−15
8) 6 ± √ 13
9) 2 ± √ 102
10) 5 ± √ 7
11) 2 ± √ 2 i
12) ) −1 ± √ 19 i
B. IV QUADRATIC FORMULA
ACTIVITY 1
1) 7 ,−2
2)2
3) 2, -3
5
4) ,−1
2
5) −1 ,−3
5
6) ,−4
2
−1 3
7) ,−
2 2
7 ± √ 73
8)
4
−5
9) 4,
2
10) 1 ,−3
−7
11) -4,
2
12)± 4
C. NATURE OF ROOTS QUADRATIC EQUATIONS
Solutions
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
2
Here, b - 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
2
Here, b - 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
2
Here, b - 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
3
Then, we have a = 2, b = 8 and c = -m .
Because the roots of the given equation are equal,
b2 - 4ac = 0
8 - 4(2)(-m3) = 0
2
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
2
Then, we have a = 1, b = (2s - 1) and c = s .
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
2
Multiply each side by (x - 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.
D. SUM AND PRODUCT OF ROOTS OF QUADRATIC EQUATIONS
Problem 1 :
Find the sum and product of roots of the quadratic equation given below.
x2 - 5x + 6 = 0
Solution :
Comparing
x2 - 5x + 6 = 0
and
ax2 + bx + c = 0
we get
a = 1, b = -5 and c = 6
Therefore,
Sum of the roots = -b/a = -(-5)/1 = 5
Product of the roots = c/a = 6/1 = 6
Problem 2 :
Find the sum and product of roots of the quadratic equation given below.
x2 - 6 = 0
Solution :
Comparing
x2 - 6 = 0
and
ax2 + bx + c = 0
we get
a = 1, b = 0 and c = -6
Therefore,
Sum of the roots = -b/a = 0/1 = 0
Product of the roots = c/a = -6/1 = -6
Problem 3 :
Find the sum and product of roots of the quadratic equation given below.
3x2 + x + 1 = 0
Solution :
Comparing
3x2 + x + 1 = 0
and
ax2 + bx + c = 0
we get
a = 3, b = 1 and c = 1
Therefore,
Sum of the roots = -b/a = -1/3
Product of the roots = c/a = 1/3
Problem 4 :
Find the sum and product of roots of the quadratic equation given below.
3x2 + 7x = 2x - 5
Solution :
First write the given quadratic equation in standard form.
3x2 +7x = 2x - 5
3x2 + 5x + 5 = 0
Comparing
3x2 + 5x + 5 = 0
and
ax2 + bx + c = 0
we get
a = 3, b = 5 and c = 5
Therefore,
Sum of the roots = -b/a = -5/3
Product of the roots = c/a = 5/3
Problem 5 :
Find the sum and product of roots of the quadratic equation given below.
3x2 -7x + 6 = 6
Solution :
First write the given quadratic equation in standard form.
3x2 -7x + 6 = 6
3x2 - 7x = 0
Comparing
3x2 - 7x = 0
and
ax2 + bx + c = 0
we get
a = 3, b = -7 and c = 0
Therefore,
Sum of the roots = -b/a = -(-7)/3 = 7/3
Product of the roots = c/a = 0/3 = 0
Problem 6 :
Find the sum and product of roots of the quadratic equation given below.
x2 + 5x + 1 = 3x2 + 6
Solution :
First write the given quadratic equation in standard form.
x2 + 5x + 1 = 3x² + 6
0 = 2x2 - 5x + 5
2x2 - 5x + 5 = 0
Comparing
2x2 - 5x + 5 = 0
and
ax2 + bx + c = 0
we get
a = 2, b = -5 and c = 5
Therefore,
Sum of the roots = -b/a = -(-5)/3 = 5/2
Product of the roots = c/a = 5/2
Problem 7 :
If the product of roots of the quadratic equation given below is 4, then find the value of m.
2x2 + 8x - m3 = 0
Solution :
Comparing
2x2 + 8x - m3 = 0
and
ax2 + bx + c = 0
we get
a = 2, b = 8 and c = -m3
Given : Product of the roots is 4.
Then,
c/a = 4
-m3/2 = 4
Multiply each side by (-2).
m3 = -8
m3 = (-2)3
m = -2
Problem 8 :
If the sum of roots of the quadratic equation given below is 0, then find the value of p.
x2 -(p + 4)x + 5 = 0
Solution :
Comparing
x2 -(p + 4)x + 5 = 0
and
ax2 + bx + c = 0
we get
a = 1, b = -(p + 4) and c = 5
Given : Sum of the roots is 0.
Then,
-b/a = 0
-[-(p + 4)] = 0
(p + 4) = 0
p + 4 = 0
p = -4
Problem 9 :
If the product of roots of the quadratic equation given below is 1, then find the value of m.
x2 + (2p - 1)x + p2 = 0
Solution :
Comparing
x2 + (2p - 1)x + p2 = 0
and
ax2 + bx + c = 0
we get
a = 1, b = (2p - 1) and c = p2
Given : Product of the roots is 1.
Then,
c/a = 1
p2/1 = 1
p2 = 1
Take square root on both sides.
√p2 = √1
p = ±1
Problem 10 :
Find the sum and product of roots of the quadratic equation given below.
1/(x+1) + 2/(x-4) = 2
Solution :
First write the given quadratic equation in standard form.
1/(x+1) + 2/(x-4) = 2
[1(x-4) + 2(x+1)] / [(x+1)(x-4)] = 2
1(x-4) + 2(x+1) = 2(x+1)(x-4)
x - 4 + 2x + 2 = 2(x2 - 4x + x - 4)
3x - 2 = 2(x2 - 3x - 4)
3x - 2 = 2x2 - 6x - 8
0 = 2x2 - 9x - 6
2x2 - 9x - 6 = 0
Comparing
2x2 - 9x - 6 = 0
and
ax2 + bx + c = 0
we get
a = 2, b = -9 and c = -6
Therefore,
Sum of the roots = -b/a = -(-9)/2 = 9/2
Product of the roots = c/a = -6/2 = -3
A. APPLICATIONS OF QUADRATIC EQUATIONS AND RATIONAL ALGEBRAIC
EQUATIONS
1. 2(7.3255)−1=13.651m2(7.3255)−1=13.651m .
2. This means that Car A (i.e. the one traveling at 40 mph) travels for
8.871 hours while Car B (i.e. the one traveling at 60 mph) travels for
5.871 hours (three hours less than Car A time!)
3. This means that Person A can paint the house in 27.0357 hours while
Person B can paint the house in 29.0357 hours (two hours more than
Person A).
