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Algebra Formulas and Solutions 50+
Total 50+ Formulas & Solutions
...

These formulas are used to simplify and manipulate expressions, making it easier to find unknown
values
...
Basic Algebraic Identity



Formula: (a+b)2=a2+2ab+b2(a + b)^2 = a^2 + 2ab + b^2(a+b)2=a2+2ab+b2
Solution: To expand (x+5)2(x + 5)^2(x+5)2, use the identity:
(x+5)2=x2+2(x)(5)+52=x2+10x+25(x + 5)^2 = x^2 + 2(x)(5) + 5^2 = x^2 + 10x +
25(x+5)2=x2+2(x)(5)+52=x2+10x+25

2
...
McKeague, Algebra for College Students" by Mark Dugopolski, "Higher Algebra" by Hall &
Knight, "College Algebra" by James Stewart, Lothar Redlin, and Saleem Watson, "Schaum's Outline of College Algebra" by
Murray Spiegel, Khan Academy (www
...
org), Paul's Online Math Notes (tutorial
...
lamar
...
Page 2 of 20


Solution: The quadratic formula is used to find the roots (solutions) of a quadratic
equation ax2+bx+c=0ax^2 + bx + c = 0ax2+bx+c=0
...

o Identify a=2a = 2a=2, b=3b = 3b=3, and c=−5c = -5c=−5
...
5x = \frac{-3 + 7}{4} =
\frac{4}{4} = 1 \quad \text{or} \quad x = \frac{-3 - 7}{4} = \frac{-10}{4} = 2
...
5
o So, the solutions are x=1x = 1x=1 and x=−2
...
5x=−2
...


3
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o Find two numbers that multiply to 666 (constant term) and add to 555 (coefficient
of xxx): 222 and 333
...
Difference of Squares



Formula: a2−b2=(a+b)(a−b)a^2 - b^2 = (a + b)(a - b)a2−b2=(a+b)(a−b)
Solution: Factor x2−16x^2 - 16x2−16
...

o Apply the formula: x2−16=(x+4)(x−4)x^2 - 16 = (x + 4)(x - 4)x2−16=(x+4)(x−4)

5
...

o Plug in the values: y=2(4)+3=8+3=11y = 2(4) + 3 = 8 + 3 = 11y=2(4)+3=8+3=11
o So, y=11y = 11y=11
...
Slope of a Line
Reference: Books,
"Elementary Algebra" by Charles P
...
khanacademy
...
math
...
edu)

Algebra Formulas and Solutions 50+
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o Use the formula: m=7−34−2=42=2m = \frac{7 - 3}{4 - 2} = \frac{4}{2} =
2m=4−27−3=24=2
o So, the slope is m=2m = 2m=2
...
Solving for an Unknown in Proportions



Formula: ab=cd\frac{a}{b} = \frac{c}{d}ba=dc
Solution: Solve for xxx in the proportion 3x=612\frac{3}{x} = \frac{6}{12}x3=126
...
Exponent Rules



Formula: am×an=am+na^m \times a^n = a^{m+n}am×an=am+n (am)n=am×n\left( a^m
\right)^n = a^{m \times n}(am)n=am×n
Solution: Simplify x3×x2x^3 \times x^2x3×x2:
o Apply the first exponent rule: x3×x2=x3+2=x5x^3 \times x^2 = x^{3+2} =
x^5x3×x2=x3+2=x5
o Simplified expression: x5x^5x5
...
Absolute Value Equations



Formula: ∣x∣=aimpliesx=aorx=−a|x| = a \quad \text{implies} \quad x = a \quad \text{or}
\quad x = -a∣x∣=aimpliesx=aorx=−a
Solution: Solve ∣x∣=4|x| = 4∣x∣=4
...
Sum and Difference of Cubes




Formula:
o Sum of cubes: a3+b3=(a+b)(a2−ab+b2)a^3 + b^3 = (a + b)(a^2 - ab +
b^2)a3+b3=(a+b)(a2−ab+b2)
o Difference of cubes: a3−b3=(a−b)(a2+ab+b2)a^3 - b^3 = (a - b)(a^2 + ab +
b^2)a3−b3=(a−b)(a2+ab+b2)
Solution: For example, simplify x3+8x^3 + 8x3+8:
o Notice that 8=238 = 2^38=23, so this is a sum of cubes
...
McKeague, Algebra for College Students" by Mark Dugopolski, "Higher Algebra" by Hall &
Knight, "College Algebra" by James Stewart, Lothar Redlin, and Saleem Watson, "Schaum's Outline of College Algebra" by
Murray Spiegel, Khan Academy (www
...
org), Paul's Online Math Notes (tutorial
...
lamar
...
Page 4 of 20
o

Apply the sum of cubes formula: x3+8=(x+2)(x2−2x+4)x^3 + 8 = (x + 2)(x^2 2x + 4)x3+8=(x+2)(x2−2x+4)

This is the factored form
...
Completing the Square




Formula: To complete the square of the quadratic ax2+bx+cax^2 + bx + cax2+bx+c,
transform it into the form (x+d)2(x + d)^2(x+d)2 by adding and subtracting the square of
half the coefficient of xxx
...

o Start with x2+6x=−5x^2 + 6x = -5x2+6x=−5
...


12
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o Use the distributive property: 3(x+4)=3x+123(x + 4) = 3x + 123(x+4)=3x+12
This is the simplified expression
...
Rational Expressions



Formula: If ab=cd\frac{a}{b} = \frac{c}{d}ba=dc, then a⋅d=b⋅ca \cdot d = b \cdot
ca⋅d=b⋅c
...

o Cross-multiply: 6(x+2)=4(5)6(x + 2) = 4(5)6(x+2)=4(5)
o Expand both sides: 6x+12=206x + 12 = 206x+12=20
o Solve for xxx: 6x=20−12=8⇒x=86=436x = 20 - 12 = 8 \quad \Rightarrow \quad x
= \frac{8}{6} = \frac{4}{3}6x=20−12=8⇒x=68=34

14
...
McKeague, Algebra for College Students" by Mark Dugopolski, "Higher Algebra" by Hall &
Knight, "College Algebra" by James Stewart, Lothar Redlin, and Saleem Watson, "Schaum's Outline of College Algebra" by
Murray Spiegel, Khan Academy (www
...
org), Paul's Online Math Notes (tutorial
...
lamar
...
Page 5 of 20




Formula:
o If ∣x∣ ...

o If ∣x∣>a|x| > a∣x∣>a, then x<−ax < -ax<−a or x>ax > ax>a
...


15
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o Use the property log⁡b(xy)=log⁡b(x)+log⁡b(y)\log_b(xy) = \log_b(x) +
\log_b(y)logb(xy)=logb(x)+logb(y):
log⁡2(8)+log⁡2(4)=log⁡2(8×4)=log⁡2(32)\log_2(8) + \log_2(4) = \log_2(8
\times 4) = \log_2(32)log2(8)+log2(4)=log2(8×4)=log2(32)
o Now simplify log⁡2(32)\log_2(32)log2(32)
...


16
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Solution: Solve the system of equations: 2x+y=102x + y = 102x+y=10 x−y=1x - y =
1x−y=1
o Solve the second equation for xxx: x=y+1x = y + 1x=y+1
o Substitute x=y+1x = y + 1x=y+1 into the first equation: 2(y+1)+y=102(y + 1) + y
= 102(y+1)+y=10

Reference: Books,
"Elementary Algebra" by Charles P
...
khanacademy
...
math
...
edu)

Algebra Formulas and Solutions 50+
Total 50+ Formulas & Solutions
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17
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Solution: Solve the system of equations: 3x+2y=163x + 2y = 163x+2y=16 2x−y=32x - y
= 32x−y=3
o Multiply the second equation by 2 to match the coefficients of yyy: 4x−2y=64x 2y = 64x−2y=6
o Now add the two equations: (3x+2y)+(4x−2y)=16+6(3x + 2y) + (4x - 2y) = 16 +
6(3x+2y)+(4x−2y)=16+6
o Simplify: 7x=22⇒x=2277x = 22 \quad \Rightarrow \quad x =
\frac{22}{7}7x=22⇒x=722
o Substitute x=227x = \frac{22}{7}x=722 into one of the original equations to find
yyy: 3(227)+2y=16⇒667+2y=163\left(\frac{22}{7}\right) + 2y = 16 \quad
\Rightarrow \quad \frac{66}{7} + 2y = 163(722)+2y=16⇒766+2y=16
o Multiply through by 7 to eliminate the fraction:
66+14y=112⇒14y=46⇒y=4614=23766 + 14y = 112 \quad \Rightarrow \quad 14y
= 46 \quad \Rightarrow \quad y = \frac{46}{14} =
\frac{23}{7}66+14y=112⇒14y=46⇒y=1446=723
o So, the solution is x=227x = \frac{22}{7}x=722 and y=237y =
\frac{23}{7}y=723
...
Binomial Theorem



Formula: (a+b)n=∑k=0n(nk)an−kbk(a + b)^n = \sum_{k=0}^{n} \binom{n}{k} a^{nk} b^k(a+b)n=k=0∑n(kn)an−kbk
Solution: To expand (x+2)3(x + 2)^3(x+2)3 using the binomial theorem, we have:
(x+2)3=(30)x3⋅20+(31)x2⋅21+(32)x1⋅22+(33)x0⋅23(x + 2)^3 = \binom{3}{0}x^3 \cdot
2^0 + \binom{3}{1}x^2 \cdot 2^1 + \binom{3}{2}x^1 \cdot 2^2 + \binom{3}{3}x^0
\cdot 2^3(x+2)3=(03)x3⋅20+(13)x2⋅21+(23)x1⋅22+(33)x0⋅23
o First, calculate the binomial coefficients:
(30)=1,(31)=3,(32)=3,(33)=1\binom{3}{0} = 1, \binom{3}{1} = 3,
\binom{3}{2} = 3, \binom{3}{3} = 1(03)=1,(13)=3,(23)=3,(33)=1

Reference: Books,
"Elementary Algebra" by Charles P
...
khanacademy
...
math
...
edu)

Algebra Formulas and Solutions 50+
Total 50+ Formulas & Solutions
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19
...

Solution: For the polynomial p(x)=x2−5x+6p(x) = x^2 - 5x + 6p(x)=x2−5x+6, we want
to check if x−2x - 2x−2 is a factor
...

o Thus, p(x)=x2−5x+6=(x−2)(x−3)p(x) = x^2 - 5x + 6 = (x - 2)(x 3)p(x)=x2−5x+6=(x−2)(x−3)
...
Law of Exponents (Power of a Power)



Formula: (am)n=am⋅n(a^m)^n = a^{m \cdot n}(am)n=am⋅n
Solution: Simplify (x2)3(x^2)^3(x2)3
...


21
...

o Apply the law of exponents: x4⋅x2=x4+2=x6x^4 \cdot x^2 = x^{4+2} =
x^6x4⋅x2=x4+2=x6
So, x4⋅x2=x6x^4 \cdot x^2 = x^6x4⋅x2=x6
...
Rationalizing Denominators
Reference: Books,
"Elementary Algebra" by Charles P
...
khanacademy
...
math
...
edu)

Algebra Formulas and Solutions 50+
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1a=1a⋅aa=aa\frac{1}{\sqrt{a}} = \frac{1}{\sqrt{a}}
\cdot \frac{\sqrt{a}}{\sqrt{a}} = \frac{\sqrt{a}}{a}a1=a1⋅aa=aa
Solution: Simplify 53\frac{5}{\sqrt{3}}35
...


23
...

o Recognize this as a difference of squares: 9x2−25y2=(3x+5y)(3x−5y)9x^2 25y^2 = (3x + 5y)(3x - 5y)9x2−25y2=(3x+5y)(3x−5y)
So, 9x2−25y2=(3x+5y)(3x−5y)9x^2 - 25y^2 = (3x + 5y)(3x 5y)9x2−25y2=(3x+5y)(3x−5y)
...
Solving Rational Equations



Formula: To solve rational equations, multiply through by the least common
denominator (LCD) to eliminate fractions
...

o The LCD is x(x+1)x(x + 1)x(x+1)
...


25
...

Solution: Solve 32x=813^{2x} = 8132x=81
...
McKeague, Algebra for College Students" by Mark Dugopolski, "Higher Algebra" by Hall &
Knight, "College Algebra" by James Stewart, Lothar Redlin, and Saleem Watson, "Schaum's Outline of College Algebra" by
Murray Spiegel, Khan Academy (www
...
org), Paul's Online Math Notes (tutorial
...
lamar
...
Page 9 of 20
o

Since the bases are the same, set the exponents equal to each other: 2x=42x =
42x=4
o Solve for xxx: x=42=2x = \frac{4}{2} = 2x=24=2

26
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Solution: Solve log⁡2(x)=5\log_2(x) = 5log2(x)=5
...


27
...

Solution: Find the 10th term of the arithmetic sequence where a1=3a_1 = 3a1=3 and
d=5d = 5d=5
...


28
...

Solution: Find the 6th term of the geometric sequence where a1=2a_1 = 2a1=2 and r=3r
= 3r=3
...
Geometric Sequence Formula (continued)



Formula: an=a1⋅rn−1a_n = a_1 \cdot r^{n-1}an=a1⋅rn−1 where ana_nan is the nth term,
a1a_1a1 is the first term, rrr is the common ratio, and nnn is the number of terms
...


Reference: Books,
"Elementary Algebra" by Charles P
...
khanacademy
...
math
...
edu)

Algebra Formulas and Solutions 50+
Total 50+ Formulas & Solutions
...


30
...

o Use the quadratic formula with a=2a = 2a=2, b=3b = 3b=3, and c=−2c = -2c=−2:
x=−3±32−4(2)(−2)2(2)=−3±9+164=−3±254x = \frac{-3 \pm \sqrt{3^2 - 4(2)(2)}}{2(2)} = \frac{-3 \pm \sqrt{9 + 16}}{4} = \frac{-3 \pm
\sqrt{25}}{4}x=2(2)−3±32−4(2)(−2)=4−3±9+16=4−3±25
o Simplify: x=−3±54x = \frac{-3 \pm 5}{4}x=4−3±5
o This gives two solutions: x=−3+54=24=12orx=−3−54=−84=−2x = \frac{-3 +
5}{4} = \frac{2}{4} = \frac{1}{2} \quad \text{or} \quad x = \frac{-3 - 5}{4} =
\frac{-8}{4} = -2x=4−3+5=42=21orx=4−3−5=4−8=−2
So, the solutions are x=12x = \frac{1}{2}x=21 and x=−2x = -2x=−2
...
Factorizing a Perfect Square Trinomial



Formula: A perfect square trinomial is in the form a2+2ab+b2a^2 + 2ab +
b^2a2+2ab+b2, and it factors as: (a+b)2(a + b)^2(a+b)2
Solution: Factor x2+6x+9x^2 + 6x + 9x2+6x+9
...

o Factor as: x2+6x+9=(x+3)2x^2 + 6x + 9 = (x + 3)^2x2+6x+9=(x+3)2
So, the factored form is (x+3)2(x + 3)^2(x+3)2
...
Solving Systems of Linear Equations (Graphical Method)




Formula:
o The graphical method involves plotting the equations as lines on a graph
...

Solution: Solve the system: x+y=5x + y = 5x+y=5 x−y=1x - y = 1x−y=1

Reference: Books,
"Elementary Algebra" by Charles P
...
khanacademy
...
math
...
edu)

Algebra Formulas and Solutions 50+
Total 50+ Formulas & Solutions
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o The lines intersect at the point (3,2)(3, 2)(3,2)
...

o

33
...

Solution: Find the sum of the first 10 terms of the arithmetic series where a1=2a_1 = 2a1
=2 and the common difference d=4d = 4d=4
...


34
...

Solution: Find the sum of the first 5 terms of a geometric series where a1=3a_1 = 3a1=3
and r=2r = 2r=2
...


35
...
McKeague, Algebra for College Students" by Mark Dugopolski, "Higher Algebra" by Hall &
Knight, "College Algebra" by James Stewart, Lothar Redlin, and Saleem Watson, "Schaum's Outline of College Algebra" by
Murray Spiegel, Khan Academy (www
...
org), Paul's Online Math Notes (tutorial
...
lamar
...
Page 12 of 20
o
o
o
o



PPP is the initial amount,
kkk is the rate of growth (positive) or decay (negative),
ttt is the time elapsed,
eee is Euler's number (approximately 2
...

Solution: If a population of bacteria grows at a rate of 5% per hour, and initially there are
100 bacteria, find the population after 3 hours
...
05⋅3A = 100 \cdot e^{0
...
05⋅3
o Calculate: A=100⋅e0
...
1618=116
...
15} \approx 100
\cdot 1
...
18A=100⋅e0
...
1618=116
...


36
...

o log⁡2(16)=4\log_2(16) = 4log2(16)=4 because 16=2416 = 2^416=24,
o log⁡2(4)=2\log_2(4) = 2log2(4)=2 because 4=224 = 2^24=22
...


37
...

o Use the formula for the x-coordinate of the vertex:
xvertex=−(−4)2(2)=44=1x_{\text{vertex}} = \frac{-(-4)}{2(2)} = \frac{4}{4} =
1xvertex=2(2)−(−4)=44=1
o Substitute x=1x = 1x=1 into the equation to find the y-coordinate:
y=2(1)2−4(1)+1=2−y = 2(1)^2 - 4(1) + 1 = 2 -y=2(1)2−4(1)+1=2−

Reference: Books,
"Elementary Algebra" by Charles P
...
khanacademy
...
math
...
edu)

Algebra Formulas and Solutions 50+
Total 50+ Formulas & Solutions
...
Finding the Vertex of a Parabola (continued)



Formula: The vertex of the parabola y=ax2+bx+cy = ax^2 + bx + cy=ax2+bx+c is given
by: xvertex=−b2ax_{\text{vertex}} = \frac{-b}{2a}xvertex=2a−b
Solution: Find the vertex of the parabola y=2x2−4x+1y = 2x^2 - 4x + 1y=2x2−4x+1
...

o Substitute x=1x = 1x=1 into the equation to find the y-coordinate:
y=2(1)2−4(1)+1=2−4+1=−1y = 2(1)^2 - 4(1) + 1 = 2 - 4 + 1 = 1y=2(1)2−4(1)+1=2−4+1=−1
So, the vertex of the parabola is (1,−1)(1, -1)(1,−1)
...
Direct Variation



Formula: If yyy varies directly with xxx, then y=kxy = kxy=kx, where kkk is the
constant of variation
...

o First, find the constant of variation kkk: k=yx=62=3k = \frac{y}{x} = \frac{6}{2}
= 3k=xy=26=3
o Now, use the equation y=kxy = kxy=kx to find yyy when x=5x = 5x=5:
y=3⋅5=15y = 3 \cdot 5 = 15y=3⋅5=15
So, when x=5x = 5x=5, y=15y = 15y=15
...
Inverse Variation



Formula: If yyy varies inversely with xxx, then y=kxy = \frac{k}{x}y=xk, where kkk is
the constant of variation
...

o First, find the constant of variation kkk: k=y⋅x=4⋅2=8k = y \cdot x = 4 \cdot 2 =
8k=y⋅x=4⋅2=8
o Now, use the equation y=kxy = \frac{k}{x}y=xk to find yyy when x=6x = 6x=6:
y=86=43y = \frac{8}{6} = \frac{4}{3}y=68=34
So, when x=6x = 6x=6, y=43y = \frac{4}{3}y=34
...
Completing the Square


Formula: To complete the square for a quadratic equation ax2+bx+c=0ax^2 + bx + c =
0ax2+bx+c=0, follow these steps:

Reference: Books,
"Elementary Algebra" by Charles P
...
khanacademy
...
math
...
edu)

Algebra Formulas and Solutions 50+
Total 50+ Formulas & Solutions
...
If a≠1a \neq 1a =1, divide through by aaa
...
Move the constant term to the other side
...
Add (b2)2\left(\frac{b}{2}\right)^2(2b)2 to both sides
...
Factor the left-hand side as a perfect square
...

o Start by moving the constant term to the other side: x2+6x=7x^2 + 6x =
7x2+6x=7
o Add (62)2=9\left(\frac{6}{2}\right)^2 = 9(26)2=9 to both sides:
x2+6x+9=7+9x^2 + 6x + 9 = 7 + 9x2+6x+9=7+9 (x+3)2=16(x + 3)^2 =
16(x+3)2=16
o Now, take the square root of both sides: x+3=±4x + 3 = \pm 4x+3=±4
o Solve for xxx: x=−3±4x = -3 \pm 4x=−3±4 So, the solutions are:
x=−3+4=1orx=−3−4=−7x = -3 + 4 = 1 \quad \text{or} \quad x = -3 - 4 = 7x=−3+4=1orx=−3−4=−7
Therefore, x=1x = 1x=1 or x=−7x = -7x=−7
...
Cubic Formula




Formula: Solving cubic equations can be complex, and typically, they involve using a
numerical or graphical method
...

Solution: Solve x3−6x2+11x−6=0x^3 - 6x^2 + 11x - 6 = 0x3−6x2+11x−6=0
...
Try x=1x =
1x=1: 13−6(1)2+11(1)−6=1−6+11−6=01^3 - 6(1)^2 + 11(1) - 6 = 1 - 6 + 11 - 6 =
013−6(1)2+11(1)−6=1−6+11−6=0
o Since x=1x = 1x=1 is a root, factor x−1x - 1x−1 from the cubic polynomial using
synthetic division: x3−6x2+11x−6=(x−1)(x2−5x+6)x^3 - 6x^2 + 11x - 6 = (x 1)(x^2 - 5x + 6)x3−6x2+11x−6=(x−1)(x2−5x+6)
o Now, factor the quadratic x2−5x+6x^2 - 5x + 6x2−5x+6:
x2−5x+6=(x−2)(x−3)x^2 - 5x + 6 = (x - 2)(x - 3)x2−5x+6=(x−2)(x−3)
o So, the factored form is: (x−1)(x−2)(x−3)=0(x - 1)(x - 2)(x - 3) =
0(x−1)(x−2)(x−3)=0
o The solutions are x=1x = 1x=1, x=2x = 2x=2, and x=3x = 3x=3
...
Sine and Cosine Rule (Law of Sines and Cosines)


Formula:

Reference: Books,
"Elementary Algebra" by Charles P
...
khanacademy
...
math
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edu)

Algebra Formulas and Solutions 50+
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o Law of Cosines: c2=a2+b2−2ab⋅cos⁡Cc^2 = a^2 + b^2 - 2ab \cdot \cos
Cc2=a2+b2−2ab⋅cosC, where aaa, bbb, and ccc are the sides of the triangle, and
CCC is the included angle
...

o Apply the formula: c2=72+102−2(7)(10)⋅cos⁡(60∘)c^2 = 7^2 + 10^2 - 2(7)(10)
\cdot \cos(60^\circ)c2=72+102−2(7)(10)⋅cos(60∘) c2=49+100−140⋅12c^2 = 49 +
100 - 140 \cdot \frac{1}{2}c2=49+100−140⋅21 c2=49+100−70=79c^2 = 49 + 100
- 70 = 79c2=49+100−70=79 c=79≈8
...
89c=79≈8
...
89
...
Finding the Slope of a Line




Formula: The slope mmm of a line passing through two points (x1,y1)(x_1, y_1)(x1,y1)
and (x2,y2)(x_2, y_2)(x2,y2) is given by: m=y2−y1x2−x1m = \frac{y_2 - y_1}{x_2 x_1}m=x2−x1y2−y1
Solution: Find the slope of the line passing through points (3,4)(3, 4)(3,4) and (7,10)(7,
10)(7,10)
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45
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Solution: Find the equation of the line with slope m=3m = 3m=3 passing through the
point (2,5)(2, 5)(2,5)
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Standard Form of a Linear Equation


Formula: The standard form of the equation of a line is: Ax+By=CAx + By =
CAx+By=C where AAA, BBB, and CCC are constants, and AAA is non-negative
...
McKeague, Algebra for College Students" by Mark Dugopolski, "Higher Algebra" by Hall &
Knight, "College Algebra" by James Stewart, Lothar Redlin, and Saleem Watson, "Schaum's Outline of College Algebra" by
Murray Spiegel, Khan Academy (www
...
org), Paul's Online Math Notes (tutorial
...
lamar
...
Page 16 of 20


Solution: Convert the equation y−5=3(x−2)y - 5 = 3(x - 2)y−5=3(x−2) into standard
form
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47
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o Apply the formula: Sum of the roots=−−42=2\text{Sum of the roots} = -\frac{4}{2} = 2Sum of the roots=−2−4=2
So, the sum of the roots is 2
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Product of the Roots of a Quadratic Equation
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

Formula: For the quadratic equation ax2+bx+c=0ax^2 + bx + c = 0ax2+bx+c=0, the
product of the roots is given by: Product of the roots=ca\text{Product of the roots} =
\frac{c}{a}Product of the roots=ac
Solution: Find the product of the roots of the equation 2x2−4x+3=02x^2 - 4x + 3 =
02x2−4x+3=0
...


49
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o Express 16 as a power of 2: 2x=242^x = 2^42x=24
o Therefore, x=4x = 4x=4
...


50
...
McKeague, Algebra for College Students" by Mark Dugopolski, "Higher Algebra" by Hall &
Knight, "College Algebra" by James Stewart, Lothar Redlin, and Saleem Watson, "Schaum's Outline of College Algebra" by
Murray Spiegel, Khan Academy (www
...
org), Paul's Online Math Notes (tutorial
...
lamar
...
Page 17 of 20
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

Formula: If the equation is of the form log⁡b(x)=y\log_b(x) = ylogb(x)=y, then solve
for xxx as: x=byx = b^yx=by
Solution: Solve log⁡2(x)=5\log_2(x) = 5log2(x)=5
...


51
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The sum is: Sn=2n−12−1=2n−1S_n = \frac{2^n - 1}{2 - 1} = 2^n - 1Sn=2−12n−1=2n−1
Solution: Find the sum of the first 6 terms of the powers of 2
...


52
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Solution: Solve x2+5x+6=0x^2 + 5x + 6 = 0x2+5x+6=0 by factoring
...


53
...
McKeague, Algebra for College Students" by Mark Dugopolski, "Higher Algebra" by Hall &
Knight, "College Algebra" by James Stewart, Lothar Redlin, and Saleem Watson, "Schaum's Outline of College Algebra" by
Murray Spiegel, Khan Academy (www
...
org), Paul's Online Math Notes (tutorial
...
lamar
...
Page 18 of 20


Solution: Simplify 35⋅3234\frac{3^5 \cdot 3^2}{3^4}3435⋅32
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54
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Solution: Solve the system of equations: y=2x+3y = 2x + 3y=2x+3 x+y=7x + y =
7x+y=7
o Substitute y=2x+3y = 2x + 3y=2x+3 into the second equation: x+(2x+3)=7x + (2x
+ 3) = 7x+(2x+3)=7
o Simplify: 3x+3=73x + 3 = 73x+3=7 3x=43x = 43x=4 x=43x = \frac{4}{3}x=34
o Substitute x=43x = \frac{4}{3}x=34 into y=2x+3y = 2x + 3y=2x+3:
y=2(43)+3=83+3=83+93=173y = 2\left(\frac{4}{3}\right) + 3 = \frac{8}{3} + 3
= \frac{8}{3} + \frac{9}{3} = \frac{17}{3}y=2(34)+3=38+3=38+39=317
So, the solution is x=43x = \frac{4}{3}x=34 and y=173y = \frac{17}{3}y=317
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Solving Systems of Equations (Elimination Method)



Formula: The elimination method involves adding or subtracting the equations to
eliminate one variable
...


Let me know if you would
Reference: Books,
"Elementary Algebra" by Charles P
...
khanacademy
...
math
...
edu)

Algebra Formulas and Solutions 50+
Total 50+ Formulas & Solutions
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"Elementary Algebra" by Charles P
...

2
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3
...

4
...

5
...


Online Resources
1
...
khanacademy
...

2
...
math
...
edu)
A reliable resource for algebraic and calculus-based problem-solving
...
Wolfram MathWorld (mathworld
...
com)
Detailed mathematical explanations and derivations for various algebraic topics
...
Brilliant
...
brilliant
...

5
...
purplemath
...


Research Papers and Journals
1
...

2
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Historical Reference
For classical algebraic techniques, Isaac Newton’s works on polynomials, symmetry, and series
are foundational
...
McKeague, Algebra for College Students" by Mark Dugopolski, "Higher Algebra" by Hall &
Knight, "College Algebra" by James Stewart, Lothar Redlin, and Saleem Watson, "Schaum's Outline of College Algebra" by
Murray Spiegel, Khan Academy (www
...
org), Paul's Online Math Notes (tutorial
...
lamar
...
Page 20 of 20
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

"Principia Mathematica" by Isaac Newton
Scholarly articles analyzing Newton's work in mathematics and its relevance to algebra
...
Let me know!

Reference: Books,
"Elementary Algebra" by Charles P
...
khanacademy
...
math
...
edu)



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Title: Algebra Formulas and Solutions
Description: Algebra Formulas and Solutions Algebra formulas are fundamental in solving equations and understanding mathematical relationships. These formulas are used to simplify and manipulate expressions, making it easier to find unknown values. Below is an overview of key algebra formulas along with detailed solutions to illustrate their application:

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