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When faced with a quadratic equation on the GMAT which are the best 2 ways to solve?
 1) Factoring by inspection
 2) Plugging in the answer choices

Is x^{2}+2x=24 a quadratic equation?
 No  must be x^{2}+2x24=0
 Must = 0

What are the 4 steps for factoring a quadratic equation?
 1) Bring down the first sign
 2) Look at the 2nd sign, if neg means different from the first sign
 3) Figure out multiples of the last # & choose #s which also add/sub to middle #
 4) Place the larger # on left

Regarding Linear Equations with multiple variables, how do you decide which method to use?
Use the substitution method when it is not easy to cancel out terms. But use the direct approach on all other problems (when you can cancel out terms by adding two equations together).




x^{a }∙ x^{b} = ?
x^{a+b}

y^{a} ∙ x^{b} = ?
y^{a}x^{b} (can't combine nonlike terms)



(x∙y)^{a }= ?
x^{a}∙y^{a
}You can distribute the exponent to each variable

( x )^{a} = ?
_ y
parenthesis is around x & y together
For division you can also distribute the exponent to each variable


4^{3}=2^{x}
x = ?
Factor 4 so we have same base to solve
4^{3}=2^{x} > (2^{2})^{3} = 2^{x} > 2^{6}=2^{x} > x=6

If each side of the equation has identical prime bases...
then the exponents must equal one another
(thus the goal is to make each side of the equation have the same prime #s and then equate the exponents)

2^{x}∙3^{y}∙5^{z} = 2^{5}∙3^{4}∙5^{5
}
x=? y=? z=?
x=5, y=4, z=5

Whenever you have multiplication or division of #s or variables with exponents use...
the exponent rules

Whenever you have addition & subtraction of #s or variables with exponents...
rely on your understanding of factoring to combine like terms and then use any necessary exponent rules

Can you add/subtract #s/variables with different exponents and/or different bases?
No  to add/subtract must have same base & same exponent.
(Try to change the add/sub into multiplication by factoring)

3^{5}∙3^{8}∙3^{7}∙3^{6} = ?
 Factor out 3^{5} from each #
 3^{5}(1+3^{3}3^{2}3) = 3^{5}(1+2793) = 16(3^{5})

Exponent Rules: x^{a} ∙ x^{b} =
x^{a+b}
same base, add exponents

Exponent Rules: x^{a}/x^{b} =
x^{ab
}When dividing same base, subtract exponents

Exponent Rules: (x^{a})^{b} =
x^{a∙b}
When raising a variable with an exponent to another power, multiply the exponents

x/x^{a} =
x^{1a
}Same as the exponent division rule (1 is just implied)

x^{(3)^2} =
x^{9
}In this case the parenthesis location is important. 3^{2} = 9

Exponent rule: (x∙y)^{a} =
x^{a}∙y^{a
Distribute the exponent to each variable122 = (3∙4)2 = 32∙42 = 144
}

Exponent rule: x^{(a/b)} =
(x ^{1/b})^{a} = (x^{a})^{1/b} = ^{b}√(x^{a})

Remember: If each side of the equation has identical prime bases, then the exponents must equal one another.
2^{x}∙3^{y}∙5^{z} = 2^{5}∙3^{4}∙5^{5} so x=5, y=4, z=5

For addition/subtraction of exponents rely on factoring & combining like terms
3^{5}+3^{8}3^{7}3^{6}= 3^{5} (1+3^{3}3^{2}3)=
3^{5} (1+2793) = 16(3^{5})

(^{4}√25)^{2} = (25^{1/4})^{2} = 25^{2/4} = 25^{1/2}
= √25 = 5


64 ^{2/3} =
(^{3}√64)^{2} = (4)^{2} = 16



Is it ok to have roots in the denominator?
No, not standard


What's the strategy/approach for solving linear equations with one unknown?
Isolate the unknown on one side of the equation

If you have two linear equations with two unknowns that are equivalent equations, how many solutions are there?
Infinite

If you have two linear equations with two unknowns that are NOT equivalent to eachother, how many solutions are there?
One or None

3x + 4y = 17
6x + 8y = 35
These two equations are examples of equations which aren't equal and they contradict one another. Thus there is no solution which can satisfy both.

What is the substituation approach in regards to linear equations with two unknowns?
 1) Express one of the unknowns in terms of the other using one of the equations
 2) Substitute the expression into the remaining equation, to cancel out one unknown.
 3) Then solve for the unknown
 4) Plug the solution for the one unknown back into one of the orig. equations & solve for the other unknown.

What is the direct approach in regards to linear equations with two unknowns?
 1) Make the coefficients of one of the unknowns the same, so that they cancel out when you add/subtract the equations together.
 2) Add the equations together (leaving one unknown). Solve for it.
 3) Plug the solution for the one unknown back into one of the equations and solve for the other unknown.

What are the 2 methods/approaches for solving two linear equations with two unknowns?
 1) Substitition method
 2) Direct method (cancel out an unknown)

Remember some equations can be solved by factoring. What are the steps to solving by factoring?
 1) Add/Subtract expressions to bring them all on one side and set = 0.
 2) Try to factor out the side with all the expressions
 3) Keep combining/simplifying until you see the solution(s) to make the expressions = 0.

The solutions of an equation are also called the ____ of the equation.
roots

What's the standard form for a quadratic equation?
ax^{2} + bx + c = 0

How many solutions can a quadratic equation have?
Two, One, or None.

In regards to functions f(x) is referred to as what? And x is what?
f(x) is the corresponding output to the input (x).

The set of all allowable inputs for a function is called?
The domain of the function. (i.e. the limitations/rules of what x can be).

a(n) = n^{2} + n/5, for n=0, 1, 2, 3...
is an example of what type of function?
a sequence and noted as a_{n}

An even root has how many answer choices vs. an odd root?
 even roots = 2 solutions (+ and )
 odd roots = 1 solution

(x + y)^{2} =
x^{2 }+ 2xy + y^{2}

(x  y)^{2} =
x^{2 }2xy + y^{2}

(x+ y)(x  y) =
x^{2} y^{2}


