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Example Questions
Example Question #2274 : Algebra 1
Solve for :
None of the other answers
To solve this equation, you must first eliminate the exponent from the by taking the square root of both sides:
Since the square root of 36 could be either or , there must be 2 values of . So, solve for
and
to get solutions of .
Example Question #1 : Quadratic Roots
Find the roots of .
When we factor, we are looking for two number that multiply to the constant, , and add to the middle term, . Looking through the factors of , we can find those factors to be and .
Thus, we have the factors:
.
To solve for the solutions, set each of these factors equal to zero.
Thus, we get , or .
Our second solution is, , or .
Example Question #1451 : Algebra Ii
Write a quadratic function in standard form with roots of -1 and 2.
From the zeroes we know
Use FOIL method to obtain:
Example Question #1453 : Algebra Ii
Select the quadratic equation that has these roots:
None of these.
FOIL the two factors to find the quadratic equation.
First terms:
Outer terms:
Inner terms:
Last terms:
Simplify:
Example Question #1454 : Algebra Ii
Solve for a possible root:
Write the quadratic equation.
The equation is in the form .
Substitute the proper coefficients into the quadratic equation.
The negative square root can be replaced by the imaginary term . Simplify square root 60 by common factors of numbers with perfect squares.
Simplify the fraction.
A possible root is:
Example Question #1455 : Algebra Ii
Solve for the roots (if any) of
Pull out a common factor of negative four.
The term inside the parentheses can be factored.
Set the binomials equal to zero and solve for the roots. We can ignore the negative four coefficient.
The answers are:
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