Question 197703
Q: Can you use measures of central tendancy (mean,median,mode) in quadratic functions?


A: The mean, median, and mode are totally different concepts compared to the quadratic equation. I don't see you using them in these problems.


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Q: Do you have to find the GCF for all of the problems?


A: You don't have to, but it helps simplify things (sometimes). 



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As for the other questions, it's probably best to show you examples...


# 1 Square Root Method:


Example: Let's solve {{{x^2=81}}}



{{{x^2=81}}} Start with the given equation.



{{{x=""+-sqrt(81)}}} Take the square root of both sides to "undo" the square.



{{{x=sqrt(81)}}} or {{{x=-sqrt(81)}}} Break up the "plus/minus" to form two equations.



{{{x=9}}} or {{{x=-9}}} Evaluate the square root of 81 to get 9.




So the solutions are {{{x=9}}} or {{{x=-9}}}




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# 2 Graphing



Example: Let's solve {{{x^2+5x+6=0}}}. If we graph {{{y=x^2+5x+6}}}, we get:



{{{ drawing(500, 500, -10, 10, -10, 10,
 graph( 500, 500, -10, 10, -10, 10,x^2+5x+6)

)}}}


Graph of {{{y=x^2+5x+6}}}



From the graph, we see that the curve intersects with the x-axis at {{{x=-3}}} and {{{x=-2}}}. So the solutions are {{{x=-3}}} or {{{x=-2}}}




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# 3 Factoring



Example: Let's solve {{{x^2+6x+8=0}}}



First, we need to factor {{{x^2+6x+8}}}



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Looking at the expression {{{x^2+6x+8}}}, we can see that the first coefficient is {{{1}}}, the second coefficient is {{{6}}}, and the last term is {{{8}}}.



Now multiply the first coefficient {{{1}}} by the last term {{{8}}} to get {{{(1)(8)=8}}}.



Now the question is: what two whole numbers multiply to {{{8}}} (the previous product) <font size=4><b>and</b></font> add to the second coefficient {{{6}}}?



To find these two numbers, we need to list <font size=4><b>all</b></font> of the factors of {{{8}}} (the previous product).



Factors of {{{8}}}:

1,2,4,8

-1,-2,-4,-8



Note: list the negative of each factor. This will allow us to find all possible combinations.



These factors pair up and multiply to {{{8}}}.

1*8
2*4
(-1)*(-8)
(-2)*(-4)


Now let's add up each pair of factors to see if one pair adds to the middle coefficient {{{6}}}:



<table border="1"><th>First Number</th><th>Second Number</th><th>Sum</th><tr><td  align="center"><font color=black>1</font></td><td  align="center"><font color=black>8</font></td><td  align="center"><font color=black>1+8=9</font></td></tr><tr><td  align="center"><font color=red>2</font></td><td  align="center"><font color=red>4</font></td><td  align="center"><font color=red>2+4=6</font></td></tr><tr><td  align="center"><font color=black>-1</font></td><td  align="center"><font color=black>-8</font></td><td  align="center"><font color=black>-1+(-8)=-9</font></td></tr><tr><td  align="center"><font color=black>-2</font></td><td  align="center"><font color=black>-4</font></td><td  align="center"><font color=black>-2+(-4)=-6</font></td></tr></table>



From the table, we can see that the two numbers {{{2}}} and {{{4}}} add to {{{6}}} (the middle coefficient).



So the two numbers {{{2}}} and {{{4}}} both multiply to {{{8}}} <font size=4><b>and</b></font> add to {{{6}}}



Now replace the middle term {{{6x}}} with {{{2x+4x}}}. Remember, {{{2}}} and {{{4}}} add to {{{6}}}. So this shows us that {{{2x+4x=6x}}}.



{{{x^2+highlight(2x+4x)+8}}} Replace the second term {{{6x}}} with {{{2x+4x}}}.



{{{(x^2+2x)+(4x+8)}}} Group the terms into two pairs.



{{{x(x+2)+(4x+8)}}} Factor out the GCF {{{x}}} from the first group.



{{{x(x+2)+4(x+2)}}} Factor out {{{4}}} from the second group. The goal of this step is to make the terms in the second parenthesis equal to the terms in the first parenthesis.



{{{(x+4)(x+2)}}} Combine like terms. Or factor out the common term {{{x+2}}}



So {{{x^2+6x+8}}} factors to {{{(x+4)(x+2)}}}.



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Now let's start solving {{{x^2+6x+8=0}}}



{{{x^2+6x+8=0}}} Start with the given equation.



{{{(x+4)(x+2)=0}}} Factor (see steps above)



{{{x+4=0}}} or {{{x+2=0}}} Set each factor equal to zero



{{{x=-4}}} or {{{x=-2}}} Solve for "x" in each equation



So the solutions are {{{x=-4}}} or {{{x=-2}}} 




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# 4 Quadratic Formula



Example: Let's solve {{{x^2-22x+117=0}}}




{{{x^2-22x+117=0}}} Start with the given equation.



Notice we have a quadratic in the form of {{{Ax^2+Bx+C}}} where {{{A=1}}}, {{{B=-22}}}, and {{{C=117}}}



Let's use the quadratic formula to solve for "x":



{{{x = (-B +- sqrt( B^2-4AC ))/(2A)}}} Start with the quadratic formula



{{{x = (-(-22) +- sqrt( (-22)^2-4(1)(117) ))/(2(1))}}} Plug in  {{{A=1}}}, {{{B=-22}}}, and {{{C=117}}}



{{{x = (22 +- sqrt( (-22)^2-4(1)(117) ))/(2(1))}}} Negate {{{-22}}} to get {{{22}}}. 



{{{x = (22 +- sqrt( 484-4(1)(117) ))/(2(1))}}} Square {{{-22}}} to get {{{484}}}. 



{{{x = (22 +- sqrt( 484-468 ))/(2(1))}}} Multiply {{{4(1)(117)}}} to get {{{468}}}



{{{x = (22 +- sqrt( 16 ))/(2(1))}}} Subtract {{{468}}} from {{{484}}} to get {{{16}}}



{{{x = (22 +- sqrt( 16 ))/(2)}}} Multiply {{{2}}} and {{{1}}} to get {{{2}}}. 



{{{x = (22 +- 4)/(2)}}} Take the square root of {{{16}}} to get {{{4}}}. 



{{{x = (22 + 4)/(2)}}} or {{{x = (22 - 4)/(2)}}} Break up the expression. 



{{{x = (26)/(2)}}} or {{{x =  (18)/(2)}}} Combine like terms. 



{{{x = 13}}} or {{{x = 9}}} Simplify. 



So the solutions are {{{x = 13}}} or {{{x = 9}}}