Question 169895
your equation is:
{{{(1/3)x^2 + (2/9)x - 1 = 0}}}
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quadratic formula is x = {{{(-b +- root(2,b^2-4ac))/(2a)}}}
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standard form of the quadratic equation is {{{ax^2 + bx + c = 0}}}
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you can make your calculations easier by multiplying each side of the equation by 9.
this will remove the denominators.
since the right side of the equation was 0, it will remain 0.
doing that, your formula becomes:
{{{3x^2 + 2x - 9 = 0}}}
comparing to the standard form, you get
a = 3
b = 2
c = -9
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plugging these values into the quadratic formula, you get:
x = {{{(-(2) +- root(2,(2)^2-4*(3)*(-9)))/(2*(3))}}}
your answer is:
x = 1.430500874
or:
x = -2.097167541
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plugging these values into your original equations proves these values are good if the equation stays true.
it does.
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it is much easier to see if you have an equation that factors easily.
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take (x-1)*(x+2) = 0
these factors are chosen so that the resulting equation factors easier when using the quadratic equation to solve it.
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the resulting equation becomes x^2 +x-2=0
use the quadratic formula to solve this:
a = 1
b = 1
c = -2
quadratic formula of x = {{{(-b +- root(2,b^2-4ac))/(2a)}}}
becomes: x = {{{(-(1) +- root(2,(1)^2-4*(1)*(-2)))/(2*(1))}}}
this becomes: x = {{{(-(1) +- root(2,9))/(2*(1))}}}
which becomes: x = {{{(-(1) +- 3)/(2)}}}
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x = (-1+3)/2 = 1
or:
x = (-1-3)/2 = -2
if x = 1, then (x-1) = 0
if x = -2, then (x+2) = 0
these are the factors you started with in creating this problem so the answers come out good.