Question 118217
My directions say to use the quadratic formula to solve.
3x^2 – 2x = 15x – 10 
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The quadratic formula is based on the form ax^2 + bx + c = 0
:
Put your problem in that format:
3x^2 - 2x = 15x - 10
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3x^2 - 2x - 15x = -10; subtract 15x from both sides
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3x^2 -17x = -10; combined like terms
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3x^2 -17x + 10 = 0; added 10 to both sides, this is format we want
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In this problem a=3; b=-17; c=10
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The quadratic formula:
{{{x = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}}
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Substitute our values for a, b, and c
{{{x = (-(-17) +- sqrt(-17^2 - 4 * 3 * 10 ))/(2*3) }}}
:
{{{x = (17 +- sqrt(289 - 120))/(6) }}}; it's +17, minus a minus is plus
:
{{{x = (17 +- sqrt(169))/6 }}}
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{{{x = (17 +- 13)/6}}}; found the square root of 169
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Two solutions:
{{{x = (17 + 13)/6}}};
{{{x = 30/6}}}
x = +5
and
{{{x = (17 - 13)/6}}};
{{{x = 4/6}}}
{{{x = +2/3}}}
:
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Check both solution in the original equation to make sure we did not make a mistake somewhere.
x = 5
3x^2 – 2x = 15x – 10 
3(5^2) - 2(5) = 15(5) - 10
3(25) - 10 = 75 - 10; equality reigns, a good solution
and
x = {{{2/3}}}
3(2/3)^2 - 2(2/3) = 15(2/3) - 10
3(4/9) - (4/3) = 10 - 10
(12/9) - (4/3) = 0; checks OK too
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Could you follow this OK, Any questions?