Question 148962
I need some help with this problem. I have tried to work it and cannot seem to come up with the answer All the directions say is to complete the square to solve the following equations
x^2+6x=7 ( i know that you need to get every thing on 1 side but then what)
x^2+6x-7=0 
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If they want you to solve by completing the square you can leave 7 on the right
And find the value that will complete the square
:
x^2 + 6x + __ = 7
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To find the term that will complete the square, 
divide the coefficient of x by 2 and square that.  6/2 = 3, 3^2 = 9, so we have:
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x^2 + 6x + 9 = 7 + 9; we have to add 9 to both sides to preserve equality
:
x^2 + 6x + 9 = 16; we know that left side is (x+3)(x+3) or (x+3)^2, so we have:
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(x + 3)^2 = 16
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Find the square root of both sides;
x + 3 = +/-{{{sqrt(16)}}}
x + 3 = +/-4
Two solutions:
x = -3 + 4
x = +1
and
x = -3 - 4
x = -7
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You can confirm that: FOIL (x-1)(x+7) = x^2 - 1x + 7x - 7 = x^2 + 6x - 7
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and with this problem the same directions but would you us the quadratic formula to solve? 2x^2 - 5x - 3 = 0
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you could use the quadratic formula, but this will factor to: (2x + 1)(x - 3)
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If you insist on using the quadratic formula:
{{{x = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}} 
In your equation: a=2; b=-5; c=-3
{{{x = (-(-5) +- sqrt( -5^2-4*2*-3 ))/(2*2) }}} 
{{{x = (5 +- sqrt(25 - (-24) ))/(4) }}}
{{{x = (5 +- sqrt(25 + 24 ))/(4) }}}
{{{x = (5 +- sqrt(49))/(4) }}}
Two solutions
x = {{{(5 + 7)/4}}}
x = +3
and
x = {{{(5 - 7)/4}}}
x = {{{-1/2}}}
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Did this help you? Any questions?