Question 173959
Hi, Hope I can help,
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Can you please help me find the solution for {{{x^2+20x}}}
your help is greatly appreciated.
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I believe this is the equation, {{{x^2+20x = 0 }}}
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For this equation, we can just factor this polynomial, and find the solutions
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We can factor out an "x"
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{{{x^2+20x = 0 }}} = {{{ x(x+20) = 0 }}}
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There are two solutions, so we have two equations, we have to let both of the factors equal "0", then solve for "x"
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(x), {{{ x = 0 }}}, {{{ x = 0 }}} is our first answer
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(x+20), {{{ x+20 = 0 }}} = {{{ x + 20 - 20 = 0 - 20 }}} = {{{ x = -20 }}}, this is our second answer
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{{{ x = 0 }}} and {{{ x = -20 }}}, are your two solutions
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You can always use the quadratic equation
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{{{x^2+20x = 0 }}}, you can also say {{{x^2+20x + 0 = 0 }}}
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{{{x^2+20x + 0 = 0 }}}, quadratic equations are given in the form {{{ax^2+bx + c = 0 }}}
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Quadratic equation = {{{x = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}}
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{{{x^2+20x + 0 = 0 }}}, {{{ax^2+bx + c = 0 }}}
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Our equation
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a = 1
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b = 20
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c = 0
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Now just replace the letters with the numbers
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{{{x = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}} = {{{x = (-(20) +- sqrt( (20)^2-4*(1)*(0) ))/(2*(1)) }}} = {{{x = (-20 +- sqrt( 400-0 ))/2 }}} = {{{x = (-20 +- sqrt( 400))/2 }}} = {{{x = (-20 +- 20)/2 }}}
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{{{x = (-20 + 20)/2 }}} = {{{x = 0/2 }}} = {{{ x = 0 }}}
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{{{x = (-20 - 20)/2 }}} = {{{ x = (-40)/2 }}} = {{{ x = -20 }}}
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{{{ x = 0 }}} and {{{ x = -20 }}} are your two solutions
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Hope I helped, Levi