Question 168786
Hi, Hope I can help,
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If you are wanting us to solve this quadratic equation, here is how you do it.
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{{{ y^2-6y+ 9 }}}
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The easiest way you can solve this is by factoring, in this case we can
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First we need to write the "y"'s in parentheses
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It is only "y" squared so there will be two "y"'s
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( y   )(y   )
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Next, we need to find the rest of the factors of this quadratic
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If the factors of the third term, in this case "9", add up to the middle term( in this case (-6)), it can be factored
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factors of "9" added together
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(9) + (1) = 10
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(-9) + (-1) = (-10)
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(3) + (3) = 6
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(-3) + (-3) = (-6)
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The last pair of factors add up to the middle term
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Now we will put the factors in the parentheses with the "y"'s
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( y   )(y   ) = (y - 3 )(y - 3 )
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This is the factored quadratic
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( y - 3 )(y - 3 ) = 0, you can check by using the FOIL method ( First, Inner, Outer, Last )
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{{{ ( y - 3 )(y - 3 ) }}} = (F) {{{ ( highlight(y) - 3 )(highlight(y) - 3 ) }}} = (O) {{{ ( highlight(y) - 3 )(y - highlight(3) ) }}} = (I) {{{ ( y - highlight(3) )(highlight(y) - 3 ) }}} = (L) {{{ ( y - highlight(3) )(y - highlight(3) ) }}}, Remember the negative signs on the "3's"
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{{{ y^2 - 3y - 3y + 9 }}}, combine the like terms, {{{ (y^2) highlight(- 3y - 3y) + 9 }}} (forget about the multiplication sign, it is just being weird) = {{{ y^2 - 6y + 9 }}} ( True )
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 now we take each of the factors and have them equal "0", then solve for "y"
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First factor = ( y - 3 )
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{{{ ( y - 3 ) = 0 }}}, removing the parentheses
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{{{  y - 3  = 0 }}}, moving (-3) to the right side
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{{{  y - 3 + 3  = 0 + 3 }}} = {{{ y = 3 }}}
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{{{ y = 3 }}}, the other factor is the same, so we would get the same answer.
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your answer is {{{ y = 3 }}}, you can check by replacing "y" with "3" in the original equation
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{{{ y^2-6y+ 9 = 0 }}} = {{{ (3)^2-6(3)+ 9 = 0 }}} = {{{ (9)-(18)+ 9 = 0 }}} = {{{ (-9) + 9 = 0 }}} = {{{ 0 = 0 }}} ( True )
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You can also use the quadratic formula, for every quadratic equation, make sure your equation is in the form {{{ ax^2 + bx + c = 0 }}} ( in this case "y" ) it is
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{{{ y^2-6y+ 9 }}}
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a = 1
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b = -6
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c = 9
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{{{y = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}} = {{{y = (-(-6) +- sqrt( (-6)^2-4*(1)*(9) ))/(2*(1)) }}} = {{{y = ( 6 +- sqrt( (36)-36 ))/(2)) }}} = {{{y = ( 6 +- sqrt (0))/(2)) }}} = {{{y = ( 6 +- 0)/(2)) }}} = {{{y = ( 6 /2) }}} = {{{y = 3 }}}
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Your answer is {{{ y = 3 }}}
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Hope I helped, Levi