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arlik [135]
2 years ago
7

Solve the equation by completing the square.  ^x2- 12x = -28

Mathematics
2 answers:
vova2212 [387]2 years ago
5 0

Answer:

x = 6 + 2√2  and  x = 6 - 2√2

Step-by-step explanation:

Please write the " ^ " symbol between x and 2 here:  x^2 - 12x = -28.

Take half the coefficient of x; here that is -6.

Square this, obtaining 36.

Add 36 to, and then subtract 36 from, x^2 - 12x:  x^2 - 12x + 36 - 36.

Rewrite  x^2 - 12x + 36 as the square of a binomial:  (x - 6)^2.

Then rewrite  x^2 - 12x = -28  as (x - 6)^2 - 36 = -28.

Simplifying, (x - 6)^2 = 8.

Taking the square root of both sides, x - 6 = plus or minus 2√2.


Then the solutions are x = 6 + 2√2  and  x = 6 - 2√2

Olegator [25]2 years ago
3 0

x^2 -12 x = -28

(-12/2) ^ 2  =36

x^2 -12 x + 36 = -28 + 36

x^2 -12x + 36 = 8

(x-6)^2 = 8

take the square root of each side

x-6 = sqrt (8)    x-6 = -sqrt (8)  

you get a positive and a negative when taking                                                                 the square root

x= 6 + sqrt (8)  

x = 6 - sqrt (8)

sqrt (8) = 2 sqrt (2)

x= 6 + 2 sqrt (2)

x = 6 - 2 sqrt (2)

Answer:

x= 6 + 2 sqrt (2)

x = 6 - 2 sqrt (2)



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Step-by-step explanation:

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Step 2: Subtract 120 from both sides.

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Now that we know the value of x, we can solve for y in any of the equations, but let's use the equation "y = −5x + 8"

\mathrm{So\:it\:would\:look\:like\:this:\ y =  -5 \left(  \dfrac{ 53  }{ 30  }    \right)  +8}

\mathrm{Now\:lets\:solve\:for\:"y"\:then}

y =  -5 \left(  \dfrac{ 53  }{ 30  }    \right)  +8}

\mathrm{Express\: -5 \times   \dfrac{ 53  }{ 30  }\:as\:a\:single\:fraction}

y =   \dfrac{ -5 \times  53  }{ 30  }  +8

\mathrm{Multiply\:-5 \:and\:53\:to\:get\:-265 }

y =   \dfrac{ -265  }{ 30  }  +8

\mathrm{Simplify\:  \dfrac{ -265  }{ 30  }    \:,by\:dividing\:both\:-265\:and\:30\:by\:5} }

y =   \dfrac{ -265 \div  5  }{ 30 \div  5  }  +8

\mathrm{Simplify}

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\mathrm{Multiples\:of\:1: \:1,2,3,4,5,6}

\mathrm{Multiples\:of\:6: \:6,12,18,24,30,36,42,48}

\mathrm{Convert\:8\:to\:fraction\:\dfrac{ 48  }{ 6  }}

y =  - \dfrac{ 53  }{ 6  }  + \dfrac{ 48  }{ 6  }

\mathrm{Since\: - \dfrac{ 53  }{ 6  }\:have\:the\:same\:denominator\:,\:add\:them\:by\:adding\:their\:numerators}

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\mathrm{Add\: -53 \: and\: 48\: to\: get\:  -5}

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\mathrm{The\:solution\:is\:the\:ordered\:pair\:(\dfrac{ 53  }{ 30  }, - \dfrac{ 5  }{ 6  })}

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