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notka56 [123]
3 years ago
7

‏Good morning, what is the

Mathematics
1 answer:
kifflom [539]3 years ago
3 0

Heh- i saw u like 5 times already today

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Solution<span><span> 
1. x =(12-√228)/2=6-√ 57 = -1.550</span><span> 
2. x =(12+√228)/2=6+√ 57 = 13.550
 H3LL0 this is How to get this^ is down there i suggest you take some notesL0L  

</span></span>The first term is, <span> <span>x2</span> </span><span> its coefficient is </span><span> 1 </span>.
<span>The middle term is, </span><span> -12x </span><span> its coefficient is </span><span> -12 </span>.
<span>The last term, "the constant", is </span><span> -21 
</span>
Multiply the coefficient of the first term by the constant <span> <span> 1</span> • -21 = -21
</span>

Find two factors of  -21  whose sum equals the coefficient of the middle term, which is  <span> -12 </span>.

<span><span>  -21   +   1   =   -20</span><span>    
  -7   +   3   =   -4</span><span>    
  -3   +   7   =   4</span><span>     
  -1   +   21   =   20

</span></span>Find the Vertex of   <span>y = x2-12x-21
</span>For any parabola,<span>Ax2+Bx+C,</span><span>the </span> x <span>-coordinate of the vertex is given by </span> -B/(2A) is 6.0000  <span>Plugging into the parabola formula </span>  6.0000. <span> for </span> x <span> we can calculate the </span> y -coordinate :<span> 
 </span><span> y = 1.0 * 6.00 * 6.00 - 12.0 * 6.00 - 21.0 
</span><span>or </span>  y = -57.000 <span>Root plot for : </span><span> y = x2-12x-21</span>
<span>Axis of Symmetry (dashed) </span> {x}={ 6.00} 
<span>Vertex at </span> {x,y} = { 6.00,-57.00}  
 x -Intercepts (Roots) :
Root 1 at  {x,y} = {-1.55, 0.00} 
Root 2 at<span>  {x,y} = {13.55, 0.00} 
</span>
read the squar <span> 2.2 </span>    Solving  <span> x2-12x-21 = 0</span><span> by Completing The Square</span><span> .

 </span><span>Add </span><span> 21 </span><span> to both side of the equation : </span>
  <span> x2-12x = 21 </span>Now the clever bit:<span> Take the coefficient of </span> x <span>, which is </span><span> 12</span><span> , divide by two, giving </span><span> 6</span><span> , and finally square it giving </span><span> 36</span><span> 
</span>Adding <span> 36</span>  has completed the left hand side into a perfect square :
  <span> <span>x2-12x+36</span> </span> =
   (x-6) • (x-6)  =
  <span>(x-6)2 
</span>Since
  <span> x2-12x+36 = 57</span><span> and</span>
  <span> x2-12x+36 = (x-6)2 
</span>then, according to the law of transitivity,
  <span> (x-6)2 = 57 
</span>We'll refer to this Equation as <span> Eq. #2.2.1  
</span>The <span>Square Root Principle </span><span>says that When two things are equal, their square roots are equal.
</span>the square root of <span><span>(x-6)2 </span> </span><span> is</span>
  <span> <span>(x-6)2/2</span> =
  <span>(x-6)1</span> =
   <span>x-6
</span></span>Now, applying the Square Root Principle to  Eq. #2.2.1  we get:
  <span> x-6 = <span>√<span> 57 
</span></span></span>Add <span> 6 </span><span> to both sides to obtain:</span>
  <span> x = 6 + √<span> 57 
</span></span>Since a square root has two values, one positive and the other negative
  <span> x2 - 12x - 21 = 0</span>
   has two solutions:
  <span>x = 6 + √<span> 57 </span></span>
   or
  <span>x = 6 - √<span> 57 
</span></span>using the Quadratic Formula

<span> 2.3 </span>    Solving   <span> x2-12x-21 = 0</span> by the Quadratic Formula<span> .

 </span>According to the Quadratic Formula,<span>  x  ,</span> the solution for  <span><span> Ax2+Bx+C </span> = <span>0 </span> </span>, where  A, B  and  C  are numbers, often called coefficients, is given by :
<span>                                     
            - B  ±  √ B2-4AC
  x =   ————————
                      2A 

</span>  In our case,  <span>A   =     1
                      B   =   -12
                      C   =  -21 

</span>Accordingly,  <span>B2  -  4AC   =
                     144 - (-84) =
                     228

</span>Applying the quadratic formula :

               <span>12 ± √<span> 228 </span>
   x  =    ——————
</span>

<span>                      2
</span> <span>The prime factorization of </span> 228   is
  <span> 2•2•3•19</span><span>  
</span>To be able to remove something from under the radical, there have to be <span> 2 </span><span> instances of it because we are taking a </span>square<span> i.e.</span><span> second </span>root
<span>√<span> 228 </span>  =  √<span> 2•2•3•19 </span>  =
                ±  <span>2 </span>• √<span> 57 </span></span>

<span>  √<span> 57 </span></span><span>  , rounded to 4 decimal digits, is </span><span> <span> 7.5498
</span></span> So now we are looking at:
           x  =  ( 12 ± 2 •  7.550 ) / 2

Two real solutions:

<span> x =(12+√228)/2=6+√<span> 57 </span>= 13.550 </span>

or:

<span> x =(12-√228)/2=6-√<span> 57 </span>= -1.550 </span>
 

6 0
3 years ago
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