Question 259986
give exact and approximate solutions to three decimal places: y^2-8y+16=1
<pre><font size = 4 color = "indigo"><b>
{{{y^2-8y+16=1}}}

Subtract 1 from both sides to get 0 on the right side:

{{{y^2-8y+15=0}}}

You can factor that as

{{{(y-3)(y-5)=0}}}

Set each equal to 0:

{{{y-3=0}}}
{{{y=3}}}

{{{y-5=0}}}
{{{y=5}}}

But since you mentioned decimal approximating that
made me think you were supposed to do it using
the quadratic formula:

You would use {{{y}}} for {{{x}}}, {{{1}}} for {{{a}}},

{{{-8}}} for {{{b}}} and {{{15}}} for {{{c}}}

{{{y^2-8y+15=0}}}

 {{{x = (-b +- sqrt( b^2-4*a*c ))/(2*a) }}} 

 {{{x = (-(-8) +- sqrt( (-8)^2-4*(1)*(15) ))/(2*(1)) }}}

  {{{x = (-(-8) +- sqrt(64-60 ))/2 }}}

  {{{x = (8 +- sqrt(4))/2 }}}

  x = (8±2)/2 

Using the +,

  {{{x = (8 + 2)/2 }}}
  {{{x = 10/2 }}}  
  {{{x=5}}}

Using the -,

  {{{x = (8 - 2)/2 }}}
  {{{x = 6/2 }}}  
  {{{x=3}}}

Edwin</pre>