25p^2 - 10p = 2 #solve the quadratic equation by completing the square and applying the square root property

Question

25p^2 - 10p = 2 #solve the quadratic equation by completing the square and applying the square root property 

Answer 1

Hello again!

 

25p^2 - 10p   =   2

 

25(p^2 - 2/5p    )  = 2                <--- factors out 25

 

     [half of 2/5]^2 = 1/5*1/5 = 1/25

 

25(p^2 - 2/5p + 1/25) = 2 + 1      <---  adding 1/25 INSIDE the parenthesis actually adds 1 to both sides

                                                                                               because of the  distributive property

 

 25( p    -    1/5  )(p    -      1/5   )  = 3

 

  (p - 1/5)^2 = 3/25

 

  p - 1/5 =    +/-    square-root(3/25)   =   +/- square-root(3)/5  = +/-  rad3/5

 

 p = 1/5 +/- rad3/5 =  (1 +/-  rad3)/5  <--- same denominator of 5, tops can be combined

 

 CHECK:

 

  [positive case]  P = (1 + rad3)/5

 

          squaring it via FOIL method:    ( 1 + 2rad3 + 3)/25 = (4 + 2rad3)/25

 

      multiplying by 25  gives   4 + 2rad3   <-- quadratic term on left side complete    

 

10p = 10( 1 + rad3)/5  = 2(1 + rad3)  = 2 + 2rad3

 

So the left hand side is  4 + 2rad3 - (2 + 2rad3) = 2rad3 - 2 - 2rad3  = 2 which is the same as the right side.

 

Check of the negative case is again left for you.

 

THis problem was much easier because the square root of the positive number, so no imaginaries 

Answer 2

25p² - 10p = 2

25(p² - 0.4p) = 2

25(p² - 0.4p + (0.2)²) = 2 - 25(0.2)²

25(p - 0.2)² = 2 - 1

 

Can you continue?