Question 109283
1. Solve by completing the square.  
	
{{{x^2 – 4x + 4 = 0}}}.......{{{-4x}}} write as {{{-2x - 2x}}}, then group first two terms and second two terms

{{{x^2 + 2x - 2x + 4 = 0}}}.......


({{{x^2 – 2x}}}) - ({{{2x - 4}}}) = {{{0}}}.......factor out in first group {{{x}}}, and in second group {{{2}}}

x(x – 2) - 2(x - 2) = 0.......factor out {{{x-2}}}

(x – 2)(x - 2) = 0.......solution



2. Solve by completing the square.  
	
{{{x^2 = 5x + 2}}} 

write this first in standard form:

{{{x^2 - 5x - 2 = 0}}}

coefitients are: {{{a = 1}}}, {{{b = -5}}}, and {{{c = -2}}}

now apply the quadratic formula:

{{{x[1,2]=(-b +- sqrt (b^2 -4*a*c )) / (2*a)}}}

{{{x[1,2]=(5 +- sqrt ((-5)^2 -4*1*(-2) )) / (2*1)}}}

{{{x[1,2]=(5 +- sqrt (25 + 8 )) / 2}}}

{{{x[1,2]=(5 +- sqrt (33)) / 2}}}

{{{x[1,2]=(5 +- 5.74) / 2}}}


{{{x[1]=(5 + 5.74) / 2}}}

{{{x[1]=5.37 }}}


{{{x[2]=(5 - 5.74) / 2}}}

{{{x[2]=-0.74 / 2}}}

{{{x[2]= -0.37}}}



3. Solve by using the quadratic formula.  

{{{x^2 }}}= –{{{3x + 8 }}}

{{{x^2 + 3x - 8 = 0}}}

coefitients are: {{{a = 1}}}, {{{b = 3}}}, and {{{c = -8}}}

now apply the quadratic formula:

{{{x[1,2]=(-b +- sqrt (b^2 -4*a*c )) / (2*a)}}}

{{{x[1,2]=(-3 +- sqrt ((3)^2 -4*1*(-8) )) / (2*1)}}}

{{{x[1,2]=(-3 +- sqrt (9 + 32 )) / 2}}}

{{{x[1,2]=(-3 +- sqrt (41) ) / 2}}}

{{{x[1,2]=(-3 +- 6.4) / 2}}}


{{{x[1]=(-3 + 6.4) / 2}}}

{{{x[1]= 3.4 / 2}}}

{{{x[1]= 1.7}}}


{{{x[2]=(-3 - 6.4) / 2}}}

{{{x[2]=( - 9.4) / 2}}}

{{{x[2]=( -4.7) / 2}}}



4. Solve by using the quadratic formula.  

{{{3x^2 = 11x + 4}}}

{{{3x^2 - 11x - 4 = 0}}}

coefitients are: {{{a = 3}}}, {{{b = -11}}}, and {{{c = -4}}}

now apply the quadratic formula:

{{{x[1,2]=(-b +- sqrt (b^2 -4*a*c )) / (2*a)}}}

{{{x[1,2]=(-11 +- sqrt ((-11)^2 -4*3*(-4) )) / (2*3)}}}


{{{x[1,2]=(-11 +- sqrt (121 + 48 )) / 6}}}

{{{x[1,2]=(-11 +- sqrt (169 )) / 6}}}

{{{x[1,2]=(-11 +- 13) / 6}}}


{{{x[1]=(-11 + 13) / 6}}}

{{{x[1]= 2 / 6}}}

{{{x[1]= 1 / 3}}}


{{{x[2]=(-11 - 13) / 6}}}

{{{x[2]= - 24 / 6}}}

{{{x[2]= - 4 }}}



5. Solve.    {{{(x + 1)2 = 3}}}....divide both sides by {{{2}}}

{{{x+1 = 3/2}}}...move {{{1}}} to the right

{{{x = 3/2 - 1}}}....{{{1=2/2}}}, so 

{{{x = 3/2 - 2/2}}}

{{{x = 1/2}}}