Question 950712
1. 

{{{(sqrt(2x^2 -1))=x}}}.......square both sides

{{{((sqrt(2x^2 -1)))^2=x^2}}}

{{{2x^2-1=x^2}}}

{{{2x^2-x^2-1=0}}}

{{{x^2-1=0}}}

{{{(x-1)(x+1)=0}}}

solutions:
{{{(x-1)=0}}}=>{{{x=1}}}
{{{(x+1)=0}}}=>{{{x=-1}}}

answer is: 
B.{-1,1} 

{{{drawing( 600, 600, -10, 10, -10, 10,
circle(-1,-1,.12),circle(1,1,.12),
 graph( 600, 600, -10, 10, -10, 10, x,-sqrt(2x^2 -1), sqrt(2x^2 -1))) }}}



2.

{{{sqrt(2x^2 + 5x + 6)=x}}}

{{{(sqrt(2x^2 + 5x + 6))^2=x^2}}}

{{{2x^2 + 5x + 6=x^2}}}

{{{2x^2-x^2 + 5x + 6=0}}}

{{{x^2 + 5x + 6=0}}}

{{{x^2+ 2x+ 3x + 6=0}}}

{{{(x^2+ 2x)+ (3x + 6)=0}}}

{{{x(x+ 2)+ 3(x + 2)=0}}}

{{{(x+ 3)(x + 2)=0}}}

solutions:

{{{(x+ 3)=0}}}=>{{{x=-3}}}
{{{(x + 2)=0}}}=>{{{x=-2}}}

answer:

B.{-2,-3} 

{{{drawing( 600, 600, -10, 10, -10, 10,
circle(-2,-2,.12),circle(-3,-3,.12),
 graph( 600, 600, -10, 10, -10, 10, x,-sqrt(2x^2 + 5x + 6), sqrt(2x^2 + 5x + 6))) }}}


3. 

{{{sqrt(x + 3) = 2sqrt(x)}}}

{{{(sqrt(x + 3))^2 =( 2sqrt(x))^2}}}

{{{x + 3 = 4x}}}

{{{3 = 4x-x}}}

{{{3 = 3x}}}

{{{x=1}}}

answer:

F.1 

since {{{x=1}}}, {{{sqrt(1 + 3) = 2sqrt(1)}}}=>{{{sqrt(4) = 2*1}}}=>{{{2 = 2}}}

so, these lines intersect at ({{{1}}},{{{2}}})


{{{drawing( 600, 600, -10, 10, -10, 10,
circle(1,2,.12),
 graph( 600, 600, -10, 10, -10, 10, sqrt(x + 3), 2*sqrt(x))) }}}


4.

{{{sqrt(2x^2 + x -12) = x}}}

{{{(sqrt(2x^2 + x -12))^2 = x^2}}}

{{{2x^2 + x -12 = x^2}}}

{{{2x^2 -x^2+ x -12 = 0}}}

{{{x^2+ x -12 = 0}}}

{{{x^2+ 4x-3x -12 = 0}}}

{{{(x^2+ 4x)-(3x +12) = 0}}}

{{{x(x+ 4)-3(x +4) = 0}}}

{{{(x+4)(x-3)=0}}}

solutions:
{{{(x+4)=0}}}=>{{{x=-4}}}
{{{(x-3)=0}}}=>{{{x=3}}}

answer is:
B.3 

{{{drawing( 600, 600, -10, 10, -10, 10,
circle(3,-3,.12),circle(-4,-4,.12),
 graph( 600, 600, -10, 10, -10, 10, x,sqrt(2x^2 + x -12), -sqrt(2x^2 + x -12))) }}}

5.

{{{sqrt(2x^2 + 6x + 4) = x + 1}}}

{{{(sqrt(2x^2 + 6x + 4))^2 = (x + 1)^2}}}

{{{2x^2 + 6x + 4= x^2+2x + 1}}}

{{{2x^2 + 6x + 4-x^2-2x - 1=0}}}

{{{x^2 + 4x + 3=0}}}

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

{{{(x^2 +x)+ (3x + 3)=0}}}

{{{x(x +1)+ 3(x + 1)=0}}}

{{{(x +3)(x + 1)=0}}}


solutions:

{{{(x +3)=0}}}=>{{{x=-3}}}
{{{(x + 1)=0}}}=>{{{x=-1}}}

answer:
B.{-3, -1} 

{{{drawing( 600, 600, -10, 10, -10, 10,
circle(-3,-2,.12),circle(-1,0,.12),
 graph( 600, 600, -10, 10, -10, 10, x+1,-sqrt(2x^2 + 6x + 4),sqrt(2x^2 + 6x + 4) )) }}}