Q) how to solve by completing the square method ? a) #2x^2+16x+5# b) #6+4x-x^2#

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Answer 1 · 2017-12-25T13:13:15

a) #2(x+2)^2-3#

b) #10-(x-2)^2#

a) #2x^2+16x+5#

#=>2[x^2+8x+5/2]#

#(color(red)a+color(blue)b)^2=a^2+color(green)2color(red)acolor(blue)b+b^2#

#=>2[color(red)x^2+color(green)2*color(blue)4color(red)x+color(blue)4^2-4^2+5/2]#

#=>2[(color(red)x^2+color(green)2*color(blue)4color(red)x+color(blue)4^2)-16+5/2]#

#=>2[(x+4)^2-32/2+5/2]#

#=>2[(x+4)^2-27/2]#

#=>2(x+4)^2-cancel2*27/cancel2#

#=>2(x+4)^2-27#

b) #6+4x−x^2#

#=>-1*[x^2-4x-6]#

#=>-1*[color(red)x^2-color(green)2*color(blue)2color(red)x+color(blue)2^2-2^2-6]#

#=>-1*[(color(red)x^2-color(green)2*color(blue)2color(red)x+color(blue)2^2)-4-6]#

#=>-1*[(color(red)x-color(blue)2)^2-10]#

#=>-(x-2)^2+10#

#=>10-(x-2)^2#

Answer 2 · 2017-12-25T13:37:48

Assuming that we are to solve the equations completing the square method.

a) #2x^2+16x+5=0#

#=>x^2+8x+5/2=0#

#=>x^2+2*x*4+4^2-4^2+5/2=0#

#=>(x+4)^2-16+5/2=0#

#=>(x+4)^2-27/2=0#

#=>(x+4)^2=27/2#

#=>x+4=pmsqrt27/2#

#=>x=-4pm(3sqrt3)/2#

(b) #6+4x-x^2=0#

#=>6+2^2-2^2+2*x*2-x^2=0#

#=>10-(2^2-2*x*2+x^2)=0#

#=>(x-2)^2 =10#

#=>x-2 =pmsqrt10#

#=>x=2pmsqrt10#