How do you solve #8x^2 + 14x + 5 = 0# by completing the square?

3 Answers
May 8, 2018

Answer:

#x=-5/4" or "x=-1/2#

Explanation:

#"using the method of "color(blue)"completing the square"#

#• " the coefficient of the "x^2" term must be 1"#

#"factor out 8"#

#rArr8(x^2+7/4x+5/8)=0#

#• "add/subtract "(1/2"coefficient of the x-term")^2" to"#
#x^2+7/4x#

#8(x^2+2(7/8)xcolor(red)(+49/64)color(red)(-49/64)+5/8)=0#

#rArr8(x+7/8)^2+8(-49/64+5/8)=0#

#rArr8(x+7/8)^2-9/8=0#

#rArr8(x+7/8)^2=9/8#

#"divide both sides by 8"#

#rArr(x+7/8)^2=9/64#

#color(blue)"take the square root of both sides"#

#sqrt((x+7/8)^2)=+-sqrt(9/64)larrcolor(blue)"note plus or minus"#

#rArrx+7/8=+-3/8#

#"subtract "7/8" from both sides"#

#rArrx=-7/8+-3/8#

#rArrx=-7/8-3/8=-10/8=-5/4#

#"or "x=-7/8+3/8=-4/8=-1/2#

May 8, 2018

Answer:

#x= -10/8=-5/4#
#x=-4/8=-1/2#

Explanation:

Given: #color(green)(y=0=8x^2+14x+5)#

Write as:
#color(green)(0=8(x^2+14/8x)+5 color(white)("ddd")->color(white)("ddd")0=8(x^2+7/4x)+5)#

The 'perfect' square related to #x^2+7/4x " is " (x+7/8)^2#

but #8(x+7/8)^2# introduces the value of #color(red)(8xx(7/8)^2)# that is not in the original equation. We put it in so we have to take it out.

#color(green)(0=8(x^2+7/4x)+5 )#

#color(green)(0= 8(x+7/8)^2color(red)(-[8xx(7/8)^2 ] )+5 )#

#color(green)(0=8(x+7/8)^2-9/8 )#
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#color(blue)("Check")#

#8(x+7/8)^2-9/8 #
#8(x^2+7/4x +49/64)-9/8#

#8x^2+14x+49/8-9/8#

#8x^2+14x +40/8#

#8x^2+14x+5# as required!
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#color(green)(0=8(x+7/8)^2-9/8color(white)("dd") ->color(white)("dd")+9/8=8(x+7/8)^2#

#color(green)( color(white)("ddddddddddddddddddd")->color(white)("dddd") 9/64=(x+7/8)^2)#

#color(green)(color(white)("ddddddddddddddddddd")->color(white)("d.d")+-3/8=color(white)("d")x+7/8) #

#color(green)(color(white)("ddddddddddddddddddd")->color(white)("ddd.dd")x=-7/8+-3/8)#

#x= -10/8=-5/4#
#x=-4/8=-1/2#

Tony B

May 8, 2018

Answer:

#x=-1/2 or x=-5/4#

Explanation:

We have,

#8x^2+14x+5=0#

#=>8x^2+14x=-5#

#=>8x^2+14x+K=K-5...to(A)#

Now we have to find #K#,such that,# (8x^2+14x+K)#is a square.

Here,

#I^(st)term=8x^2#

#II^(nd) term=14x#

#III^(rd)term=K#

Formula for #color(red)(III^(rd)term=(II^(nd)term)^2/(4xxI^(st)term)...to(psi)#

#i.e. K=(14x)^2/(4xx8x^2)=(196x^2)/(32x^2)=49/8#

So,from#(A)#,we get

#8x^2+14x+49/8=49/8-5#

#(sqrt8x)^2+2(sqrt8x)(7/sqrt8)+(7/sqrt8)^2=9/8#

#(sqrt8x+7/sqrt8)^2=(3/sqrt8)^2#

#=>sqrt8x+7/sqrt8=+-3/sqrt8#

#=>8x+7=+-3#

#=>8x+7=3 or 8x+7=-3#

#=>8x=3-7 or 8x=-3-7#

#=>8x=-4 or 8x=-10#

#=>x=-1/2 or x=-5/4#

#=>x=-0.5 or x=-1.25#

Note: Formula #color(red)((psi)# for #color(red)(III^(rd)term#, is always true.