How do you solve using the completing the square method #2x^2 + 8x - 25 = 0#?

1 Answer
Aug 24, 2016

Answer:

#=> x~~-6.06 " and " 2.06# to 2 decimal places

Explanation:

Given:#" "2x^2+8x-25=0# ....................Equation(1)

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#color(blue)("Completing the square")#

write as #y=2(x^2+4x)-25=0#

Introduce the correction k to compensate for the changes we are going to make. They introduce an error.

#2(x^2+4x)-25+k=0 larr" at this stage "k=0#

Take the power of 2 outside the brackets

#2(x+4x)^2-25+k=0 #

Remove the #x# from #4x#

#2(x+4)^2-25+k=0#

Halve the 4

#2(x+2)^2-25+k=0#............................. Equation(2)

From the part: #color(red)(2)(xcolor(blue)(+2))^2# you get the constant #color(red)(2)color(blue)(xx2)^2=8# which is the error.

so #8+k=0=> k=-8# Substituting into equation(2) gives

#2(x+2)^2-25-8=0............................. Equation(2_a)#

#2(x+2)^2-33=0#
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#color(blue)("Determine "x_("intercepts"))#

Write as#(x+2)^2=33/2#

Square root both sides

#x+2=sqrt(33/2)#

#=>x=-2+-sqrt(33/2)#

#=> x~~-6.06 " and " 2.06# to 2 decimal places

Tony B