How do you solve #-4x^2+8x= -3#?

1 Answer
Apr 18, 2018

Answer:

#x=1+-1/2sqrt7#

Explanation:

#"rearrange the equation in "color(blue)"standard form"#

#•color(white)(x)ax^2+bx+c=0#

#rArr-4x^2+8x+3=0larrcolor(blue)"in standard form"#

#"consider the a-c method of factorising"#

#"there are no whole numbers of - 12 which sum to + 8"#

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

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

#"factor out "-4#

#rArr-4(x^2-2x-3/4)=0#

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

#rArr-4(x^2+2(-1)xcolor(red)(+1)color(red)(-1)-3/4)=0#

#rArr-4(x-1)^2-4(-1-3/4)=0#

#rArr-4(x-1)^2+7=0#

#"subtract 7 and divide by "-4#

#rArr(x-1)^2=7/4#

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

#rArrx-1=+-sqrt(7/4)larrcolor(blue)"note plus or minus"#

#"add 1 to both sides"#

#rArrx=1+-1/2sqrt7larrcolor(red)"exact solutions"#