Given:#" "(x+3)/4=9/(x+3)#
#color(red)("Two methods shown for starting this off")#
#color(blue)("Starting off by using short cut method and missing out steps")#
Write as #(x+3)^2=36 larr" Using cross multiplication"#
#" "color(green)(ul(bar(|color(white)(2/2)x^2+6x-27=0color(white)(2/2)|)))#
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
#color(blue)("Starting off by using first principles")#
The cross multiplication comes from this
Multiply both sides by #color(blue)((x+3))# 'getting rid of it' on the right hand side
#color(brown)((x+3)/4color(blue)(xx(x+3)) = 9/(x+3)color(blue)(xx(x+3)))#
#((x+3)^2)/4=9xx(x+3)/(x+3)#
But #(x+3)/(x+3) = 1#
#((x+3)^2)/4=9#
Multiply both sides by #color(blue)(4)#
#color(brown)(((x+3)^2)/4color(blue)(xx4)=9color(blue)(xx4))#
#(x+3)^2xx4/4=36#
But #4/4=1#
#(x+3)^2=36# ........................Equation(1)
The above is the same as in the cross multiplication.
'................................................................
Consider: #(x+3)^2=color(blue)((x+3))color(brown)((x+3)#
Multiply everything in the right hand bracket by everything in the left
#color(brown)( color(blue)(x)(x+3)" "color(blue)(+3)(x+3)) #
#x^2+3x" "+3x+9#
#x^2+6x+9# Putting this back into Equation(1)
#color(brown)((x+3)^2=36)color(blue)(" "->" " x^2+6x+9=36)" "........Equation(1_a)#
Subtract 36 from both sides
#" "color(green)(ul(bar(|color(white)(2/2)x^2+6x-27=0color(white)(2/2)|)))#
'~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
The only whole number factors of 27 are, 3 and 9; 1 and 27
#3xx9=27" and "9-3=6#
#=> (x+9)(x-3) =0#
#=> x=-9" and "x=+3#
#color(brown)("If you can not spot the factors after a few attempts in a quadratic use the equation")##color(brown)("or complete the square.")#
#ax^2+bx+c=0 -> x=(-b+-sqrt(b^2-4ac))/(2a)#
#a=1; b=6; c=-27#
#=>x=(-6+-sqrt(6^2-4(1)(-27)))/(2(1))#
#=> x=-6/2+-sqrt(144)/2#
#=>x=-3+-6#
#x=-9" and " +3#
