How do you solve #x^{2} + 5x + 4= x#?

2 Answers
Mar 16, 2018

Given: #x^2 + 5x + 4= x#

Subtract x from both sides:

#x^2 + 4x + 4= 0#

Factor:

#(x+2)(x+2) = 0#

This above shows that there is a repeated root at #x = -2#

Mar 16, 2018

#x=-2#

Explanation:

#"Rearrange the equation into standard form"#

#rArrx^2+4x+4=0larrcolor(blue)"in standard form"#

#"the factors of + 4 which sum to + 4 are + 2 and + 2"#

#rArrx^2+2x+2x+4=0larrcolor(blue)"split the middle term"#

#rArrcolor(red)(x)(x+2)color(red)(+2)(x+2)=0larrcolor(blue)"factor by grouping"#

#rArr(x+2)(color(red)(x+2))=0#

#rArr(x+2)^2=0rArrx=-2#