Solve `|x^2+4x+3|=x+1`

Answer

x=-1

Explanation

`|x^2+4x+3|=x+1`
or |(x+1)(x+3)|=x+1
|(x+1)(x+3)=(x+1)(x+3)| , when `(x+1)(x+3) ge 0 `
`or x le-3 or x ge -1`
Hence , given equation educes reduces to (x+1)(x+3)=x+1
`rArr x=-1(x=-2 " is rejected as" x le -3 or x ge -1)`
|(x+1)(x+3)|=-(x+1)(x+3) when `(x+1)(x+3) lt 0 `
`or -3 lt x lt -1`
Hence , given equation redues to -(x+1)(x+3)=x+1
`rArr x=-4 " which is rejected as " -3 lt x lt -1`