Consider the equation 2+x2+4x+3=m,m∈RSet of all real values of m so that the given equation have three solutions isSet of all real values of m so that given equation have four distinct solutions, isSet of all real values of m so that the given equation have two solutions is

Given equation is 2+x2+4x+3=m∴ x2+4x+3=m−2Given equation is meaningful if m≥2 .                          (1)∴     x2+4x+3=±(m−2)∴     x2+4x+5−m=0 and x2+4x+1+m=0∴     x=−4±16−4(5−m)2 and         x=−4±16−4(1+m)2 ∴ x=−2±m−1 and x=−2±3−mFor four distinct solutions m−1≥0 and 3−m≥0∴ m∈(2,3)                           [ using (1)]For two solutions  m−1≥0 and 3−m≤0⇒m≥3or    m−1≤0 and 3−m≥0⇒m=2 [using(1)] ∴ m∈{2}∪(3,∞)From three solutions, exactly one equation must give equal roots.∴    m=1 and 3−m≥0⇒ no values of m( as m≥2) or     m=3 and m≥1⇒m=3Alternate method:Let us draw the graph of f(x)=x2+4x+3Now equation is l(x + 1) (x + 3)l = m - 2From the graph(1) for three solutions m−2=1⇒m=3(2)for four solutions01⇒ m=2 or m>3 Given equation is 2+x2+4x+3=m∴ x2+4x+3=m−2Given equation is meaningful if m≥2 .                          (1)∴     x2+4x+3=±(m−2)∴     x2+4x+5−m=0 and x2+4x+1+m=0∴     x=−4±16−4(5−m)2 and         x=−4±16−4(1+m)2 ∴ x=−2±m−1 and x=−2±3−mFor four distinct solutions m−1≥0 and 3−m≥0∴ m∈(2,3)                           [ using (1)]For two solutions  m−1≥0 and 3−m≤0⇒m≥3or    m−1≤0 and 3−m≥0⇒m=2 [using(1)] ∴ m∈{2}∪(3,∞)From three solutions, exactly one equation must give equal roots.∴    m=1 and 3−m≥0⇒ no values of m( as m≥2) or     m=3 and m≥1⇒m=3Alternate method:Let us draw the graph of f(x)=x2+4x+3Now equation is l(x + 1) (x + 3)l = m - 2From the graph(1) for three solutions m−2=1⇒m=3(2)for four solutions01⇒ m=2 or m>3 Given equation is 2+x2+4x+3=m∴ x2+4x+3=m−2Given equation is meaningful if m≥2 .                          (1)∴     x2+4x+3=±(m−2)∴     x2+4x+5−m=0 and x2+4x+1+m=0∴     x=−4±16−4(5−m)2 and         x=−4±16−4(1+m)2 ∴ x=−2±m−1 and x=−2±3−mFor four distinct solutions m−1≥0 and 3−m≥0∴ m∈(2,3)                           [ using (1)]For two solutions  m−1≥0 and 3−m≤0⇒m≥3or    m−1≤0 and 3−m≥0⇒m=2 [using(1)] ∴ m∈{2}∪(3,∞)From three solutions, exactly one equation must give equal roots.∴    m=1 and 3−m≥0⇒ no values of m( as m≥2) or     m=3 and m≥1⇒m=3Alternate method:Let us draw the graph of f(x)=x2+4x+3Now equation is l(x + 1) (x + 3)l = m - 2From the graph(1) for three solutions m−2=1⇒m=3(2)for four solutions01⇒ m=2 or m>3