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JAMB 1987 · UME · Q21

Find the values of mm which make the quadratic function x2+2(m+1)x+m+3x^2 + 2(m + 1)x + m + 3 a perfect square.

Worked solution (try it first)
  1. A perfect square has equal roots, so b2=4acb^2 = 4ac: [2(m+1)]2=4(m+3)[2(m + 1)]^2 = 4(m + 3).
  2. Divide both sides by 4: (m+1)2=m+3(m + 1)^2 = m + 3, so m2+2m+1=m+3m^2 + 2m + 1 = m + 3.
  3. Rearrange: m2+m−2=0m^2 + m - 2 = 0, which factorises as (m+2)(m−1)=0(m + 2)(m - 1) = 0.
  4. So m=1m = 1 or m=−2m = -2, option C.

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