For what values of $'p'$ the equation $\displaystyle \left( 1+p \right) { x }^{ 2 }+2\left( 1+2p \right) x+\left( 1+p \right) =0$ has coincident roots ?

For the quadratic equation to have coincident roots, its discriminant must be zero, meaning b^2 - 4ac = 0. Substituting the coefficients from (1+p)x^2 + 2(1+2p)x + (1+p) = 0 gives [2(1+2p)]^2 - 4(1+p)(1+p) = 0. Expanding both parts results in 4(1 + 4p + 4p^2) - 4(1 + 2p + p^2) = 0, which simplifies to 4 + 16p + 16p^2 - 4 - 8p - 4p^2 = 0. Combining like terms gives 12p^2 + 8p = 0, and factoring out 4p yields 4p(3p + 2) = 0, so the values of p are 0 and -2/3.

• Thanks
• Wrong explanation