Showing that there are real roots.

**acedurza** · January 24th 2013, 07:39 PM

(p+1)x2+(2p+3)x+(p+2)=0Need help solving it and showing that the equation has real roots for all values of p. Explanation as well please. Thanks

**Prove It** · January 24th 2013, 07:52 PM

Originally Posted by acedurza

(p+1)x2+(2p+3)x+(p+2)=0Need help solving it and showing that the equation has real roots for all values of p. Explanation as well please. Thanks

I assume this is $\displaystyle \begin{align*} \left( p + 1 \right) x^2 + \left( 2p + 3 \right)^2 x + \left( p + 2 \right) = 0 \end{align*}$ . This is a quadratic of the form $\displaystyle \begin{align*} a\, x^2 + b\, x + c = 0 \end{align*}$ . These quadratics only have real solutions when the discriminant $\displaystyle \begin{align*} \Delta = b^2 - 4ac \end{align*}$ is nonnegative. So start by evaluating the discriminant.

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