Math Problem Solver

Solve by factoring:

Factor the nonzero side of the equation completely. $$\begin{aligned} x^{2} + 2 x + 1 &= 0 \\ \left(x + 1\right)^{2} &= 0\end{aligned}$$

The product of any number and zero is zero, so the equation will hold true if any factors of the quadratic are zero. Set the factors equal to zero and solve. Since the quadratic is square, there is only one factor to set equal to zero and only one solution. $$\begin{aligned} x + 1 &= 0\\ x &= -1\end{aligned}$$

Solve by quadratic formula:

Given a quadratic equation \(a x^{2} + b x + c = 0\), where \(a\), \(b\), \(c\) are constants and \(a \ne 0\), the solutions are given by the quadratic formula: \[x = \frac{-b \pm \sqrt{b^{2} - 4ac}}{2a}\] In this case \(a=1\), \(b=2\), and \(c=1\).

The discriminant is the quantity under the square root sign in the quadratic formula, and its sign determines the number of solutions to the quadratic equation when the coefficients are real. The discriminant is:\[b^{2}-4ac = 2^{2} - 4 \cdot 1=0 = 0\] The discriminant is equal to zero, so this quadratic equation has one real solution.

The solution is: \[x = \frac{-2 \pm \sqrt{0}}{2 \cdot 1} = -1\]