Mathematico | Mathematico

The complete square form of the quadratic

x^{2} + b x + c

(x + c)^{2} + b

In simplest form, the roots are given by

m \pm \sqrt{n}, m, n \in N

Given that $b c \neq 0$ , find the value of $m + n$ .

Hint

You could complete the square on $x^{2} + b x + c$ , or you could expand $(x + c)^{2} + b$ and compare coefficients.

Hint

The condition $b c \neq 0$ implies that $b \neq 0$ and $c \neq 0$ .

Solution

We'll stay true to the topic of study and complete the square

\begin{aligned} x^{2} + b x + c & = {(x + \frac{b}{2})}^{2} - \frac{b^{2}}{4} + c \\ = {(x + \frac{b}{2})}^{2} - \frac{b^{2} - 4 c}{4} \end{aligned}

Comparing this with $(x + c)^{2} + b$ we see that

\frac{b}{2} = c \Rightarrow b = 2 c

and

\begin{aligned} - \frac{b^{2} - 4 c}{4} & = b \\ b^{2} - 4 c & = - 4 b \\ 4 c^{2} - 4 c & = - 8 c \\ c^{2} - c & = - 2 c \\ c^{2} + c & = 0 \\ c (c + 1) & = 0 \\ c & \in {0, - 1} \end{aligned}

However $b c \neq 0$ , so $c = - 1$ and $b = 2 c = - 2$ .

Putting these into the complete square form to find the roots, we get

\begin{aligned} (x - 1)^{2} - 2 & = 0 \\ (x - 1)^{2} & = 2 \\ x & = 1 \pm \sqrt{2} \end{aligned}