Solve quadratic equation by completing square

Question

We have a equation: $2x^2+7x+3$
I tried to find the vertex of the parabola by this formula: $a(x-h)^2+k$
but I could not get it.
I got this: but it is not right.
$2(x^2+ \frac{7}{2}x+\frac{49}{4})+3-\frac{49}{2}$
What's the problem? Is it possible to solve it by completing square?

Answer 1

Completing the square, it is :

$$2x^2 + 7x + 3 = 2\bigg(x^2+ \frac{7}{2}x + \frac{49}{16}\bigg) - \frac{25}{8} = 2\bigg(x + \frac{7}{4}\bigg)^2 - \frac{25}{8}$$

Thus, the vertex can be found at the point where the "completed square term" equals zero :

$$x+\frac{7}{4} = 0 \Leftrightarrow x = -\frac{7}{4}$$

This is truly the vertex point, as it can be cross-validated by the fact that a vertex of a parabola can be found at :

$$x_v = -\frac{b}{2a} \Rightarrow x_v = -\frac{7}{2\cdot 2}$$

Now, the other coordinate of the vertex is the remaining constant term on the right, thus $-25/8$.

This means that the given vertex coordinates, are :

$$(h,k) = \bigg(-\frac{7}{4},-\frac{25}{8}\bigg)$$

Answer 2

It's possible.

\begin{align*} 2x^2 + 7x + 3 & = 2\left(x^2 + \frac{7}{2} x \right) + 3 = 2\left(\left(x + \frac{7}{4}\right)^2 - \left(\frac{7}{4}\right)^2 \right) + 3 \\ & = 2 \left(x + \frac{7}{4}\right)^2 - \frac{49}{8} + 3 = 2 \left(x + \frac{7}{4}\right)^2 - \frac{25}{8} \end{align*} So we have the vertex at $(h,k) = \left(- \frac{7}{4}, - \frac{25}{8}\right)$.

Answer 3

In completing the square, you want to follow a plan like below.

$$ax^2+bx+c = 0$$

$$x^2+\frac{b}{a}x+\frac{c}{a} = 0 \implies x^2+\frac{b}{a}x = -\frac{c}{a}$$

You want to reach a quadratic trinomial on the LHS so you can solve it. This is done by adding $\big(\frac{b}{2a}\big)^2$ to both sides.

$$x^2+\frac{b}{a}x+\bigg(\frac{b}{2a}\bigg)^2 = -\frac{c}{a}+\frac{b^2}{4a^2}$$

$$\bigg(x+\frac{b}{2a}\bigg)^2 = \frac{b^2-4ac}{4a^2}$$

After which you can take the square root of both sides and solve for $x$. Can you use this general plan to solve the equation you’ve given?