Solving quadratic equations by completing the square.

Question

Graphing $y=ax^2+ bx + c$ by completing the square

1. Add and subtract the square of half the coefficent of $x$.

2. Group the perfect square trinomial.

3. Write the trinomial as a square of a binomial.

Rewrite $y = x^2 + 6x + 8$ into $y = a(x-h)^2 + k$.

I've tried solving this but I get a bit confused at the step where I have to "write the trinomial as a square of a binomial". Not exactly sure how to do that.

Answer 1

Hint $\rm\,\ \ X^2\! + 2b\, X\! + c\ =\ \overbrace{(X^2\! + 2b\,X\! + b^2)}^{\rm complete\ \ the\ \ \color{#c00}{square}}\! -\! b^2\!+c\ =\ \overbrace{(X + b)^2}^{\rm \color{#c00}{square}} - b^2\!+c $

Answer 2

$$x^2+6x+8=(x+3)^2-1$$

and now try difference of squares...

In general:

$$x^2\pm ax=\left(x\pm\frac a2\right)^2-\frac{a^2}4$$

Answer 3

To complete the square for the equation:

$y = x² + 6x + 8$

Start by:

1) $x^2+6x+8$

2) $x^2+\dfrac 62x +8$ *divide the bx by 2

3) $(x^2+\dfrac 62x + (\dfrac 62)^2)+8$ *square the bx term to form a perfect square

4) $(x^2+ 3x + (36/4))+8$

5) $(x^2+3x + 9)+8-9$ *add the inverse of the c term

6) $(x^2+3x+9)-1$ *factor the perfect square

7) $(x+3)^2-1$

8) $y=(x+3)^2-1$