Expressing a Quadratic in $a(x + b)^2 + c$ form

Question

I have question as follows: Express $2x^2 - 6x + 7$ in the form $a(x + b)^2 + c$.

My approach is to complete the square so I did this:

$(2x^2 + 6/2)^2 - (6/2)^2 + 7 = 0$

$(2x^2 - 3)^2 - 9 + 7 = 0$

$(2x^2 - 3)^2 - 2 = 0$

I got a bit lost here and not sure how to proceed forward.

Answer 1

Try it with $$2\left(x^2-3x+\frac{7}{2}\right)=2\left(x^2-2\cdot \frac{3}{2}x+\frac{9}{4}-\frac{9}{4}+\frac{7}{2}\right)$$

Answer 2

Hint: $$2x^2 - 6x + 7=ax^2+2abx+ab^2+c$$ $$a=2$$ $$-6=2ab$$ $$7=ab^2+c$$

Answer 3

$2x^2-6x+7 = 2(x^2-3x)+7=2(x^2-3x+(\frac 32)^2) -2(\frac 32)^2+7 =2(x-\frac 32)^2+\frac 52$

Answer 4

Let's investigate completing the square for the general case:

$$ax^2+bx+c$$

Factor out the $a$ to obtain

$$a\left(x^2+\frac{b}{a}x+\frac{c}{a}\right)\tag{*}$$

Let's complete the square on the bracketed term. We Need to write this as $(x+B)^2$, so start with

$$\left(x+\frac{b}{a}\right)^2=x^2+2\frac{b}{a}x+\frac{b^2}{a^2}.$$

But wait ! We've got 2 lots of $b/a$ when we expand it, so what we really should use instead is

$$\left(x+\frac{b}{2a}\right)^2=x^2+\frac{b}{a}x+\frac{b^2}{4a^2}.$$

Now that's better, but this square introduces the unwanted term $b^2/(4a^2)$, so we will need to remove that,

$$\left(x+\frac{b}{2a}\right)^2-\frac{b^2}{4a^2}=x^2+\frac{b}{a}x.$$ That's exactly what we want, except that we need a $c/a$ term, so let's add that back in,

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

This is precisely what is required, so just sub this back into (*) and we get

$$a\left(\left(x+\frac{b}{2a}\right)^2-\frac{b^2}{4a^2}+\frac{c}{a}\right).$$

Expanding, we have

$$a\left(x+\frac{b}{2a}\right)^2+c-\frac{b^2}{4a}\equiv A(x+B)^2+C,$$

where $A=a$, $$B=\frac{b}{2a},$$ and $$C=c-\frac{b^2}{4a}.$$

Answer 5

To proceed, assume $y=2x^2-3$ then we get $y^2-2=0$ as follows: $$ assume:\quad y = 2x^2-3 \\ y^2-2=0 \\ y=\pm\sqrt{2} \\ (1)\qquad y=\sqrt{2} \\ 2x^2-3=\sqrt{2} \\ x=\pm\sqrt{\frac{3+\sqrt{2}}{2}} \\ (2) \qquad y=\sqrt{-2}=\sqrt{2}i \\ 2x^2-3=\sqrt{2}i \\ x=\pm\sqrt{\frac{3+\sqrt{2}i}{2}} $$