Show that the solutions to the equation $ax^2 + 2bx + c =0$ are given by $x = -\frac{b}{a} \pm \sqrt{\frac{b^2-ac}{a^2}}$

Question

Show that the solutions to the equation $ax^2 + 2bx + c =0$ are given by $x = -\frac{b}{a} \pm \sqrt{\frac{b^2-ac}{a^2}}$

Hint: Start by dividing the whole equation by $a$

At first I have tried solving the equation without using the hint provided in my exercise and directly applying completing square, I get $x = -\frac{b}{a} \pm \sqrt{\frac{b^2-c}{a}}$. So if I am to use the hint, I obtain the appropriate answer. But I wonder if I am asked the same question in my exam where the hint will not be provided then how am I supposed to answer.

I would like to know how one should approach this kind of question and how do I realise when to divide the whole equation with in this case $a$ or is there any other ways so that I can avoid dividing the whole equation by $a$.Thanks in advance for any help you are able to provide!

EDIT: Here's my steps. Please see where have I done wrong.

\begin{align} ax^2+2bx+c&=0 \\ a\left[\left(x+\frac{b}{a}\right)^2-\frac{b^2}{a^2}\right] + c&=0 \\ a\left(x+\frac{b}{a}\right)^2-\frac{b^2}{a} + c&=0 \\ \left(x+\frac{b}{a}\right)^2&=\left(\frac{b^2}{a}-c\right)\left(\frac{1}{a}\right)\\ \left(x+\frac{b}{a}\right)^2&=\frac{a(b^2-c)}{a^2}\\ \left(x+\frac{b}{a}\right)^2&=\frac{b^2-c}{a} \\ x+\frac{b}{a}&=\pm\sqrt{\frac{b^2-c}{a}} \\ \implies x&=-\frac{b}{a}\pm\sqrt{\frac{b^2-c}{a}}\\ \end{align}

Answer 1

As you did, complete the square by first taking out a factor of $a$, where $a\neq 0$ since the equation is a quadratic.

\begin{align} ax^2+2bx+c&=0 \\ a(x^2+2\frac{b}{a}+\frac{c}{a})&=0 \\ a\left[(x+\frac{b}{a})^2-\frac{b^2}{a^2}+\frac{c}{a}\right]&=0 \\ (x+\frac{b}{a})^2&=\frac{b^2-ac}{a^2} \\ x+\frac{b}{a}&=\pm\sqrt{\frac{b^2-ac}{a^2}} \\ \implies x&=-\frac{b}{a}\pm\sqrt{\frac{b^2-ac}{a^2}}\\ \end{align}

Maybe you made a mistake somewhere?

Answer 2

HINT: Complete the square:

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

$$a\left(x^2+\frac{2b}{a}x\right)+c=0$$

Try the rest from here.

Answer 3

As JCAA said, you should assume $a\ne 0$, otherwise, it is not a quadratic equation, but a linear equation. Also, division by zero is undefined. We just could divide both sides by $a$ because of that.

$ax^2+2bx+c=0 \implies x^2+2\dfrac{b}{a}x+\dfrac{a}{c}=0 \implies x^2+2\dfrac{b}{a}x=-\dfrac{c}{a}$

Completing the square,

$x^2+2\dfrac{b}{a}x + \color{red}{\dfrac{b^2}{a^2}} =-\dfrac{c}{a}+ \color{red}{\dfrac{b^2}{a^2}}\\\\ \left(x+\dfrac{b}{a}\right)^2 = \dfrac{b^2-ac}{a^2} \\\boxed{x=-\dfrac{b}{a}\pm \sqrt{\dfrac{b^2-ac}{a^2}}} $

Answer 4

You are given the equation and the answers and asked to verify. Additionally, you don't want to divide the equation by the leading coefficient. You could:

• Verify the answers directly by plugging them in. Then argue that these must be the only solutions.
• Solve the equation in a way that avoids dividing.

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The first method would look something like:

By the fundamental theorem of algebra, the given equation has at most two roots. We show that they are indeed the given solutions. $$\dots\text{(some god awful amount of algebra)}\dots$$ $\dots$completing our proof.

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To avoid dividing, your method was good. An alternative, if you are familiar with the quadratic formula, would be to plug your coefficients in and simplify, taking care not to confuse yourself with overloaded variables.

$$\text{Let } a=a, b=2b, c=c. \text{ Then,}$$

\begin{align} x &= \frac{-2b \pm \sqrt{(2b)^2 - 4ac}}{2a}\\ &= -\frac{b}{a} \pm \frac{\sqrt{4b^2 - 4ac}}{2a}\\ &= -\frac{b}{a} \pm \sqrt{\frac{4b^2 - 4ac}{4a^2}}\\ &= -\frac{b}{a} \pm \sqrt{\frac{b^2 - ac}{a^2}} \end{align}

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When there's no time pressure the approach you choose to solve a problem doesn't need to be a final decision. Often there is more than one way to get where you're going. Even when you do get the right answer it can be helpful to examine what you did.

On an exam though, it will help to have some experience solving similar problems and leaning towards techniques you've used before.