Advanced Proof of Quadratic Formula

The Three Important Identities

In this post you wanna dive in and find resourful identities that you thought little of in developnment of quadratic formula.

The general formula of quadratic equation is given by

${\color{Red} \mathbf{ax^{2}+bx+c=0}}$

To get more explanation about the parameters find insightful peek in this basic proof of quadratic formula

The algebraic identities you wanna learn here

1.

2.

3.

Now given the three identities, let’s see how to combine them to produce a path towards the quadratic formula. By dividing a through the equation, it inevitabley becomes

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

Suppose the roots of a quadratic equation are give by r₁ and r₂ and that they are building blocks towards quadratic formula advancement. Now, form a quadratic equation using the roots. Notice that a is divided through the general quadratic equation above so as to make roots the same $\mathbf{\left ( x-r_{1}\right )\left ( x-r_{2} \right )=x^{2}-\left ( r_{1}+r_{2} \right )x+r_{1}r_{2}}$

It is crystal clear that both equations have their first term a perfect square-x squared. And that both of them equals to zero. Since they are equivalent to zero and their first term same, let’s equate them.

$\mathbf{x^{2}+\frac{b}{a}x+\frac{c}{a}=x^{2}-\left ( r_{1}+r_{2} \right )x+r_{1}r_{2}}$

By observation, it is established that the sum of the roots of the quadratic equation is

$\mathbf{r_{1}+r_{2}=-\frac{b}{a}}$

And the product is

$\mathbf{r_{1}r_{2}=\frac{c}{a}}$

Step 1

In a special case when the first two identities (1. & 2.) are combined, then

${\color{Magenta} \mathbf{\left ( a+b \right )^{2}={\left ( a-b \right )^{2}+4ab}$

The relationship of the combined identities becomes

$\mathbf{\left ( r_{1}-r_{2} \right )^{2}=\left ( r_{1}+r_{2} \right )^{2}-4r_{1}r_{2}}$

Express the left side linearly by taking square roots both sides

$\mathbf{\left ( r_{1}-r_{2} \right )=\pm \sqrt{\left ( r_{1}+r_{2} \right )^{2}-4r_{1}r_{2}}$

Substitute the sum and product of the general equation into the combined identity . It relates as

$\mathbf{\left ( r_{1}-r_{2} \right )=\pm \sqrt{\left ( -\frac{b}{a} \right )^{2}-4\frac{c}{a}}$

Step 2

By rearranging, make 4r₁r₂ the subject of the formula

$\mathbf{4r_{1}r_{2}=\left ( r_{1}+r_{2} \right )^{2}-\left ( r_{1}-r_{2} \right )^{2}}$

Let’s use the difference of two squares to find r₁ from this equation. Divide both sides by 4r₂ , so r₁ becomes

$\mathbf{r_{1}=\frac{\left ( r_{1}+r_{2} \right )^{2}-\left ( r_{1}-r_{2} \right )^{2}}{4r_{2}}}$

The difference of two squares property

${\color{Magenta} \mathbf{a^{2}-b^{2}=\left ( a+b \right )\left (a-b \right )}}$

Exploit the difference of two squares

$\mathbf{r_{1}=\frac{[\left ( r_{1} +r_{2}\right )+\left ( r_{1} -r_{2}\right )][\left ( r_{1} +r_{2}\right )-\left ( r_{1}-r_{2} \right )]}{4r_{2}}}$

Now, after applying difference of two squares, r₁ simplifies to

$\mathbf{r_{1}=\frac{\left ( r_{1}+r_{2} \right ) + \left ( r_{1}-r_{2} \right )}{2}}$

Plug in the parameters given that r₁+r₂and r₁-r₂ are known $\mathbf{r_{1}=\frac{-\frac{b}{a}\pm \sqrt{\left ( \frac{b}{a} \right )^{2}-4\frac{c}{a}}}{2}}$

When

$\mathbf{r_{1}=-r_{2}-\frac{b}{a}}$

it reduces to

$\mathbf{r_{1}=\frac{-b+\sqrt{b^{2}-4ac}}{2a}}$

and we get

$\mathbf{r_{2}=\frac{-b-\sqrt{b^{2}-4ac}}{2a}}$

When

$\mathbf{r_{1}=\frac{-b-\sqrt{b^{2}-4ac}}{2a}}$

the other root is

$\mathbf{r_{2}=\frac{-b+\sqrt{b^{2}-4ac}}{2a}}$

And by combining the solutions we obtain

$\mathbf{x=\frac{-b\pm \sqrt{b^{2}-4ac}}{2a}}$

Proof quadratic formula by converting first term into a perfect square

Now, given the general quadratic equation, you are Required↗ to make the first term a perfect square. ${\color{Black} \mathbf{ax^{2}+bx+c=0}}$

Converting the first term into a perfect square is essential so that the quadratic equation resembles the form of this identity

${\color{Magenta} \mathbf{\left ( a+b \right )^{2}=a^{2}+2ab+b^{2}}}$

Multiply the equation by a to boths sides

$\mathbf{a\left ( ax^{2}+bx+c \right )=0}$

The equation becomes

$\mathbf{a^{2}x^{2}+abx+ac=0}$

Rewrite and make the coefficients explict as in general form of quadratic equation

$\mathbf{{\color{Red} a}x^{2}+{\color{Red} b}x+{\color{Red} c}=0}$

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$\mathbf{\left ( ax \right )^{2}+b\left ( ax \right )+ac=0}$

It is time now to complete the square. Take half the coeffient of ax, square the result and add to both sides of the equation or add and subtract the result if all the parameters are on the same side in this case LHS. But you can choose to keep all elements in one side of the equation at initial and intermediate stages as I have done. The coefficient of ax is b . Let’s quickily complete the square

$\mathbf{\left ( ax \right )^{2}+b\left ( ax \right )+\left ( \frac{1}{2}\cdot b \right )^2-\left ( \frac{1}{2}\cdot b \right )^2+ac=0}$

Transform the left side into a perfect square

$\mathbf{\left ( ax+\frac{b}{2} \right )^{2}=\left ( \frac{b}{2} \right )^{2}-ac}$

Find the LCM on the right side and simplify the fraction

$\mathbf{\left ( ax+\frac{b}{2} \right )^{2}=\frac{b^{2}-4ac}{4}}$

Take square roots on both sides to transform the equation into linear

$\mathbf{\sqrt{\left (ax+\frac{b}{2} \right )^{2}}=\pm \sqrt{\frac{b^{2}-4ac}{4}}}$

It reduces to

$\mathbf{ax+\frac{b}{2}=\pm \sqrt{\frac{b^{2}-4ac}{4}}}$

Take all parameters to the right side to make x the subject of the formula

$\mathbf{ax=-\frac{b}{2}\pm \frac{\sqrt{b^{2}-4ac}}{2}}$

Divide both sides by a to find x as

$\mathbf{x=\frac{-b\pm \sqrt{b^{2}-4ac}}{2a}}$