Proofs......omg!

I often wonder how we get a root of a quadratic equation by putting a formula

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

But can anyone tell me the proof of this, this might be interesting, isn't it? Also, I would be pleased to know the proof of the GP series to infinity. Hey, you can also share some more interesting proofs which I would love to learn.

#Proofs

Easy Math Editor

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Comments

Let the equation be $ax^2 + bx + c = 0$

Then $x^2 + \frac{bx}{a} + \frac{c}{a} = 0$

$x^2 + \frac{bx}{a} + \frac{b^2}{4a^2} - \frac{b^2}{4a^2} + \frac{c}{a} = 0$

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

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

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

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

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

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

My advice. Get NCERT Math books of higher classes. It'll contain most if not all of the basic proofs.

Siddhartha Srivastava - 7 years, 2 months ago

Thanks for your advice! And ye, thanks for the proof, also.

Kartik Sharma - 7 years, 2 months ago

Um the quad equation proof can be done using squares.

Joshua Ong - 7 years, 2 months ago

tell me the whole procedure, how?

Kartik Sharma - 7 years, 2 months ago

also, the GP series proof, any?

Kartik Sharma - 7 years, 2 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$