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.

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TopNewestLet 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 · 3 years, 3 months ago

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– Kartik Sharma · 3 years, 3 months ago

Thanks for your advice! And ye, thanks for the proof, also.Log in to reply

Um the quad equation proof can be done using squares. – Joshua Ong · 3 years, 3 months ago

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– Kartik Sharma · 3 years, 3 months ago

tell me the whole procedure, how?Log in to reply

– Kartik Sharma · 3 years, 3 months ago

also, the GP series proof, any?Log in to reply