Hello! here is my tutorial on quadratic equation:)
consider the quadratic equation\[ax^2+bx+c=0 \space a \neq 0\]dividing both sides by 'a'. we get \[x^2+\frac{b}{a}x+\frac{c}{a}=0\]\[x^2+\frac{2b}{2a}x+\left( \frac{b}{2a} \right)^2-\left( \frac{b}{2a} \right)^2+\frac{c}{a}=0\]\[\left( x+\frac{b}{2a} \right)^2-\left( \frac{b}{2a} \right)^2+\frac{c}{a}=0\]\[\left( x+\frac{b}{2a} \right)^2=\left( \frac{b}{2a} \right)^2-\frac{c}{a}\]\[x+\frac{b}{2a}={\sqrt{b^2-4ac}\over4a^2}\]\[x=\frac{-b \pm \sqrt{b^2-4ac}}{2a}\] \[b^2-4ac=D \space is \space called \space discriminant.\]
typo i think it's 4a, not 4a^2 :D Nice job tho.
a) If D>0 then there are two distinct roots given\[\alpha={-b + \sqrt{b^2-4ac}\over2a}\]\[\beta={-b-\sqrt{b^2-4ac}\over2a}\]b) if D=0 then the roots are equal & real\[\alpha+\beta={-b \over a}\]c) if D<0 then there are no real roots sum of the roots alpha+beta=-b/a & the product of the roots alpha*beta=c/a Therefore the quadratic equation is\[x^2-(\alpha+\beta)x+(\alpha \beta)=0\]
im interested to know which part of the quadratic equation tells you if a polynomial is factorable. @jiteshmeghwal9
sir ! i think it's the middle term.
Have an Example in ur tutorial @jiteshmeghwal9 ;)
nyss
sorry i meant quadratic formula not equation
\[\Huge{\color{gold}{\star \star}{\color{blue}{Keep \space It \space Up \space Brother}}}\]
@lgbasallote sir. am i right??
How should i know? I was asking you o.O
Lol no. Ive seen so many questions about it but idk the answer so i was hoping you did since you were making a tutorial heh. Im not as sadistic as tou think :p
Btw you did not answer my question...i was asking which part of the quadratic FORMULA states whether a quadratic equation is factorable :(
What do you mean?
sir the discriminant tells whether the given equation is factorable
& i have given the tutorial of solving a quadratic equation
\[b^2-4ac=d \space is \space called \space "DISCRIMINANT"\]
@lgbasallote sir!
good job!!
thanx!
i see thanks
yw:)
if d is a perfect square, then the equation is factorable.
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