For a quadratic equation, $ax^2 + bx + c = 0$, why is
$ax^2 + bx + c = a(x-\alpha)(x-\beta)$ where alpha, beta are the roots of the equation? Why not just $(x-\alpha)(x-\beta)$?
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6Well, what do you get if you multiply $(x - \alpha)(x - \beta)$? You don't get any terms that look like $ax^2$. – Jun 15 '14 at 04:11
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In a quadratic equation where the coefficient of the first term ($x^2$) is unity, following holds:
- the sum of the roots is equal to the coefficient of $x$ with its sign changed;
- the product of the roots is equal to the third term.
Your equation is $\hspace{12 pt}ax^2+bx+c=0\hspace{12 pt}$coefficient of the first term ($x^2$) is NOT $1$.
If you write the equation as $\hspace{12 pt}a(x^2+\frac{b}{a}x+\frac{c}{a})=0$,
the coefficient of the first term of the quadratic expression in the bracket ($x^2$) is $1$.
Let the roots of the quadratic equation in the bracket be $\alpha,\beta$.
Then it can be written as $\hspace{12 pt}a(x-\alpha)(x-\beta)=0$,
where $\hspace{12 pt}\alpha+\beta=-\frac{b}{a}\hspace{12 pt}$ and $\hspace{12 pt}\alpha.\beta=\frac{c}{a}$
Vikram
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Here, α and β are not the roots of the original equation ax²+bx+c=0 ? – ADITYA DAS Aug 29 '23 at 06:59
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If so then if α and β are the roots of ax²+bx+c=0, then this equation is not equal to a(x-α)(x-β) – ADITYA DAS Aug 29 '23 at 07:00