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Find the Quadratic Equation whose roots are $2+\sqrt3$ and $2-\sqrt3$.

Some basics:

  • The general form of a Quadratic Equation is $ax^2+bx+c=0$

  • In Quadratic Equation, $ax^2+bx+c=0$, if $\alpha$ and $\beta$ are the roots of the given Quadratic Equation, Then, $$\alpha+\beta=\frac{-b}{a}, \alpha\beta=\frac{c}{a}$$

I am here confused that how we can derive a Quadratic Equations from the given roots

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up vote 2 down vote accepted

Here $$-\frac ba=\alpha+\beta=2+\sqrt3+2-\sqrt3=4$$ and
$$\frac ca=\alpha\beta=(2+\sqrt3)(2-\sqrt3)=2^2-3=1$$

So, the quadratic equation becomes $$x^2-4x+1=0$$

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@Alpha, I think your confusion arises due to $a=1$ – lab bhattacharjee Jan 22 '13 at 17:18

It seems to me that the sum and product relations for roots is much more specialized knowledge than is needed for this problem, although perhaps the problem was intended to be an application of these relations.

Just work backwards from "the solution":

$$x = 2 \pm \sqrt{3} \; \implies \; (x-2)=\pm\sqrt{3} \; \implies \; (x-2)^2 = \left(\pm \sqrt{3}\right)^2$$

$$\implies \; (x-2)^2 = 3,$$

and now you have a quadratic equation whose solution is $x = 2 \pm \sqrt{3}.$

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$$ 0 = \left(x-\alpha\right)\left(x-\beta\right)=x^2 -\left(\alpha+\beta\right)x + \alpha \beta $$

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