Derive the Quadratic Equation Find the Quadratic Equation whose roots are $2+\sqrt3$ and $2-\sqrt3$.
Some basics:


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*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

This appears to be a standard question from the R.D. Sharma textbook used in Indian schools. A common trick I use for this type of sums is
$$x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$
Comparing with the given roots, we get
$$x = \frac{4 \pm \sqrt{4 - 4c}}{2}$$
In this step we find the value of $a$ to be $1$, $b$ to be $-4$.
To find $c$, simply multiply the roots and divide by $a$ to get c.
$$c = \frac{(2 + \sqrt 3)(2 - \sqrt 3)}{1}$$
$$ c = \frac{4 - 3}{1}$$
$$c = 1$$
We know that the standard form of a quadratic equation is
$$ax^2 + bx + c = 0$$
Hence substituting $a$, $b$ and $c$ into the standard form yields
$$x^2 - 4x + 1 = 0$$
