Derive the Quadratic Equation

Question

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

Answer 1

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 $$x^2-4x+1=0$$

Answer 2

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

Answer 3

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 $x = 2 \pm \sqrt{3}.$

Answer 4

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

$$x^2 - 4x + 1 = 0$$