# Irrational roots of quadratic equation

The irrational roots of quadratic equation are of the form $\alpha + \sqrt{\beta}$
and $\alpha - \sqrt{\beta}$.These two are conjugate roots. Here $\alpha$ is a rational number where as $\sqrt{\beta}$ is an irrational number.
Proof for irrational roots of quadratic equation :
As $\alpha + \sqrt{\beta}$ is a root of the quadratic equation, $ax^{2}$ + bx + c = 0 so it satisfies the equation.
So put x = $\alpha + \sqrt{\beta}$ in $ax^{2}$ + bx + c = 0
$a(\alpha + \sqrt{\beta})^{2} + b(\alpha + \sqrt{\beta}$) + c = 0

$a (\alpha^{2} + \beta^{2} + 2.\alpha.\sqrt{\beta}) + b.\alpha + b.\sqrt{\beta}$ + c = 0

$a.\alpha^{2} + a.\beta^{2} + 2.a.\alpha.\sqrt{\beta} + b.\alpha + b.\sqrt{\beta}$ + c = 0

$a.\alpha^{2} + a.\beta^{2} + b.\alpha + c + 2.a.\alpha.\sqrt{\beta} + b.\sqrt{\beta}$ = 0

$a.\alpha^{2} + a.\beta^{2} + b.\alpha + c + (2.a.\alpha + b)\sqrt{\beta}$ = 0

$a.\alpha^{2} + a.\beta^{2} + b.\alpha + c + (2.a.\alpha + b)\sqrt{\beta} = 0 + 0.\sqrt{\beta}$

∴ $a.\alpha^{2} + a.\beta^{2} + b.\alpha$ + c = 0 and 2.a.$\alpha$ + b = 0
Now put x = $\alpha - \sqrt{\beta}$ in the equation $ax^{2}$ + bx + c = 0, we get
$a(\alpha - \sqrt{\beta})^{2} + b(\alpha - \sqrt{\beta}$) + c = 0

$a (\alpha^{2} + \beta^{2} - 2.\alpha.\sqrt{\beta}) + b.\alpha - b.\sqrt{\beta}$ + c = 0

$a.\alpha^{2} + a.\beta^{2} - 2.a.\alpha.\sqrt{\beta} + b.\alpha - b.\sqrt{\beta}$ + c = 0

$a.\alpha^{2} + a.\beta^{2}+ b.\alpha + c - 2.a.\alpha.\sqrt{\beta}- b.\sqrt{\beta}$ = 0

$a.\alpha^{2} + a.\beta^{2} + b.\alpha + c - (2.a.\alpha + b)\sqrt{\beta}$ = 0

$a.\alpha^{2} + a.\beta^{2} + b.\alpha + c - (2.a.\alpha + b)\sqrt{\beta} = 0 - 0.\sqrt{\beta}$

From the above we can see that the equation $ax^{2}$ + bx + c = 0 is satisfied by $\alpha - \sqrt{\beta}$ when $\alpha + \sqrt{\beta}$ is a surd of the equation. Now its clear that if one root of the equation $ax^{2}$ + bx + c = 0 is $\alpha + \sqrt{\beta}$ then the other irrational root is $\alpha - \sqrt{\beta}$. Thus, ($\alpha +\sqrt{\beta}$) and ($\alpha - \sqrt{\beta}$) are conjugates roots. So quadratic equation irrational roots are occur in pairs.

## Examples on Irrational roots of quadratic equation

1) Find the quadratic equation with rational coefficients which has 1 + $\sqrt{3}$ as a root.
Solution : As we know that, irrational roots of quadratic equation are occurs in pair. If one root is 1 + $\sqrt{3}$ then the other root must be 1 - $\sqrt{3}$.
$\alpha = 1 + \sqrt{3}$ and $\beta = 1 - \sqrt{3}$
∴ sum of roots = $\alpha + \beta = 1 + \sqrt{3} + 1 - \sqrt{3}$ = 2
Product of roots = $\alpha . \beta = (1 + \sqrt{3}) (1 - \sqrt{3})$ = 1- 3 = -2
$x^{2}$ + (sum of roots).x + Product of roots = 0
$x^{2} - (\alpha + \beta).x + \alpha . \beta$ = 0
$x^{2}$ - 2x - 2 = 0
∴ Required quadratic equation is $x^{2}$ - 2x - 2 = 0

2) Find the quadratic equation with rational coefficients which has
5 - 2$\sqrt{5}$ as a root.
Solution : As we know that, irrational roots of quadratic equation are occurs in pair. If one root is 5 - 2$\sqrt{5}$ then the other root must be
5 + 2$\sqrt{5}$.
$\alpha = 5 - 2\sqrt{5}$ and $\beta = 5 + 2\sqrt{5}$
∴ sum of roots = $\alpha + \beta = 5 - 2\sqrt{5} + 5 + 2 \sqrt{5}$ = 10
Product of roots = $\alpha . \beta = (5 - 2\sqrt{5}) (5 + 2\sqrt{5})$ = 25 -20 = 5
$x^{2}$ + (sum of roots).x + Product of roots = 0
$x^{2} - (\alpha + \beta).x + \alpha . \beta$ = 0
$x^{2}$ - 10x + 5 = 0
∴ Required quadratic equation is $x^{2}$ - 10x + 5 = 0

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