Question

If m and n are rational numbers and if m-√3 is a root of the equation x^2-6x+4+n=0 find m and n

Answer 1

m=3, n=2

Answer 2

The given quadratic equation is $x^2 - 6x + 4 + n = 0$.
The coefficients of this equation are $1$, $-6$, and $4+n$.
It is given that $m$ and $n$ are rational numbers.
Since $n$ is rational, $4+n$ is also a rational number.
Therefore, all coefficients of the quadratic equation ($1$, $-6$, $4+n$) are rational.

A fundamental property of quadratic equations with rational coefficients is that if an irrational root of the form $a - \sqrt{b}$ (where $a$ is rational and $\sqrt{b}$ is irrational) exists, then its conjugate $a + \sqrt{b}$ must also be a root.

Given one root is $m - \sqrt{3}$. Since $m$ is rational and $\sqrt{3}$ is irrational, $m - \sqrt{3}$ is an irrational root.
Therefore, the other root must be $m + \sqrt{3}$.

Let the roots be $\alpha = m - \sqrt{3}$ and $\beta = m + \sqrt{3}$.

For a quadratic equation $ax^2 + bx + c = 0$:

1. Sum of roots $= -\frac{b}{a}$
2. Product of roots $= \frac{c}{a}$

In our equation $x^2 - 6x + 4 + n = 0$, we have $a=1$, $b=-6$, and $c=4+n$.

1. Sum of roots:
$\alpha + \beta = (m - \sqrt{3}) + (m + \sqrt{3}) = 2m$
From the equation, sum of roots $= -\frac{-6}{1} = 6$
Equating the two expressions for the sum of roots:
$2m = 6$
$m = 3$

2. Product of roots:
$\alpha \cdot \beta = (m - \sqrt{3})(m + \sqrt{3})$
Using the identity $(A-B)(A+B) = A^2 - B^2$:
$\alpha \cdot \beta = m^2 - (\sqrt{3})^2 = m^2 - 3$
From the equation, product of roots $= \frac{4+n}{1} = 4+n$
Equating the two expressions for the product of roots:
$m^2 - 3 = 4 + n$

Now, substitute the value of $m=3$ into this equation:
$3^2 - 3 = 4 + n$
$9 - 3 = 4 + n$
$6 = 4 + n$
$n = 6 - 4$
$n = 2$

Both $m=3$ and $n=2$ are rational numbers, which is consistent with the problem statement.

The final values are $m=3$ and $n=2$.