Question

Suppose a and b are rational and $\alpha$, $\beta$ be the roots of $x^2+2ax+b=0$, then the equation with rational coefficients on of whose roots is $\alpha+\beta+\sqrt{\alpha^2+\beta^2}$ is

• A. $x^2+4ax+2b=0$
• B. $x^2+4ax-2b=0$
• C. $x^2-4ax+2b=0$
• D. $x^2-4ax-2=0$

Answer 1

Answer: (A)

x^2+4ax+2b=0

Answer 2

The given quadratic equation is $x^2+2ax+b=0$.
Let its roots be $\alpha$ and $\beta$.
According to Vieta's formulas:
Sum of roots: $\alpha + \beta = -2a$
Product of roots: $\alpha \beta = b$

We are given that $a$ and $b$ are rational.

We need to find an equation with rational coefficients, one of whose roots is $\gamma = \alpha+\beta+\sqrt{\alpha^2+\beta^2}$.

First, let's simplify the term $\sqrt{\alpha^2+\beta^2}$:
We know that $\alpha^2+\beta^2 = (\alpha+\beta)^2 - 2\alpha\beta$.
Substitute the values of $\alpha+\beta$ and $\alpha\beta$:
$\alpha^2+\beta^2 = (-2a)^2 - 2(b)$
$\alpha^2+\beta^2 = 4a^2 - 2b$

Now substitute this back into the expression for $\gamma$:
$\gamma = (\alpha+\beta) + \sqrt{4a^2 - 2b}$
$\gamma = -2a + \sqrt{4a^2 - 2b}$

Let this root be $X_1 = -2a + \sqrt{4a^2 - 2b}$.
Since $a$ and $b$ are rational, $-2a$ is rational.
For a quadratic equation to have rational coefficients, if one root is of the form $p+\sqrt{q}$ (where $p$ is rational and $\sqrt{q}$ is irrational), then the other root must be its conjugate, $p-\sqrt{q}$. If $\sqrt{q}$ is rational, then both roots are rational, and the rule still holds.

So, let the other root be $X_2 = -2a - \sqrt{4a^2 - 2b}$.

Now, we form the quadratic equation $x^2 - (X_1+X_2)x + X_1X_2 = 0$.

Calculate the sum of the roots ($X_1+X_2$):
$X_1+X_2 = (-2a + \sqrt{4a^2 - 2b}) + (-2a - \sqrt{4a^2 - 2b})$
$X_1+X_2 = -2a - 2a + \sqrt{4a^2 - 2b} - \sqrt{4a^2 - 2b}$
$X_1+X_2 = -4a$

Calculate the product of the roots ($X_1X_2$):
$X_1X_2 = (-2a + \sqrt{4a^2 - 2b})(-2a - \sqrt{4a^2 - 2b})$
This is in the form $(A+B)(A-B) = A^2 - B^2$, where $A=-2a$ and $B=\sqrt{4a^2 - 2b}$.
$X_1X_2 = (-2a)^2 - (\sqrt{4a^2 - 2b})^2$
$X_1X_2 = 4a^2 - (4a^2 - 2b)$
$X_1X_2 = 4a^2 - 4a^2 + 2b$
$X_1X_2 = 2b$

Substitute the sum and product of roots into the quadratic equation formula:
$x^2 - (-4a)x + (2b) = 0$
$x^2 + 4ax + 2b = 0$

All coefficients ($1$, $4a$, $2b$) are rational since $a$ and $b$ are rational.

Comparing this equation with the given options:
(A) $x^2+4ax+2b=0$
(B) $x^2+4ax-2b=0$
(C) $x^2-4ax+2b=0$
(D) $x^2-4ax-2=0$

The derived equation matches option (A).

The final answer is \boxed{\text{x^2+4ax+2b=0}}.

Explanation of the solution:

1. Identify Given Information: The roots $\alpha, \beta$ of $x^2+2ax+b=0$ are related by Vieta's formulas: $\alpha+\beta = -2a$ and $\alpha\beta = b$. $a, b$ are rational.
2. Simplify the Target Root: The desired root is $\gamma = \alpha+\beta+\sqrt{\alpha^2+\beta^2}$. Use the identity $\alpha^2+\beta^2 = (\alpha+\beta)^2 - 2\alpha\beta$ to express $\gamma$ in terms of $a$ and $b$. This gives $\gamma = -2a + \sqrt{4a^2-2b}$.
3. Form the Conjugate Root: Since the coefficients of the new quadratic equation must be rational, if one root is $p+\sqrt{q}$ (where $p$ is rational and $\sqrt{q}$ might be irrational), the other root must be its conjugate, $p-\sqrt{q}$. So, the other root is $-2a - \sqrt{4a^2-2b}$.
4. Calculate Sum and Product of New Roots:
   • Sum: $(-2a + \sqrt{4a^2-2b}) + (-2a - \sqrt{4a^2-2b}) = -4a$.
   • Product: $(-2a + \sqrt{4a^2-2b})(-2a - \sqrt{4a^2-2b}) = (-2a)^2 - (4a^2-2b) = 4a^2 - 4a^2 + 2b = 2b$.
5. Construct the Equation: Use the general form $x^2 - (\text{sum of roots})x + (\text{product of roots}) = 0$. Substituting the calculated values: $x^2 - (-4a)x + 2b = 0$, which simplifies to $x^2+4ax+2b=0$.