Finding the Quadratic Equation with Squared Roots

The problem asks us to find a new quadratic equation whose roots are the squares of the roots of the given quadratic equation: $6x^2 + 13x + 7 = 0$.

Understanding the Original Equation and Its Roots

Let the roots of the given quadratic equation $6x^2 + 13x + 7 = 0$ be $\alpha$ and $\beta$. For a general quadratic equation $ax^2 + bx + c = 0$, Vieta's formulas give us the sum and product of the roots:

• Sum of roots: $\alpha + \beta = -\frac{b}{a}$
• Product of roots: $\alpha \beta = \frac{c}{a}$

In our given equation, $a=6$, $b=13$, and $c=7$. So, the sum and product of the roots $\alpha$ and $\beta$ are:

• Sum of roots: $\alpha + \beta = -\frac{13}{6}$
• Product of roots: $\alpha \beta = \frac{7}{6}$

Calculating the Sum and Product of the New Roots

We need to find the quadratic equation whose roots are $\alpha^2$ and $\beta^2$. Let the new roots be $r_1 = \alpha^2$ and $r_2 = \beta^2$. The new quadratic equation will be of the form $x^2 - (\text{sum of new roots})x + (\text{product of new roots}) = 0$.

First, let's find the sum of the new roots, which is $\alpha^2 + \beta^2$. We can express this in terms of $\alpha + \beta$ and $\alpha \beta$ using the identity $\alpha^2 + \beta^2 = (\alpha + \beta)^2 - 2\alpha \beta$.

Substitute the values we found for $\alpha + \beta$ and $\alpha \beta$:

$\alpha^2 + \beta^2 = \left(-\frac{13}{6}\right)^2 - 2\left(\frac{7}{6}\right)$

$\alpha^2 + \beta^2 = \frac{169}{36} - \frac{14}{6}$

To subtract the fractions, we find a common denominator, which is 36:

$\alpha^2 + \beta^2 = \frac{169}{36} - \frac{14 \times 6}{6 \times 6} = \frac{169}{36} - \frac{84}{36}$

$\alpha^2 + \beta^2 = \frac{169 - 84}{36} = \frac{85}{36}$

So, the sum of the new roots is $\frac{85}{36}$.

Next, let's find the product of the new roots, which is $\alpha^2 \beta^2$. This can be written as $(\alpha \beta)^2$.

Substitute the value of $\alpha \beta$:

$\alpha^2 \beta^2 = \left(\frac{7}{6}\right)^2 = \frac{49}{36}$

So, the product of the new roots is $\frac{49}{36}$.

Forming the New Quadratic Equation

The general form of a quadratic equation with roots $r_1$ and $r_2$ is $x^2 - (r_1 + r_2)x + r_1 r_2 = 0$. Substituting the sum ($\alpha^2 + \beta^2 = \frac{85}{36}$) and product ($\alpha^2 \beta^2 = \frac{49}{36}$) of the new roots, we get:

$x^2 - \left(\frac{85}{36}\right)x + \frac{49}{36} = 0$

To clear the denominators and get integer coefficients, we multiply the entire equation by 36:

$36 \left(x^2 - \frac{85}{36}x + \frac{49}{36}\right) = 36 \times 0$

$36x^2 - 36 \times \frac{85}{36}x + 36 \times \frac{49}{36} = 0$

$36x^2 - 85x + 49 = 0$

Comparing with the Options

The derived quadratic equation is $36x^2 - 85x + 49 = 0$. Let's compare this with the given options:

1. $36x^2 - 87x + 49 = 0$
2. $36x^2 - 85x + 49 = 0$
3. $36x^2 - 85x - 49 = 0$
4. $36x^2 + 87x - 49 = 0$

Our derived equation matches option 2.

Revision Table: Key Quadratic Concepts

Additional Information: Transforming Roots of Quadratic Equations

This problem is an example of transforming the roots of a quadratic equation. If you know the equation with roots $\alpha$ and $\beta$, you can find the equation with roots related to $\alpha$ and $\beta$ (like $\alpha+k, \beta+k$; $k\alpha, k\beta$; $1/\alpha, 1/\beta$; $\alpha^2, \beta^2$; etc.) by finding the sum and product of the new roots in terms of the sum and product of the original roots.

• For roots $\alpha+k, \beta+k$: New sum = $(\alpha+k) + (\beta+k) = (\alpha+\beta) + 2k$. New product = $(\alpha+k)(\beta+k) = \alpha\beta + k(\alpha+\beta) + k^2$.
• For roots $k\alpha, k\beta$: New sum = $k\alpha + k\beta = k(\alpha+\beta)$. New product = $(k\alpha)(k\beta) = k^2 \alpha\beta$.
• For roots $1/\alpha, 1/\beta$: New sum = $1/\alpha + 1/\beta = (\alpha+\beta)/(\alpha\beta)$. New product = $(1/\alpha)(1/\beta) = 1/(\alpha\beta)$.

Knowing these transformations and how to use Vieta's formulas is key to solving such problems efficiently.