Equal Roots Condition

The first equation is $x^2 - 3px + q = 0$. For this quadratic equation to have equal roots, its discriminant must be zero.

Discriminant, $\Delta = b^2 - 4ac$. Here, $a=1$, $b=-3p$, $c=q$.

So, $(-3p)^2 - 4(1)(q) = 0$.

This simplifies to $9p^2 - 4q = 0$, which gives $q = \frac{9p^2}{4}$.

When roots are equal, the root is $x = -b / (2a)$.

Therefore, the equal root is $x = -(-3p) / (2 \times 1) = \frac{3p}{2}$.

Common Root Substitution

The problem states that the two equations share one common root. Since the first equation has equal roots, the common root must be $\frac{3p}{2}$.

This common root must also satisfy the second equation, $x^2 - 2kx + t = 0$. Substitute $x = \frac{3p}{2}$ into the second equation:

$(\frac{3p}{2})^2 - 2k(\frac{3p}{2}) + t = 0$

$\frac{9p^2}{4} - 3pk + t = 0$

Solving for $t$, we get $t = 3pk - \frac{9p^2}{4}$.

Calculating $(t + q)^2$

We need to find the value of $(t + q)^2$. We have the values for $t$ and $q$:

• $q = \frac{9p^2}{4}$
• $t = 3pk - \frac{9p^2}{4}$

First, find the sum $t + q$:

$t + q = (3pk - \frac{9p^2}{4}) + (\frac{9p^2}{4})$

$t + q = 3pk$

Now, square the sum:

$(t + q)^2 = (3pk)^2$

$(t + q)^2 = 9p^2k^2$