PQ is parallel to SR in a trapezium PQRS. It is given that PQ > SR and the diagonals PR and QS intersect at O. If PO = 3x − 15, OQ = x + 9, OR = x − 5 and OS = 5 and x has two values x₁ and x₂, then the value of ($x_1^2-x_2^2$) is:

The correct option is 11.

In a trapezium $PQRS$, side $PQ$ is parallel to side $SR$ ($PQ\parallel SR$). The diagonals $PR$ and $QS$ intersect at point $O$.

Consider triangles $△POQ$ and $△ROS$:

1. $\angle POQ=\angle ROS$ (Vertically opposite angles)

2. $\angle OPQ=\angle ORS$ (Alternate interior angles, since $PQ\parallel SR$)

3. $\angle OQP=\angle OSR$ (Alternate interior angles, since $PQ\parallel SR$)

Therefore, by AA similarity criterion, $△POQ\sim △ROS$.

Since the ratios of corresponding sides of similar triangles are equal, we have:

$\frac{PO}{OR}=\frac{OQ}{OS}$

Substitute the given values: $PO=3x-15$, $OQ=x+9$, $OR=x-5$, and $OS=5$:

$\frac{3x-15}{x-5}=\frac{x+9}{5}$

Cross-multiplying both sides gives:

$5(3x-15)=(x+9)(x-5)$

$15x-75=x^2-5x+9x-45$

$15x-75=x^2+4x-45$

Rearranging all terms to one side to form a standard quadratic equation:

$x^2+4x-15x-45+75=0$

$x^2-11x+30=0$

Factoring the quadratic equation:

$(x-6)(x-5)=0$

Thus, the two roots are:

$x_1=6$ and $x_2=5$

Now, we evaluate the expression $x_1^2-x_2^2$:

$x_1^2-x_2^2=6^2-5^2=36-25=11$

Hence, the value of $x_1^2-x_2^2$ is 11.