Question

As shown in the diagram below, the diagonals of parallelogram QRST intersect at E.
If QE = x^2 + 6x, SE = x + 14 and TE = 6x - 1, determine TE algebraically.

Answer 1

Answer:

TE = 11

Step-by-step explanation:

x² + 6x = x + 14

x² + 5x - 14 = 0

D = 25 - 4(- 14) = 81 = 9²

$x_{1}$ = $\frac{-5 +\sqrt{81} }{2}$ = 2

$x_{2}$ = $\frac{-5-9}{2}$ = - 7 (N/A)

TE = 6(2) - 1 = 11