Here is the solution to Question 19. Question 19: Given that $9x^2 - 30x - 40 + k$ is a perfect square, find the value of $k$. Step 1: Recall the form of a perfect square trinomial. A perfect square trinomial is of the form $(ax \pm b)^2 = a^2x^2 \pm 2abx + b^2$. Comparing $9x^2 - 30x - 40 + k$ with this form, we can identify $a^2x^2 = 9x^2$ and $2abx = -30x$. Step 2: Determine the values of $a$ and $b$. From $a^2x^2 = 9x^2$, we get $a = 3$. From $2abx = -30x$, substitute $a=3$: $2(3)bx = -30x$ $6bx = -30x$ $6b = -30$ $b = -5$ (or we can consider $b=5$ and the form $(ax-b)^2$) Step 3: Determine the constant term of the perfect square. If $a=3$ and $b=5$, the perfect square is $(3x - 5)^2$. Expanding $(3x - 5)^2$: $$(3x - 5)^2 = (3x)^2 - 2(3x)(5) + 5^2$$ $$= 9x^2 - 30x + 25$$ Step 4: Equate the constant term of the given expression to the constant term of the perfect square. The given expression is $9x^2 - 30x - 40 + k$. For this to be a perfect square, its constant term must be $25$. So, we set $-40 + k = 25$. Step 5: Solve for $k$. $$-40 + k = 25$$ $$k = 25 + 40$$ $$k = \boxed{65}$$