I realize many students have a challenge with presenting the solution of proving question. I am going to illustrate the correct way using the following question.
Show that is always positive for all real values of x.
=> what is always +ve? the values of y not x ( x can be of any values)
=> The graph will not touch x-axis; it is "hanging" above the x-axis
=> No roots, hence discriminant is less than 0
TIPS FOR SHOWING/PROVING QUESTION
- Have an end in your mind. Be clear of the underlying concepts that the question is asking you to prove. Work towards that. For example, in the question, we want to prove that the discriminant
is less than 0. <== this is the end.
- Sketch to have a clearer idea.
I need help to solve the following ; Given tha Q is the range 180digrees <Q<270digrees and CCos Q=-40/41 and is angle in the range 90 degrees <& <180 Degrees such that sin & = 7/25
a) Find the values of
sin Q-tan&
1-sin Q +cos &
b) Solve all tthe angles of x i.e. between O degrees to 360 degrees for the equation sin Qx tan & = 1tanx
Patricia:
You would need to draw 2 triangles with one showing angle Q and the other showing angle &.
Fill up the sides.
These 2 triangles would be able to answer sin Q-tan&
1-sin Q +cos &
As for part (b), I do not quite get your question.
can i have self assessiment questions on quadratic equestions for a beginner
show that 22/(3x²-4x+5) is positive for all real values of x and find its greatest value.
You can calculate the value of discriminant b^2 - 4ac.
It should be less than 0 which means there is no real roots and the quadratic expression is always positive for all real values of x.
To find its greatest value of 22/(3x²-4x+5), find the smallest value of 3x²-4x+5 by completing the square or finding stationary value of (3x²-4x+5).