The equation , where k is a constant, has no real roots.
The question indicates that the discriminant is negative since we are told there are no roots – see discriminants. It follows that
At this point we cannot divide both sides by k as we do not know what the sign of k is and so don’t know whether to reverse the inequality or not. You could consider k positive and negative separately but it is easier to factorise here:
This quadratic in k can easily be sketched (by considering the roots) to see that the quadratic is only strictly negative between 0 and 3/4 (not inclusive).
Question of the Week – Week 5
Questions that involve a change of base are quite rare but it is still possible that they will come up in your exam. It should be obvious that a change of base is required because the base is 3 on one of the log expressions whereas it is x on the other. The change of base formula is given (in the Edexel formula booklet) as:
Choosing a=x, x=81 and b=3 in this formula gives:
Substituting this into the original expression gives:
The final log expression can be evaluated as the power of 3 that gives 81 and so the final answer is 4.
Question of the Week – Week 4
The first three terms of a geometric sequence are
where k is a constant. Show that
Given that k is not an integer, show that .
Since the sequence is a geometric one the ratio between each of the terms must be the same i.e. . It follows that . Expanding gives and so as required.
Question of the Week – Week 3
Prove that for all positive x and y:
The trick is to spot the square root on the left and hence consider the square of the right. First note that . Secondly, and are both positive and so . By square rooting both sides, it follows that .
Question of the Week – Week 2
Solution – Although it may not be obvious at first, the above equation is actually a quadratic. It can be written as
and then as
Letting , this can be factorised to
It follows that or giving the solutions or .
Question of the Week – Week 1
Do you know the correct way to solve the following inequality?
Most students will instantly multiply both sides by but it is important to consider if is positive or not. This is because if is negative, the inequality sign will be reversed.