Questions of the Week - StudyWell

The expression for y in the first equation can be substituted into the second then simplified to give:

Rearranging and completing the square to solve the resulting quadratic as follows:

Simplifying the surd gives with corresponding y solutions given by . Note that these y solutions are best found by substituting x into the most basic of the original equations.

Consider , for example, x has to be greater than 3 or less than -3. Similarly, if , 5-c has to be greater than root 12 or less than minus root 12. Subtracting 5 suggests that or . Multiplying both sides of the inequalities by -1 reverses the sign of the inequality (click here for more on inequalities). Finally, we have or .

Solution:

First think about reading the logs – see the logarithms for more details. can actually be read as ‘the power of 2 that gives 32’ – we know this to be 5 since . Similarly, and so 

Multiply both sides of the equation by to get

.

Note that there is no rule for simplifying a log that is squared – see log rules. Square rooting gives giving the solutions and .

If we let , we can write the equation as . We are not able to factorise this quadratic and so we complete the square (you could also use the quadratic formula):

It follows that

See surds for the final stage of working – make sure to do the full calculations for yourself.

This question was used in a recent exam and students did not like question 2 – they were not used to seeing two surds in the denominator. The same principles of rationalising the denominator apply – multiplying the top and bottom of the fraction with the same expression where the minus sign is replaced with a plus eliminates the unwanted surds.

When you are confronted with a question that looks unusual, you should always try to make it look as familiar as possible so you can use the usual methods to tackle it. Start by writing:

You can then isolate the inside of the brackets and complete the square, i.e.

and it follows that

and so q=-1 and p=-2.

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Rearranging to get all trigonometric terms on one side:

or

Students often get stuck at this point – if they don’t know their double angle formulae very well. Using we can convert the equation into something solvable:

see the graph if cos noting that if then we need all the solutions for  . It follows that the solutions are

.

The question indicates that the discriminant is negative since we are told there are no roots – see discriminants. It follows that

or

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).

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.

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.

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 .

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 .