## Question of the Week – Week 6

The equation $kx^2+4kx+3=0$, where k is a constant, has no real roots.

Prove that $\hspace{30pt}0\textless k \hspace{3pt} \textless \hspace{3pt}\frac{3}{4}.$

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

$(4k)^2-4\times k\times 3\textless 0$

or

$16k^2-12k\textless 0$

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:

$4k(4k-3)\textless 0$

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

Simplify $\log_x(81)\times\log_3(x)$.

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:

$\log_a(x)=\frac{\log_b(x)}{\log_b(a)}.$

Choosing a=x, x=81 and b=3 in this formula gives:

$\log_x(81)=\frac{\log_3(81)}{\log_3(x)}.$

Substituting this into the original expression gives:

$\log_x(81)\times\log_3(x)=\frac{\log_3(81)}{\log_3(x)}\times\log_3(x)=\log_3(81).$

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

$7k-5, 5k-7, 2k+10,$

where k is a constant. Show that

$11k^2-130k+99=0.$

Given that k is not an integer, show that $\hspace{10pt}k=\frac{9}{11}$.

Since the sequence is a geometric one the ratio between each of the terms must be the same i.e. $\frac{5k-7}{7k-5}=\frac{2k+10}{5k-7}$. It follows that $(5k-7)(5k-7)=(2k+10)(7k-5)$. Expanding gives $25k^2-70k+49=14k^2+60k-50$ and so $11k^2-130k+99=0$ as required.

## Question of the Week – Week 3

Prove that for all positive x and y:

$\sqrt{xy}\leq \frac{x+y}{\sqrt{2}}.$

The trick is to spot the square root on the left and hence consider the square of the right. First note that $\left(\frac{x+y}{\sqrt{2}}\right)^2=\frac{(x+y)^2}{2}=\frac{x^2+2xy+y^2}{2}$. Secondly, $x^2$ and $y^2$ are both positive and so $\frac{2xy}{2}=xy\leq \frac{(x+y)^2}{2}$. By square rooting both sides, it follows that $\sqrt{xy}\leq \frac{x+y}{\sqrt{2}}$.

## Question of the Week – Week 2

Solve $2^{2x+1}-17\left(2^x\right)=-8.$

Solution – Although it may not be obvious at first, the above equation is actually a quadratic. It can be written as

$2\times 2^{2x} -17\left(2^x\right)=-8$

and then as

$2\left( 2^{x}\right)^2 -17\left(2^x\right)+8=0.$

Letting $y=2^x$, this can be factorised to

$\left( 2y-1\right)\left( y-8\right)=0.$

It follows that $y=2^x=\frac{1}{2}$ or $y=2^x=8$ giving the solutions $x=-1$ or $x=3$.

## Question of the Week – Week 1

Do you know the correct way to solve the following inequality?

$\frac{2}{x-5}\geq 6,\hspace{15pt}x\ne 5$

Most students will instantly multiply both sides by $x-5$ but it is important to consider if $x-5$ is positive or not. This is because if $x-5$ is negative, the inequality sign will be reversed.