{jatex options:inline}x^2+x^2_2kx+k^2=x^2+4kx+4k^2{/jatex}

{jatex options:inline}x^2-2kx-3k^2=0{/jatex}

{jatex options:inline}(x+k)(x-3k)=0{/jatex}

Hence {jatex options:inline}x+k=0 \rightarrow x=-k{/jatex} or {jatex options:inline}x-3k=0 \rightarrow x=3k{/jatex}.

This means that the sides of the triangle are {jatex options:inline}3k, \:4k, \: 5k{/jatex} - the triangle is just a scaled up 3-4-5 triangle.]]>

4, 9, 14, 19, 24 is an arithmetic sequence because we add 5 to each term to get the next term. The general form for the nth term in a geometric sequence is:

whereis the first term andis the difference between any two successive terms.

Thereflects the fact that to get the 1 ^{st } term we don't have to add anything: only from the 1 ^{st } term do we start adding things.

When we add up n terms, we write down an expression like,

By writing this backwards we obtain

We can now add the two sequences, gettingon the left hand side and altogether n terms all the same,on the right hand side, so

We may be asked: The 3 ^{rd } term of an arithmetic sequence is 9 and the 5 ^{th } term is 17. Find the first term, the common difference and the smallest value of n such that

and

Now solve the simultaneous equations

(1)

(2)

Subinto (1)

Solve

Non integer or negative values of n are not allowed here, because we are considering only the natural numbers, so

]]>Ifthensince that would mean dividing by 0, andsinceThese equations can also be obtained from the formby considering the denominatorwhich may not take the value 0, soand the coefficient ofin numerator and denominator. LettinggivesIn fact ifthe equation of theasymptote is given by dividing the coefficient ofin numerator and denominator to give

Example:The equation of the x asymptote isand the equation of the asymptote is given bySketching the graph can be aiding by finding the intercepts with the axes.andThe graph is sketched below.

]]>We need {jatex options:inline}\mathbf{BC}= - \mathbf{b}+ \mathbf{c} \rightarrow \mathbf{BP}= \frac{1}{2}(- \mathbf{b}+ \mathbf{c}){/jatex}.

The equation of the line {jatex options:inline}BQ{/jatex} is {jatex options:inline}\mathbf{b}+ (- \mathbf{b}+ \frac{\mathbf{c}})s=(1-s) \mathbf{b}+ \frac{s}{2} \mathbf{c}{/jatex}.

The equation of the line {jatex options:inline}AP{/jatex} is {jatex options:inline}(\mathbf{b} + \mathbf{c}{2})t= \frac{t}{2} \mathbf{b} + \frac{t}{2} \mathbf{c}{/jatex} (remembering that {jatex options:inline}\mathbf{b}, \; \mathbf{c}{/jatex} form a parallelogram, of which the triangle forms a half.

Equating the coefficients of {jatex options:inline}\mathbf{b}, \; \mathbf{c}{/jatex} to find the intersection gives

{jatex options:inline}\frac{s}{2}= \frac{t}{2}{/jatex}

From the second {jatex options:inline}s=t{/jatex} and from the first {jatex options:inline}s=t= \frac{2}{3}{/jatex}.]]>

The midpoint of BC is point P with coordinates {jatex options:inline}\frac{(2,3)+(8,5)}{2}=(5,4){/jatex}

The midpoint of AD is point Q with coordinates {jatex options:inline}\frac{(1,1)+(10,4)}{2}=(5.5,2.5){/jatex}

The gradient of QP is nbsp;{jatex options:inline}\frac{(4-2.5}{5-5.5}=\frac{1.5}{-0.5}=-3{/jatex}

The gradient of BC is nbsp;{jatex options:inline}\frac{(5-3}{8-2}=\frac{2}{6}=\frac{1}{3}{/jatex}

These gradients multiply to give {jatex options:inline}-1{/jatex} so the trapezium is isosceles.]]>

The circumference of a circle is 2%pi r. If we haven't got a whole circle or we want to find the length of just part of it's circumference, we find find the cicumference of just that part of the circle. The fraction we have got is {%theta} over {2%pi} – since we work in radians with circular measure.

Hence

The same logic can be used to find the formula for the area of a sector, illustrated above right.

These formulae may be used to answer the following questions.

Find the area of the segment shaded green below.

The area of the sector is

The area of the purple shaded triangle isusing the standard formula for the area of a triangle.

The area A is given by the formulawhereis in radians.. This is just the different between the area of the large sector with angleand the small sector with angle

]]>possible ways of making a choice of six. The order of selection doesn't matter here. So that although each of the 10 are distinct, once they are picked they can line up in any order, and the order of lining up does not matter.

Working from first principles we might explain like this. We are going to pick 6 from ten. There are 10 possible choices for the first one, 9 possible choices for the second one, 8 for the third, 7 for the fourth, 6 for the fifth and 5 for the sixth, hence 10*9*8*7*6*5 possible ways for picking six altogether. We can write:

(1)

Once we have picked our selection the order in which they are lined up does not matter. For the six we have selected there aredifferent ways of lining them up. To take this into account we can divide (1) by an extra factor of 6! to obtain

possible ways of picking 6 from ten. This is of course the same as calculated above.

In general then there areways of choosingfromwithout regard to order. It should also be noted that

]]>We might multiply this out to obtainNow we equateto this and solve for the coefficients a, b, c.

Example: Complete the square for the expression

Equating coefficients of

Equating coefficients of

Equating constant terms:

Hencein completed square form is

Having completed the square we can solve the equation

Example: Complete the square for the expressionHence solve the equation

Equating coefficients of

Equating coefficients of

Equating constant terms:

Hencein completed square form is

Now we can solve

]]>Let two of the roots be {jatex options:inline}\alpha = \sqrt{3} + \sqrt{2}, \; \beta = \sqrt{3} - \sqrt{2} {/jatex}.

{jatex options:inline}(x- \alpha )(x- \beta )=x^2-( \alpha + \beta ) x + \alpha \beta= x^2-2x \sqrt{3}+1{/jatex}

This polynomial does not have integer coefficients, but consider

{jatex options:inline}\begin{equation} \begin{aligned} (x^2-2x \sqrt{3}+1)(x^2+2x \sqrt{3}+1) &= (x^2+1-2x \sqrt{3})(x^2+1+2x \sqrt{3}) \\ &= (x^2+1)^2-(2x \sqrt{3})^2 \\ &= x^4+2x^2+1-12x^2 \\ &= x^4-10x^2+1 \end{aligned} \end{equation}{/jatex}

This is the required polynomial.]]>

Typically in coordinate geometry we are concerned with lines and normals, distances and circles.

We may be of, for the above diagram:

a)Show that AC is perpendicular to AB.

b)Find the distance AB.

a) Gradient of

Gradient of

The product of the gradients is -1 so AC and AB are at right angles.

b)

]]>